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CHAPTER

7

Solutions of Electrolytes

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 7: Solutions of Electrolytes

Faraday’s Laws, Molar Conductivity, and Weak Electrolytes

Chapter 7 *problems with an asterisk are slightly more demanding Faraday’s Laws, Molar Conductivity, and Weak Electrolytes 7.1.

A constant current was passed through a solution of cupric sulfate, CuSO 4 , for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5). Solution

7.2.

After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9) was deposited from a solution of silver nitrate. Calculate the current. Solution

7.3.

Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C 6 H 5 OH) to monobromophenol using a current of 20 000 A? Solution

7.4.

The following are the molar conductivities Λ of chloroacetic acid in aqueous solution at 25 °C and at various concentrations c:

c −4

10 M Λ Ω cm 2 mol−1 −1

625

312.5

156.3

78.1

39.1

19.6

9.8

53.1

72.4

96.8

127.7

164.0

205.8

249.2

Plot Λ against c. If Λ° = 362 Ω–1 cm2 mol–1, are these values in accord with the Ostwald dilution law? What is the value of the dissociation constant? (See also Problem 7.11.) Solution 7.5.

The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 °C is 1.26 × 10–6 Ω–1 cm–1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 Ω–1 cm2 mol–1; Cl–, 76.4 Ω–1 cm2 mol–1. Solution 7-2

Chapter 7: Solutions of Electrolytes

*7.6.

Faraday’s Laws, Molar Conductivity, and Weak Electrolytes

The electrolytic conductivity of a 0.001 M solution of Na 2 SO 4 is 2.6 × 10–4 Ω–1 cm–1. If the solution is saturated with CaSO 4 , the conductivity becomes 7.0 × 10–4 Ω–1 cm–1. Calculate the solubility product for CaSO 4 using the following molar conductivities at 1 2+ −1 −1 2 these concentrations: λ(Na+) = 50.1 Ω–1 cm2 mol–1; λ Ca = 59.5 Ω cm mol . 2 Solution

7.7.

The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 °C. The equivalent conductance for 0.100 M KCl at 25 °C is 128.96 S cm2 mol–1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte? Solution

*7.8.

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: λ(K+) = 73.5 Ω–1 cm2 mol–1; λ(Cl–) = 76.4 Ω–1 cm2 mol–1; λ (NH +4 ) = 73.4 Ω–1 cm2 mol–1; λ(OH–) = 198.6 Ω–1 cm2 mol–1. Solution

7.9.

The conductivity of a 0.0312 M solution of a weak base is 1.53 × 10–4 S cm–1. If the sum of the limiting ionic conductances for BH+ and OH– is 237.0 S cm2 mol–1, what is the value of the base constant K b ? Solution

7.10.

The equivalent conductance of KBr solutions as a function of concentration at 25 °C is given in the following table. By a linear regression analysis of suitable variables, find the value of Λ° for KBr.

c/10–3 M

0.25

0.36

0.50

0.75

1.00

1.60

2.00

5.00

10.00

Λ/S cm2 mol–1

150.16

149.87

149.55

149.12

148.78

148.02

147.64

145.47

143.15 Solution

7-3

Chapter 7: Solutions of Electrolytes

7.11.

Debye-Hückel Theory and Transport of Electrolytes

Equation 7.20 is one form of Ostwald’s dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of Λ at various concentrations to be tested by means of a straight-line plot). Explain how Λ° and K can be obtained from the plot. Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities κ for the dissociation of tetramethyl tin chloride, (CH 3 ) 4 SnCl, in ethyl alcohol solution at 25.0 °C and at various concentrations c:

c/10–4 mol dm–3

1.566

2.600

6.219

10.441

κ/10–6 Ω–1 cm–1

1.788

2.418

4.009

5.336

By the use of the linear plot you have devised, determine Λ° and K. Solution 7.12.

A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 °C? (Assume that the cell constant is 1.0 cm–1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The Λ° for acetic acid is 390.7 S cm2 mol–1 and K a = 1.81 × 10–5 mol dm–3 at 25 °C. Ignore the conductance of water.) Solution

7.13.

How far can the conductivity of water at 25 °C be lowered in theory by removing impurities? The Λ° (in S cm2 mol–1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. K w = 1.008 × 10–14. Compare your answer to the experimental value of 5.8 × 10–8 S cm–1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894). Solution

Debye-Hückel Theory and Transport of Electrolytes 7.14.

The radius of the ionic atmosphere (1/κ) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 °C (∈ = 78). Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of ∈ = 38 at a concentration of 0.10 M. Solution

7-4

Chapter 7: Solutions of Electrolytes

7.15.

Debye-Hückel Theory and Transport of Electrolytes

1 The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate K 2SO 4 are 149.9, 2 –1 2 –1 126.5, and 153.3 Ω cm mol , respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution 7.16.

The molar conductivity at 18 °C of a 0.0100 M aqueous solution of ammonia is 9.6 Ω–1 cm2 mol–1. For NH 4 Cl, Λ° = 129.8 Ω–1 cm2 mol–1 and the molar ionic conductivities of OH– and Cl– are 174.0 and 65.6 Ω–1 cm2 mol–1, respectively. Calculate Λ° for NH 3 and the degree of ionization in 0.01 M solution. Solution

7.17.

A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g. a. Calculate the transport numbers of the Li+ and Cl– ions. b. If Λ° (LiCl) is 115.0 Ω–1 cm2 mol–1, what are the molar ionic conductivities and the ionic mobilities? Solution

7.18.

A solution of cadmium iodide, CdI2 , having a molality of 7.545 × 10–3 mol kg–1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I–. Solution

7.19.

The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t– = 0.179 and the molar conductivity is 426.16 Ω–1 cm2 mol–1. Calculate the mobilities of the hydrogen and chloride ions. Solution

7.20.

If a potential gradient of 100 V cm–1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl– ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291. Solution

7-5

Chapter 7: Solutions of Electrolytes

Debye-Hückel Theory and Transport of Electrolytes

*7.21. A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl– ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3. Solution 7.22.

What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 × 10–19 C. Solution

*7.23. According to Bjerrum’s theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is = dN i N i exp(− zi zc e 2 /4π ∈0∈ rkBT ) 4π r 2 dr

where z i and z c are the charge numbers of the ion of type i and of the central ion and e, ∈0 , ∈ , and k B have their usual significance. Plot the exponential in this expression and also 4πr2 against r for a uni-univalent electrolyte in water at 25.0 °C (∈ = 78.3). Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN 1 /N 1 )dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion. By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 °C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of k B T, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 °C. Solution

7-6

Chapter 7: Solutions of Electrolytes

Thermodynamics of Ions

Thermodynamics of Ions 7.24.

The following are some conventional standard enthalpies of ions in aqueous solution at 25 °C: Ion

Δ f H˚/kJ mol–1

H+

0 +

–239.7

2+

–543.1

2+

Zn

–152.3

–

–167.4

–

–120.9

Na Ca Cl

Br

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl 2 , and ZnBr 2 , assuming complete dissociation. Solution 7.25.

One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is –1051.4 kJ mol–1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows: Ion

Δ hyd G˚ k J mol–1

H+

0

Na+

679.1

Mg2+

274.1

Al3+

–1346.4

Cl–

–1407.1

Br–

–1393.3 Solution

7-7

Chapter 7: Solutions of Electrolytes

7.26.

Thermodynamics of Ions

Calculate the ionic strengths of 0.1 M solutions of KNO 3 , K 2 SO 4 , ZnSO 4 , ZnCl 2 , and K 4 Fe(CN) 6 ; assume complete dissociation and neglect hydrolysis. Solution

7.27.

Calculate the mean activity coefficient γ ± for the Ba2+ and SO 24− ions in a saturated solution of BaSO 4 (K sp = 9.2 × 10–11 mol2 dm–6) in 0.2 M K 2 SO 4 , assuming the Debye-Hückel limiting law to apply. Solution

7.28.

The solubility of AgCl in water at 25 °C is 1.274 × 10–5 mol dm–3. On the assumption t

hat the Debye-Hückel limiting law applies,

a. Calculate ΔG° for the process AgCl(s) → Ag+(aq) + Cl–(aq). b. Calculate the solubility of AgCl in an 0.005 M solution of K 2 SO 4 . Solution 7.29.

Employ Eq. 7.114 to make plots of log γ ± against I for a uni-univalent electrolyte in water at 25 °C, with B = 0.51 mol–1 dm3/2 and B′ = 0.33 × 1010 mol–1 dm3/2 m–1, and for the following values of the interionic distance a: a = 0, 0.1, 0.2, 0.4, and 0.8 nm Solution

7.30.

Estimate the change in Gibbs energy ΔG when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution (∈ = 78) to the lipid environment of a cell membrane (∈ = 4) at 25 °C. Solution

7.31.

At 18 °C the electrolytic conductivity of a saturated solution of CaF2 is 3.86 × 10–5 Ω–1 cm–1, and that of pure water is 1.5 × 10–6 Ω–1 cm–1. 1 The molar ionic conductivities of Ca 2+ and F– are 51.1 Ω–1 cm2 mol–1 and 47.0 Ω–1 cm2 mol–1, respectively. Calculate the solubility of 2 CaF2 in pure water at 18 °C and the solubility product. Solution

7-8

Chapter 7: Solutions of Electrolytes

7.32.

Thermodynamics of Ions

What concentrations of the following have the same ionic strength as 0.1 M NaCl? CuSO 4 , Ni(NO 3 ) 2 , Assume complete dissociation and neglect hydrolysis.

Al 2 (SO 4 ) 3 ,

Na 3 PO 4 Solution

7.33.

The solubility product of PbF 2 at 25.0 °C is 4.0 × 10–9 mol3 dm–9. Assuming the Debye-Hückel limiting law to apply, calculate the solubility of PbF 2 in (a) pure water and (b) 0.01 M NaF. Solution

7.34.

Calculate the solubility of silver acetate in water at 25 °C, assuming the DHLL to apply; the solubility product is 4.0 × 10–3 mol2 dm–6. Solution

*7.35. Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 °C: ∈ = 78; ∂ ln ∈ /∂T = –0.0046 K–1. Suggest a qualitative explanation for the sign of the value you obtain. Solution *7.36. Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl– in aqueous solution at 25 °C (∈ = 78), under the following conditions: a. The electrolyte is present at infinite dilution. b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70. The ionic radii are 95 pm for Na+ and 181 pm for Cl–. Solution 7.37.

If the solubility product of barium sulfate is 9.2 × 10–11 mol2 dm–6, calculate the solubility of BaSO 4 in a solution that is 0.10 M in NaNO 3 and 0.20 M in Zn(NO 3 ) 2 ; assume the DHLL to apply. Solution

7-9

Chapter 7: Solutions of Electrolytes

7.38.

Ionic Equilibria

Silver chloride, AgCl, is found to have a solubility of 1.561 × 10–5 M in a solution that is 0.01 M in K 2 SO 4 . Assume the DHLL to apply and calculate the solubility in pure water. Solution

7.39.

The enthalpy of neutralization of a strong acid by a strong base, corresponding to the process H+(aq) + OH–(aq) → H 2 O is –55.90 kJ mol–1. The enthalpy of neutralization of HCN by NaOH is –12.13 kJ mol–1. Make an estimate of the enthalpy of dissociation of HCN. Solution

7.40.

Make use of the Debye-Hückel limiting law to estimate the activity coefficients of the ions in an aqueous 0.004 M solution of sodium sulfate at 298 K. Estimate also the mean activity coefficient. Solution

Ionic Equilibria 7.41.

A 0.1 M solution of sodium palmitate, C 15 H 31 COONa, is separated from a 0.2 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl– ions but not to palmitate ions. Calculate the concentrations of Na+ and Cl– ions on the two sides of the membrane after equilibrium has become established. (For a calculation of the Nernst potential, see Problem 8.18.) Solution

7.42.

Consider the ionizations H + + H 3 N + CH 2 COO − H 3 N + CH 2 COOH H 2 NCH 2 COOH + H +

Assume that the following acid dissociation constants apply to the ionizations: 1.5 ×10−10 M − NH 3+ − NH 2 + H + ; K a = −COOH − COO − + H + ; K a =× 4.0 10−3 M Estimate a value for the equilibrium constant for the process H 3 N + CH 3COO − H 2 NCH 2 COOH

Solution 7-10

Chapter 7: Solutions of Electrolytes

7.43.

Essay Questions

The pK values for the successive ionizations of phosphoric acid are given on p. 308. Which of the four species is predominant at the following values of the hydrogen or hydroxide concentration? a. [H+] = 0.1 M. b. [H+] = 2 × 10–3 M. c. [H+] = 5 × 10–5 M. d. [OH–] = 2 × 10–3 M. e. [OH–] = 1 M. Solution

7.44.

Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl– ions but not to P– ions. The initial concentrations are as shown below.

[K+] = 0.05 M

[K+] = 0.15 M

[Cl–] = 0.05 M

[P–] = 0.15 M

Calculate the concentrations on each side of the membrane after equilibrium has become established. (See Problem 8.26 in Chapter 8 for the calculation of the Nernst potential for this system.) Solution Essay Questions 7.45.

State Faraday’s two laws of electrolysis and discuss their significance in connection with the electrical nature of matter.

7.46.

Discuss the main ideas that lie behind the Debye-Hückel theory, as applied to the conductivities of solutions of strong electrolytes.

7.47.

Outline two important methods for determining transport numbers of ions.

7.48.

Explain why Li+ has a lower ionic conductivity than Na+ and why the value for H+ is so much higher than the values for both of these ions.

7.49.

Describe briefly the type of hydration found with the following ions in aqueous solution: Li+, Br–, H+, OH–.

7.50.

What modifications to the Debye-Hückel limiting law are required to explain the influence of ionic strength on solubilities?

7-11

Chapter 7: Solutions of Electrolytes

Solutions

Solutions 7.1.

A constant current was passed through a solution of cupric sulfate, CuSO 4 , for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5 g mol-1).

Solution: Given:= t 1= h 3600 s, = m 0.040 g, M = 63.5 g mol−1 Required: I To solve this problem we must use Eq. 7.6 and apply Faraday’s Laws of Electrolysis. Eq. 7.6 is given by, Q = It

Rearranging to solve for I we obtain, I=

Q t

where Q, is the quantity of electricity. Q is proportional to the mass of the element produced at the electrode. Faraday’s constant, given by the symbol F, relates the amount of substance deposited to the quantity of electricity, Q, passed through the solution. The charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1. Hence, m Q = zF M

Using the expression above, we can solve for the current through the solution. I=

zF m t M

The charge on copper in CuSO 4 is Cu2+, therefore z = 2

7-12

Chapter 7: Solutions of Electrolytes

2 × 96 485 C mol−1 0.040 g 63.5 g mol−1 3600 s −1 I = 0.033 765 529 3 C s I=

Solutions

where 1 C s −1 = 1 A I = 33.8 mA Back to Problem 7.1

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7-13

Chapter 7: Solutions of Electrolytes

7.2.

Solutions

After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9 g mol-1) was deposited from a solution of silver nitrate. Calculate the current.

Solution: Given: = t 45 = min 2700 = s, m 7.19 = mg 0.00719= g, M 107.9 g mol−1 Required: I This problem can be solved in a similar manner as problem 7.1, using the expression for current as, I =

zF m . t M

The charge on silver in AgNO 3 is Ag+, therefore z = 1 1× 96 485 C mol−1 0.00719 g 107.9 g mol−1 2700 s −1 I = 0.002 381 241 7 C s

I=

where 1 C s −1 = 1 A I = 2.4 mA

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7-14

Chapter 7: Solutions of Electrolytes

7.3.

Solutions

Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C 6 H 5 OH) to monobromophenol using a current of 20 000 A?

Solution: Given: m 500.00 = = kg, I 20 000 A Required: t To solve this problem, we must first outline the chemical reactions that are taking place.

(1) ( 2) ( 3)

2Br − → Br2 (g) + 2e − 2K + + 2e − → 2K(s) C6 H 5OH + Br2 (g) → C6 H 4 (Br)OH + HBr

Two moles of electrons are involved in the generation of each mole of bromine gas, which reacts with one mole of phenol, therefore z = 2. Each batch consists of 500.00 kg of phenol therefore converting to the number of moles and we can determine the time required to convert all 500.00 kg of phenol into monobromophenol. m Using Eq. 7.6 and Q = zF from problem 6.1, we can solve for t, M

7-15

Chapter 7: Solutions of Electrolytes

Q I zF m t= I M

Solutions

t=

(

) (

)

6 12.011 g mol−1 + 6 1.007 94 g mol−1 + 15.9994 g mol−1 M C6 H5OH = M C6 H5OH = 94.11304 g mol−1 3 2 × 96 485 C mol−1 500.00 ×10 g t= −1 20 000 A 94.113 04 g mol where 1 C s −1 = 1 A 1h t 51 260.165 44 s × = 3600 s t = 14.238 934 84 h

t = 14.239 h

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7-16

Chapter 7: Solutions of Electrolytes

7.4.

Solutions

The following are the molar conductivities Λ of chloroacetic acid in aqueous solution at 25 °C and at various concentrations c:

c −4

10 M Λ Ω cm 2 mol−1 −1

625

312.5

156.3

78.1

39.1

19.6

9.8

53.1

72.4

96.8

127.7

164.0

205.8

249.2

Plot Λ against c. If Λ° = 362 Ω–1 cm2 mol–1, are these values in accord with the Ostwald Dilution Law? What is the value of the dissociation constant? (See also Problem 7.11.) Solution: Given: c, Λ, = Λ° 362 Ω –1 cm 2 mol –1 Required: plot of Λ against c, are these values in accord with the Ostwald Dilution Law, K Using the data above, we can create the following plot of Λ against c.

7-17

Chapter 7: Solutions of Electrolytes

Solutions

The Ostwald dilution law is given by Eq. 7.20 as: c ( Λ/ Λ ° ) 2 K= 1 − ( Λ/ Λ ° ) To determine if the data given above follows the Ostwald Dilution Law, we see if we can calculate a fixed value for K, the dissociation constant of the solution. The results are plotted in the table below. c

Λ

K

10-4 M

Ω–1 cm2 mol–1

M

625

53.1

0.001575951

312.5

72.4

0.0015625

156.3

96.8

0.001525554

78.1

127.7

0.001501592

39.1

164

0.001467205

19.6

205.8

0.001468105

9.8

249.2

0.001490406

Since the values of K are reasonably constant, we can say that data given above follows the Ostwald Dilution Law. The value of the dissociation constant, K, can be calculated from the average of the K values obtained above. K average = 0.001513045 M K average = 1.5 ×10−3 M

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7-18

Chapter 7: Solutions of Electrolytes

Solutions

The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 °C is 1.26 × 10–6 Ω–1 cm–1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 Ω–1 cm2 mol–1; Cl–, 76.4 Ω–1 cm2 mol–1.

7.5.

Solution: Given: ∆κ= 1.26 ×10 –6 Ω –1 cm –1 , λAg+= 61.9 Ω –1 cm 2 mol –1 , λCl-= 76.4 Ω –1 cm 2 mol –1 Required: solubility The expression for molar conductivity is given by Eq. 7.9.

κ

Λ=

c

In order to solve this problem we can use the concentration as a measure of solubility. solubility =

∆κ Λ AgCl

Λ AgCl = λAg+ + λClsolubility = solubility =

∆κ λAg+ + λCl1.26 ×10 –6 Ω –1 cm –1 61.9 Ω –1 cm 2 mol –1 + 76.4 Ω –1 cm 2 mol –1

= solubility 9.110 629 067 ×10 –9 mol cm −3 = solubility 9.110 629 067 ×10 –6 mol dm −3 = 9.11×10 –6 mol dm −3 solubility

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7-19

Chapter 7: Solutions of Electrolytes

7.6.

Solutions

The electrolytic conductivity of a 0.001 M solution of Na 2 SO 4 is 2.6 × 10–4 Ω–1 cm–1. If the solution is saturated with CaSO 4 , the conductivity becomes 7.0 × 10–4 Ω–1 cm–1. Calculate the solubility product for CaSO 4 using the following molar conductivities at these concentrations:

1 2+ −1 2 −1 λ (Na+) = 50.1 Ω–1 cm2 mol–1; λ Ca = 59.5 Ω cm mol . 2 Solution: Given: cNa 2SO4 =0.001 M , κ 1 2

Na 2SO 4

=2.6 ×10 –4 Ω –1 cm −1 , κ 1 2

1 2

=7.0 × 10 –4 Ω –1 cm −1 CaSO 4

λ ( Na + ) = 50.1 Ω –1 cm 2 mol –1 , λ Ca 2+ = 59.5 Ω −1 cm 2 mol−1 Required: K s for CaSO 4 To determine the base dissociation constant for CaSO 4 , we must first realize which chemical reactions are taking place.

(1) ( 2)

Na 2SO 4 2Na + + SO 4 2− CaSO 4 Ca 2+ + SO 4 2−

The solubility product is therefore given by: K s = Ca 2+ SO 4 2− , To determine the concentrations of the species in the solution, we will determine the electrolytic and molar conductivities. We are given that the initial electrolytic conductivity of the Na 2 SO 4 solution is 2.6 × 10–4 Ω–1 cm–1 which is raised to 7.0 × 10–4 Ω–1 cm–1 upon saturation with CaSO 4 .This means the increase in electrolytic conductivity is: ∆κ =7.0 ×10 –4 Ω –1 cm −1 − 2.6 ×10 –4 Ω –1 cm −1 ∆κ = 4.4 ×10 –4 Ω –1 cm −1 The molar conductivity of the ½CaSO 4 solution can be calculated using Eq. 7.9.

7-20

Chapter 7: Solutions of Electrolytes

Solutions

κ

Λ= Λ1 2

c

CaSO4

∆κ = 2c

where c is the concentration of CaSO 4 and 2c is the concentration of ½ CaSO 4 . The molar conductivity of the ½Na 2 SO 4 solution is:

κ1 Λ1 2

Λ1 2

Λ1 2

Na 2SO4

Na 2SO4

Na 2SO4

2 = 2c

2.6 ×10 –4 Ω –1 cm −1 = 2 0.001×10 –3 mol cm −3

(

)

= 130 Ω –1 cm 2 mol−1 Na 2SO 4

The molar conductivity of the ½Na 2 SO 4 is the same as, Λ1 2

Na 2SO4

=λNa + + λ1 2

SO42−

Therefore we can solve for λ1 2

SO42−

to calculate the molar conductivity of the ½CaSO 4 .

7-21

Chapter 7: Solutions of Electrolytes

λ1 2

λ1 2

= Λ1

SO42−

2

Na 2SO4

Solutions

− λNa +

=130 Ω –1 cm 2 mol –1 − 50.1 Ω –1 cm 2 mol –1

SO42−

λ1 = 79.9 Ω –1 cm 2 mol –1 2

Λ1 2

Λ1 2

Λ1 2

SO42−

CaSO4

=λCa 2+ + λ1 2

SO42−

= 59.5 Ω −1 cm 2 mol−1 + 79.9 Ω –1 cm 2 mol –1 CaSO4

= 139.4 Ω −1 cm 2 mol−1 CaSO4

The concentration of ½CaSO 4 is therefore given by:

c=

∆κ 2Λ 1 2

c=

(

CaSO4

4.4 ×10 –4 Ω –1 cm −1

2 139.4 Ω –1 cm 2 mol−1

)

c 1.578192 253 ×10 –6 mol cm −3 = = c 1.578192 253 ×10 –3 mol dm −3 Solving for K s , Ca 2+ = c SO 4 2− = 1.0 ×10−3 mol dm −3 + c The concentration of SO 4 2- is influenced by the contributions of ½CaSO 4 and ½Na 2 SO 4 .

7-22

Chapter 7: Solutions of Electrolytes

(

Ks = c 1.0 ×10−3 mol dm −3 + c

Solutions

)

= K s 4.068 883 038 ×10−6 mol2 dm −6 = K s 4.07 ×10−6 mol2 dm −6

Back to Problem 7.6

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7-23

Chapter 7: Solutions of Electrolytes

7.7.

Solutions

The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 °C. The equivalent conductance for 0.100 M KCl at 25 °C is 128.96 S cm2 mol–1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte?

Solution: Given: G= 0.01178 S, T= 25 °C, Λ= 128.96 S cm 2 mol –1 at cKCl= 0.100 M = G 0.00824 = S, celectrolyte 0.0500 M Required: l/A, Λ electrolyte In order to determine the cell constant, we must first calculate the value for the electrolytic conductance. As we know, this can be obtained using Eq. 7.9.

κ

Λ=

c κ = cΛ

κ =

( 0.100 ×10

−3

)(

mol cm −3 128.96 S cm 2 mol –1

)

κ = 0.012896 S cm −1 We can now use Eq. 7.8 to solve for the cell constant, l/A.

7-24

Chapter 7: Solutions of Electrolytes

G (conductance) = κ

Solutions

A l

l κ = A G l 0.012896 S cm −1 = A 0.01178 S l =1.094 736 842 cm −1 A l =1.09 cm −1 A The equivalent conductance of the electrolyte in this same cell can be found using the cell constant calculated above and solve for Λ from Eq. 7.9.

κ

Λ=

c

κ =G

l A

l G A Λ electrolyte = celectrolyte

( 0.00824 S) (1.094 736 842 cm −1 )

Λ electrolyte = 0.0500 ×10−3 mol cm −3 Λ electrolyte = 180.412 631 6 S cm 2 mol−1 Λ electrolyte = 180 S cm 2 mol−1

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7-25

Chapter 7: Solutions of Electrolytes

7.8.

Solutions

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: λ(K+) = 73.5 Ω–1 cm2 mol–1; λ(Cl–) = 76.4 Ω–1 cm2 mol–1; λ (NH +4 ) = 73.4 Ω–1 cm2 mol–1; λ(OH–) = 198.6 Ω–1 cm2 mol–1.

Solution: Given: c= 0.01 M= , R 189 Ω, c= 0.01 M= , R 2460 Ω KCl NH3

λ (K+ ) = 73.5 Ω –1 cm 2 mol –1 , λ ( Cl – ) = 76.4 Ω –1 cm 2 mol –1 , λ ( NH 4+ ) = 73.4 Ω –1 cm 2 mol –1 , λ ( OH – ) = 198.6 Ω –1 cm 2 mol –1

Required: K b In order to determine the base dissociation constant for ammonia, we must first outline which chemical reactions are taking place.

(1) ( 2)

NH 4 OH NH 3 + H 2 O Kb + − NH 4 OH NH 4 + OH

The base dissociation constant is therefore given by: NH 4 + OH − Kb = [ NH 4OH ]

To solve for the concentrations of each species, we may begin by calculating the value for the electrolytic conductance of the standard KCl

κ

Λ=

c κ = cΛ Λ KCl = λK + + λClsolution in the cell. Λ KCl = 73.5 Ω –1 cm 2 mol –1 + 76.4 Ω –1 cm 2 mol –1 Λ KCl= 149.9 Ω –1 cm 2 mol –1

κ= KCl

( 0.01×10

−3

)(

mol cm −3 149.9 Ω –1 cm 2 mol –1

)

= κ KCl 0.001 499 Ω –1 cm −1

7-26

Chapter 7: Solutions of Electrolytes

Solutions

Recall that the resistance is inversely proportional to the conductance. The electrolytic conductivity of the ammonia solution is therefore,

κ NH = κ KCl × 3

RKCl RNH3

κ NH= 0.001 499 Ω –1 cm −1 × 3

189 Ω 2460 Ω

= κ NH3 1.151 670 732 ×10−4 Ω –1 cm −1 The molar conductivity of NH 4 + + OH − is: Λ NH4OH = λNH + + λOH4

Λ NH4OH =73.4 Ω –1 cm 2 mol –1 + 198.6 Ω –1 cm 2 mol –1 Λ NH4OH =272 Ω –1 cm 2 mol –1

Using Eq. 7.9, we can calculate the concentrations of NH 4 + and OH − . Λ= c= c=

κ c

κ NH

3

Λ NH4OH 1.151 670 732 ×10−4 Ω –1 cm –1 272 Ω –1 cm 2 mol –1

= c 4.234 083 572 ×10 –7 mol cm −3 = c 4.234 083 572 ×10 –4 mol dm −3 Knowing= that; c = NH 4 + OH − , we can solve for K b . NH 4 OH C initial C equilibrium

0.01 0.01 − c

NH 4 + 0 c

+

OH − 0 c

mol dm-3 mol dm-3 7-27

Chapter 7: Solutions of Electrolytes

Solutions

NH 4 + OH − c2 = Kb = 0.01 − c [ NH 4OH ] Kb =

( 4.234 083 572 ×10

–4

mol dm −3

)

2

0.01 − 4.234 083 572 ×10 –4 mol dm −3

= K b 1.872 008 786 ×10 –5 mol dm −3 K= 1.9 ×10 –5 mol dm −3 b Back to Problem 7.8

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7-28

Chapter 7: Solutions of Electrolytes

Solutions

The conductivity of a 0.0312 M solution of a weak base is 1.53 × 10–4 S cm–1. If the sum of the limiting ionic conductances for BH+ and OH– is 237.0 S cm2 mol–1, what is the value of the base constant K b ?

7.9.

Solution: Given:= c 0.0312 M, = κ 1.53 × 10 –4 S cm –1 ,= Λ° 237.0 S cm 2 mol –1 Required: K b In order to determine the base dissociation constant for the solution, we must write out the chemical reaction that is taking place. Kb + − B + H 2 O BH + OH

Since we are told we have a weak base, it is possible to apply Ostwald’s Dilution Law and introduce the degree of dissociation, α, given by Eq. 7.11.

α=

Λ Λ°

B

c (1 − α )

+

H 2O

BH + αc

+

OH − αc

The base dissociation constant is therefore given by Eq. 7.18. Kb =

cα 2 1− α

And the degree of dissociation is determined by calculating the molar conductivity of B + H 2 O using Eq. 7.9. Λ=

κ c

1.53 ×10 –4 S cm –1 Λ= 0.0312 ×10 –3 mol cm −3 Λ =4.903 846 154 S cm 2 mol –1

7-29

Chapter 7: Solutions of Electrolytes

Solutions

The degree of dissociation is therefore,

α=

4.903 846 154 S cm 2 mol –1

237.0 S cm 2 mol –1 α = 0.020 691 334

Solving for K b , we obtain the following:

( 0.0312 mol dm ) ( 0.020 691 334 ) = −3

Kb

2

1 − 0.020 691 334

= K b 1.363 992 486 ×10−5 mol dm −3 = K b 1.36 ×10−5 mol dm −3

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7-30

Chapter 7: Solutions of Electrolytes

7.10.

Solutions

The equivalent conductance of KBr solutions as a function of concentration at 25 °C is given in the following table. By a linear regression analysis of suitable variables, find the value of Λ° for KBr.

c/10–3 M

0.25

0.36

0.50

0.75

1.00

1.60

2.00

5.00

10.00

Λ/S cm2 mol–1

150.16

149.87

149.55

149.12

148.78

148.02

147.64

145.47

143.15

Solution: Given: T = 25 °C , data given above Required: Λ°KBr The relationship between Λ and c is given by the Debye-Hückel-Onsager Equation, Eq. 7.53. Λ = Λ ° − ( P + QΛ ° ) c

In order to solve for Λ°KBr , we can plot

c ×10−3

c c against Λ, where = c

(

M

)

(

Λ S cm 2 mol –1

0.01581139

150.16

0.01897367

149.87

0.02236068

149.55

0.02738613

149.12

0.03162278

148.78

0.04

148.02

0.04472136

147.64

0.07071068

145.47

)

7-31

Chapter 7: Solutions of Electrolytes

Solutions

0.1

143.15

From the linear regression, the y- intercept will be the value of Λ°KBr

Λ°KBr = 151.41268 S cm 2 mol –1 Λ°KBr = 151.41S cm 2 mol –1

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7-32

Chapter 7: Solutions of Electrolytes

7.11.

Solutions

Equation 7.20 is one form of Ostwald’s dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of Λ at various concentrations to be tested by means of a straight-line plot). Explain how Λ° and K can be obtained from the plot. Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities κ for the dissociation of tetramethyl tin chloride, (CH 3 ) 4 SnCl, in ethyl alcohol solution at 25.0 °C and at various concentrations c:

c/10–4 mol dm–3

1.566

2.600

6.219

10.441

κ/10–6 Ω–1 cm–1

1.788

2.418

4.009

5.336

By the use of the linear plot you have devised, determine Λ° and K.

Solution: Given: T = 25 °C , data above Required: Λ°, K Ostwald’s dilution law, given by Eq. 7.20 can be linearized in the following manner:

7-33

Chapter 7: Solutions of Electrolytes

Solutions

c ( Λ /Λ ° ) 2 K= 1 − ( Λ /Λ ° )

(

)

K 1 − ( Λ /Λ ° ) = c ( Λ /Λ ° ) c=

K − K ( Λ /Λ ° )

( Λ /Λ ) ( K − K ( Λ /Λ ) ) Λ ° 2

°

c=

2

°2

Λ2 K Λ°2 − K Λ°Λ c= Λ2 K Λ°2 cΛ = − K Λ° Λ

From here, we can plot cΛ against

1 and determine Λ° and K. Λ

κ

We can calculate Λ from Eq. 7.9 at each concentration given. Λ = . This leads to a table of values with the following: c c 10–4 mol dm–3

κ

Λ

10–6 Ω–1 cm–1 Ω–1 cm2 mol-1

cΛ

1/Λ

Ω-1 cm-1

Ω cm mol

-2

1.566

1.788

11.417625

1.788×10-6

0.087584

2.6

2.418

9.3

2.418×10-6

0.107527

6.219

4.009

6.446374

4.009×10-6

0.155126

10.441

5.336

5.1106216

5.336×10-6

0.195671

Now we obtain the following graph: 7-34

Chapter 7: Solutions of Electrolytes

Solutions

From the linear regression, the y- intercept will be KΛ° and the slope will be KΛ°2

7-35

Chapter 7: Solutions of Electrolytes

Solutions

= K Λ° 0.00111 Ω −1 cm −1 = K Λ°2 0.03294 Ω −2 cm mol−1 K Λ°2 K Λ° 0.03294 Ω −2 cm mol−1 Λ° = 0.00111 Ω −1 cm −1 Λ° =

= Λ° 29.675 675 68 Ω −1 cm 2 mol−1 Λ° = 30 Ω −1 cm 2 mol−1 K= K=

0.00111 Ω −1 cm −1 Λ° 0.00111 Ω −1 cm −1 29.675 675 68 Ω −1 cm 2 mol−1

= K 3.740 437 158 ×10−5 mol cm −3 = K 3.7 ×10−2 mol dm −3

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7-36

Chapter 7: Solutions of Electrolytes

7.12.

Solutions

A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 °C? (Assume that the cell constant is 1.0 cm–1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The Λ° for acetic acid is 390.7 S cm2 mol–1 and K a = 1.81 × 10–5 mol dm–3 at 25 °C. Ignore the conductance of water.)

Solution: A Given: cmax = 1500 ppm, = 1.0 cm −1 , ρacetic acid ≈ ρ water , Λ° = 390.7 S cm 2 mol –1 , K = 1.81×10 –5 mol dm –3 a l

T = 25 °C

Required: G In order to solve this problem, we must first convert the concentration from parts per million to SI units. 1500 g acid 106 g solution 1.500 g acid c= 1000 g solution

c 1500 ppm = =

(

) (

) (

M acetic acid = 2 12.011 g mol−1 + 4 1.007 94 g mol−1 + 2 15.9994 g mol−1

)

M acetic acid = 60.052 56 g mol−1 c=

1.500 g acid 60.052 56 g mol

−1

×

1 1 kg solution

c = 0.024 978119 2 mol kg −1 Since the solution has the same density as water, 1.00 kg of solution has a volume of 1.0 dm3. Therefore we can assume the solution has concentration, c = 0.0249781192 M. The conductance of a solution is given by Eq. 7.8.

7-37

Chapter 7: Solutions of Electrolytes

G (conductance) = κ

Solutions

A l

Since acetic acid is a weak acid, we may we use the Ostwald’s Dilution Law, Eq. 7.20, to solve for Λ . K=

c ( Λ /Λ ° ) 2 1 − ( Λ /Λ ° )

K − K ( Λ /Λ ° )= c ( Λ /Λ ° )

2

K − K ( Λ /Λ ° ) − c ( Λ /Λ ° ) =0 2

To solve for Λ /Λ ° , we use the quadratic equation. x=

−b ± b 2 − 4ac 2a

K ± K 2 − 4 ( −c ) K Λ /Λ = 2 ( −c ) °

K ± K 2 + 4cK Λ /Λ ° = −2c 1.81×10 –5 M ± ° Λ /Λ =

(1.81×10

–5

M ) + 4 ( 0.0249781192 M ) (1.81× 10 –5 M ) 2

−2 ( 0.0249781192 M )

= Λ /Λ ° 0.026 939 569 1 and − 0.026 903 369 1 We will disregard the negative value and take Λ /Λ ° =0.026 939 569 1 to solve for Λ .

( 0.026 939 569 1) ( Λ° ) Λ =( 0.026 939 569 1) ( 390.7 S cm 2 mol –1 )

= Λ

Λ =10.525 289 64 S cm 2 mol –1

7-38

Chapter 7: Solutions of Electrolytes

Solutions

Using Eq. 7.9, we can substitute for the value of κ and determine the conductance of the solution.

κ

Λ=

c κ = cΛ A G =κ l G=

( cΛ )

(

A l

)(

G = 0.024 978119 2 mol dm −3 10.525 289 64 S cm 2 mol –1

) (1.0 cm ) −1

G = 0.262 901 938 9 dm −3 S cm G 0.262 901 938 9 ×10−3 cm −3 S cm = G 2.629 019 389 ×10−4 S cm −2 = G 2.63 ×10−4 S cm −2 =

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7-39

Chapter 7: Solutions of Electrolytes

7.13.

Solutions

How far can the conductivity of water at 25 °C be lowered in theory by removing impurities? The Λ° (in S cm2 mol–1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. K w = 1.008 × 10–14. Compare your answer to the experimental value of 5.8 × 10–8 S cm–1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894).

Solution: 2 2 Given: Λ °KOH 274.4 S cm = = = mol –1 , Λ °HCl 426.04 S cm mol –1 , Λ °KCl 149.86 S cm 2 mol –1

T= 25 °C, K w = 1.008 ×10 –14 , κ exp = 5.8 ×10 –8 S cm –1 Required: κ The dissociations of each salt in water are given by:

(1) ( 2) ( 3)

KOH K + + OH − HCl H + + Cl− KCl K + + Cl−

By rearranging we find that, Λ °H2O = Λ °KOH + Λ °HCl − Λ °KCl = Λ °H2O 274.4 S cm 2 mol –1 + 426.04 S cm 2 mol –1 − 149.86 S cm 2 mol –1 550.58 S cm 2 mol –1 Λ °H2O =

In pure water, the only species that conduct electricity are H + and OH − ions. According to K w = H + OH − , each have a concentration of ;

K w =× 1.008 10−14 mol dm −3 = 1.003 999 203 2 ×10−7 mol dm −3 . Since this concentration is very low, we can assume that Λ H2O ≈ Λ °H2O .

7-40

Chapter 7: Solutions of Electrolytes

Solutions

κ

Λ= = κ

c

(1.003 999 203 2 ×10

−10

)(

mol cm −3 550.58 S cm 2 mol−1

)

= κ 5.527 779 329 ×10−8 S cm −1 = κ 5.528 ×10−8 S cm −1 Compared to the experimental value of 5.8 ×10 –8 S cm –1 , the conductivity determined through this process produces a very similar result. Back to Problem 7.13

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7-41

Chapter 7: Solutions of Electrolytes

7.14.

Solutions

The radius of the ionic atmosphere (1/κ) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 °C (∈ = 78). Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of ∈ = 38 at a concentration of 0.10 M.

Solution: 1 Given: = 0.964 nm,= c 0.10 M, T = 25 °C,= ∈ 78

κ

Required: see above a) Eq. 7.50 indicates that the thickness of the ionic atmosphere is inversely proportional to the square root of the concentration. 1/ 2

1 ∈0∈ kBT = κ e 2 ∑ ci zi2 L i 1 1 ∝ c κ Therefore the radius in water, where the electrolyte has c = 0.0001 M, can be obtained from the ratio of proportions.

7-42

Chapter 7: Solutions of Electrolytes

Solutions

1 1 κ 1 = κ 2 c2 c1 1 c1 1 κ 1 = c2 κ 2 c1 1 1 = c2 κ 1 κ 2 0.1 M 1 ( 0.964 nm ) = 0.0001 M κ 2 1 = 30.484 356 64 nm κ 2 1 = 30.5 nm κ 2 b) Similarly, we see from Eq. 7.50 that the thickness of the ionic atmosphere is proportional to the square root of the permittivity 1 ∝ ∈ κ The radius in water where∈ = 38 , can be obtained from the ratio of proportions.

7-43

Chapter 7: Solutions of Electrolytes

Solutions

∈2 1 1 = ∈1 κ 1 κ 2 38 1 ( 0.964 nm ) = 78 κ 2 1 = 0.672 855 072 6 nm κ 2 1 = 0.673 nm κ 2

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7-44

Chapter 7: Solutions of Electrolytes

7.15.

Solutions

1 The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate K 2SO 4 are 149.9, 2 –1 2 –1 126.5, and 153.3 Ω cm mol , respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution: Given: c = 0.001 M Λ KCl = 149.9 Ω –1 cm 2 mol –1 , Λ NaCl = 126.5 Ω –1 cm 2 mol –1 , Λ 1 2

Required: Λ 1 2

= 153.3 Ω –1 cm 2 mol –1 K 2SO 4

Na 2SO 4

1 The molar conductivity of Na 2SO 4 is given by the combination of the molar conductivities of each salt. We must also subtract the molar 2 conductivity of KCl since we are considering the solution containing only sodium and sulfate ions. Λ1 2

Λ1 2

Na 2SO 4

= Λ NaCl + Λ 1 2

K 2SO 4

− Λ KCl

126.5 –1 cm 2 mol –1 + 153.3 Ω –1 cm 2 mol –1 − 149.9 Ω –1 cm 2 mol –1 =Ω Na 2SO 4

Λ1 2

=129.9 Ω –1 cm 2 mol –1 Na 2SO 4

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7-45

Chapter 7: Solutions of Electrolytes

7.16.

Solutions

The molar conductivity at 18 °C of a 0.0100 M aqueous solution of ammonia is 9.6 Ω–1 cm2 mol–1. For NH 4 Cl, Λ° = 129.8 Ω–1 cm2 mol–1 and the molar ionic conductivities of OH– and Cl– are 174.0 and 65.6 Ω–1 cm2 mol–1, respectively. Calculate Λ° for NH 3 and the degree of ionization in 0.01 M solution.

Solution: Given: T =18 °C, cNH3 =0.0100 M, Λ NH3 =9.6 Ω –1 cm 2 mol –1 , Λ °NH4Cl = 129.8 Ω –1cm 2 mol –1 , ° λOH = 174.0 Ω –1cm 2 mol –1 , λCl° = 65.6 Ω –1cm 2 mol –1 , c = 0.01 M −

−

Required: Λ °NH3 , α In solution, ammonia reacts in following manner,

NH 3 + H 2 O NH 4 OH NH 4 OH NH 4 + + OH − As a result, we can obtain Λ °NH3 from the molar conductivity of NH 4 OH. ° ° Λ °NH4OH = Λ °NH4Cl + λOH − −λ − Cl

Λ °NH4OH = 129.8 Ω –1cm 2 mol –1 + 174.0 Ω –1cm 2 mol –1 − 65.6 Ω –1cm 2 mol –1 Λ °NH4OH = 238.2 Ω –1cm 2 mol –1 Λ °NH4OH =238 Ω –1cm 2 mol –1 The degree of dissociation is defined by Eq. 7.11 which states,

7-46

Chapter 7: Solutions of Electrolytes

α= α=

Solutions

Λ Λ° 9.6 Ω –1 cm 2 mol –1

238.2 Ω –1 cm 2 mol –1 α = 0.040 302 267 = α 4.0 ×10−2

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7-47

Chapter 7: Solutions of Electrolytes

7.17.

Solutions

A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g. a. Calculate the transport numbers of the Li+ and Cl– ions. b. If Λ° (LiCl) is 115.0 Ω–1 cm2 mol–1, what are the molar ionic conductivities and the ionic mobilities?

Solution: Given: I= 0.79 A, t= 2 h, ∆m= 0.793 g, Λ °LiCl= 115.0 Ω –1 cm 2 mol –1 Required: see above a) To solve this problem, we can use the Hittorf method. This method gives the transport numbers according to Eq. 7.75 and Eq.7.76. amount lost from cathode compartment amount lost from anode compartment = t− = t+ And amount deposited amount deposited We will use the number of moles to measure the amounts of the Li+ and Cl– ions. To determine the total amount deposited, we use Eq. 7.6. Q = It s Q = ( 0.79 A ) 2 h × 3600 h Q = 5688 A s Q = 5688 C In problem 7.1 we found that Q = zFn since the charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1. Solving for n, we can determine the total amount deposited. n=

Q zF

amount deposited =

5688 C

(1) ( 96 485 C mol−1 )

amount deposited = 0.058 952 168 7 mol The amount lost of LiCl lost in the anode compartment is given by n Cl . 7-48

Chapter 7: Solutions of Electrolytes

nLiCl =

Solutions

mLiCl M LiCl

= M LiCl 6.941 g mol−1 + 35.4527 g mol−1 M LiCl = 42.3927 g mol−1 nLiCl =

0.793 g 42.3927 g mol−1

nLiCl = 0.018 705 609 6 mol 1 The anode reaction that is occurring is Cl− → Cl2 + e − , therefore 0.058 952 168 7 mol of Cl- are removed by electrolysis. The amount lost 2 from the anode compartment is given by, n= ntotal − nLiCl net

= nnet 0.058 952 168 7 mol − 0.018 705 609 6 mol nnet = 0.040 246 559 2 mol Solving for t Cl - we obtain, 0.040 246 559 2 mol 0.058 952 168 7 mol = 0.682 698 534

= t+ tCl= − tCl−

tCl− = 0.68

The second transport number is given by1 − t+ , t− = tLi+ = 1 − t+ tLi+ = 1 − 0.682 698 534 tLi+ = 0.317 301 466 tLi+ = 0.32

7-49

Chapter 7: Solutions of Electrolytes

Solutions

b) In order to determine the molar ionic conductivities we will use Eq. 7.79 which states,

λ+°

= t+

λ°

− = and t− Λ° Λ°

To solve, we rearrange and use Λ °LiCl= 115.0 Ω –1 cm 2 mol –1 .

λCl° = Λ °tCl −

= λCl° −

−

(115.0 Ω

–1

cm 2 mol –1 ) ( 0.682 698 534 )

= λCl° − 78.510 331 41 Ω –1 cm 2 mol –1

λCl°= 79 Ω –1 cm 2 mol –1 −

λLi° = Λ °tLi +

= λLi° +

+

(115.0 Ω

–1

cm 2 mol –1 ) ( 0.317 301 466 )

= λLi° + 36.489 668 59 Ω –1 cm 2 mol –1

λLi°= 36 Ω –1 cm 2 mol –1 +

The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64.

7-50

Chapter 7: Solutions of Electrolytes

κ+

° λ= +

= Fu+ c+

u+ = uCl−

Solutions

λ+°

F 78.510 331 41 Ω –1 cm 2 mol –1 = 96 485 C

uCl− 8.137 050 465 ×10−4 Ω –1 cm 2 mol –1 C−1 = where 1 Ω –1 1 A= V −1 and 1 A 1 C s −1 = therefore, 1 Ω –1 = 1 C s −1 V −1

(

)

uCl− 8.137 050 465 ×10−4 C s −1 V −1 cm 2 mol –1 C –1 = uCl− 8.137 050 465 ×10−4 cm 2 mol –1 V −1 s −1 = uCl= 8.1×10−4 cm 2 mol –1 V −1 s −1 − uLi+ =

36.489 668 59 Ω –1 cm 2 mol –1 96 485 C

uLi+ 3.781 900 667 ×10−4 cm 2 mol –1 V −1 s −1 = 3.8 ×10−4 cm 2 mol –1 V −1 s −1 uLi= +

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7-51

Chapter 7: Solutions of Electrolytes

7.18.

Solutions

A solution of cadmium iodide, CdI2 , having a molality of 7.545 × 10–3 mol kg–1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I–.

Solution: Given: molality = 7.545 ×10 –3 mol kg –1 , mCd2+ = 0.03462 g, manode = 152.64 g, mCdI2 in anode = 0.3718 g

Required: tCd2+ , tI− When working with a Hittorf cell, we must use Eq. 7.75 and Eq. 7.76 to solve for tCd2+ and tI− . amount lost from anode compartment amount lost from cathode compartment = t− = t+ And amount deposited amount deposited

The number of coulombs of charge will be used as a measure of each amount. The anode compartment initially contained the following number of moles: = ni molality × manode kg mol kg –1 152.64 g ×10 –3 g ni = 0.001151 669 mol

= ni

( 7.545 ×10

–3

)

The anode compartment finally contained,

7-52

Chapter 7: Solutions of Electrolytes

nf =

Solutions

mCdI2 in anode M CdI2

(

M CdI2 112.411 g mol−1 + 2 126.904 47 g mol−1 =

)

M CdI2 = 366.219 94 g mol−1 nf =

0.3718 g 366.219 94 g mol−1

n f = 0.001 015 236 9 mol The number of moles lost from the anode compartment is therefore,

n= ni − n f = n 0.001151 669 mol − 0.001 015 236 9 mol = n 1.364 318 907 ×10−4 mol The total amount of Cd2+ deposited is calculated by, nCd2+ =

mCd2+ M Cd2+

M Cd2+ = 112.411 g mol−1 nCd2+ =

0.03462 g 112.411 g mol−1

= nCd2+ 3.079 769 773 ×10−4 mol

Now it is possible to determine the transport number at the anode.

1.364 318 907 ×10−4 mol 3.079 769 773 ×10−4 mol t− = 0.442 993 797 4 t− =

t− = 0.4430 7-53

Chapter 7: Solutions of Electrolytes

Solutions

The second transport number is given by1 − t − .

t + = 1 − 0.442 993 797 4 t + = 0.557 006 202 6 t + = 0.5570

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7-54

Chapter 7: Solutions of Electrolytes

Solutions

The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t– = 0.179 and the molar conductivity is 426.16 Ω–1 cm2 mol–1. Calculate the mobilities of the hydrogen and chloride ions.

7.19.

Solution: ° Given: = t+ 0.821,= t− 0.179, Λ= 426.16 Ω –1 cm 2 mol –1 HCl

Required: u+ , u− The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64. ° λ= +

κ+

= Fu+ c+

The molar ionic conductivities are given by Eq. 7.79. = t+

λ+°

λ°

− = and t− ° Λ Λ°

By rearranging and substituting equations 7.64 and 7.79, we can obtain an expression for the ionic mobility. u=

λ°

F Λ °t u= F Now it is possible to solve for u+ and u− .

7-55

Chapter 7: Solutions of Electrolytes

u+ =

( 426.16 Ω

–1

Solutions

cm 2 mol –1 ) ( 0.821)

96 485 C

u+ 0.003 626 236 Ω –1 cm 2 mol –1 C−1 = where 1 Ω –1 1 A= V −1 and 1 A 1 C s −1 = therefore, 1 Ω –1 = 1 C s −1 V −1

(

)

u+ = 0.003 626 236 C s −1 V −1 cm 2 mol –1 C−1 = u+ 3.63 ×10−3 V −1 cm 2 mol –1 s −1 u−

( 426.16 Ω =

cm 2 mol –1 ) ( 0.179 )

–1

96 485 C

= u− 7.906 165 725 ×10−4 V −1 cm 2 mol –1 s −1 = u− 7.91×10−4 V −1 cm 2 mol –1 s −1

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7-56

Chapter 7: Solutions of Electrolytes

Solutions

If a potential gradient of 100 V cm–1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl– ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291.

7.20.

Solution: Given: V 100 = = V cm –1 , cNaCl 0.01 M, Table 7.3 Required: vNa + , vCl− The ionic mobility is given in terms of the molar ionic conductivity by Eq. 7.64.

κ+

° λ= +

= Fu+ c+

u+ =

λ+° F

From Table 7.3 we are given that, ° λNa

+

uNa + =

= 50.08 S cm 2 mol−1 and λCl° − 76.31S cm 2 mol−1 50.08 S cm 2 mol−1 96 485 C mol−1

uNa + 5.190 444 11×10−4 S cm 2 C −1 = where 1 S =1 Ω –1 and 1 Ω –1 =1 A V −1 and 1 A =1 C s −1 therefore, 1S = 1 C s −1 V −1

(

)

uNa + 5.190 444 11×10−4 C s −1 V −1 cm 2 C−1 = −4

−1

= uNa + 5.190 444 11×10 V cm s uCl− =

2

−1

76.31S cm 2 mol−1 96 485 C mol−1

= uCl− 7.909 001 399 ×10−4 V −1 cm 2 s −1

7-57

Chapter 7: Solutions of Electrolytes

Solutions

From section 7.5 we know that, speed = uV. The velocities in a gradient of 100 V cm–1 are thus, vNa + =

(5.190 444 11×10

−4

)(

V −1 cm 2 s −1 100 V cm –1

)

vNa + 5.190 444 11×10−2 cm s −1 = 5.19 ×10−2 cm s −1 v= Na + vCl− =

( 7.909 001 399 ×10

−4

)(

V −1 cm 2 s −1 100 V cm –1

)

= vCl− 7.909 001 399 ×10−2 cm s −1 v= 7.91×10−2 cm s −1 Cl−

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7-58

Chapter 7: Solutions of Electrolytes

7.21.

Solutions

A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl– ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3.

Solution: Given: = cLiCl 0.01 = M, A 5= cm 2 , I 1 A , Table 7.3 Required: vLi+ , vCl− In order to determine the speeds of the ions, we must find the potential gradient. Once we have this information, we can proceed in a similar manner as was done in problem 7.20. The potential gradient can be calculated using Ohm’s Law (Eq.7.7) in conjunction with Eq. 7.9. R=

V κ and Λ = I c

The specific conductivity of a 0.01 M solution is calculated according to:

κ = cΛ We determine the molar conductivity of LiCl using the data for the ionic conductivities of Li+ and Cl- found in Table 7.3. Λ °LiCl = λLi° + + λCl° − = Λ °LiCl 38.66 S cm 2 mol−1 + 76.31S cm 2 mol−1 Λ °LiCl = 114.97 S cm 2 mol−1

Hence, = κ

( 0.01×10

−3

)(

mol cm −3 114.97 S cm 2 mol−1

)

= κ 1.1497 ×10−3 S cm −1

Let us consider the fact that the resistance is inversely proportional to κ, and we must factor in the 5 cm2 of surface area.

7-59

Chapter 7: Solutions of Electrolytes

R=

Solutions

1 1.1497 ×10−3 S cm −1 × 5 cm 2

= R 173.958 423 9 Ω cm −1 The potential gradient required to produce a current of 1 A is therefore, = V

(1 A ) (173.958 423 9 Ω cm −1 )

where 1 Ω =1 V A –1 V = 173.958 423 9 V cm −1 The ionic mobilities can be calculated using Eq. 7.64. ° λ= +

u+ =

κ+

= Fu+ c+

λ+° F

From Table 7.3 we know that,

7-60

Chapter 7: Solutions of Electrolytes

Solutions

uLi+ =

38.66 S cm 2 mol−1 96 485 C mol−1

uLi+ 4.006 840 442 ×10−4 S cm 2 C−1 = where 1 S =1 Ω –1 and 1 Ω –1 =1 A V −1 and 1 A =1 C s −1

λ

° Li +

therefore, 1S = 1 C s −1 V −1

= 38.66 S cm mol and λ 76.31S cm mol uLi+ 4.006 840 442 ×10−4 C s −1 V −1 cm 2 C−1 = 2

−1

° Cl−

2

−1

(

)

uLi+ 4.006 840 442 ×10−4 V −1 cm 2 s −1 = uCl− =

76.31S cm 2 mol−1 96 485 C mol−1

= uCl− 7.909 001 399 ×10−4 V −1 cm 2 s −1

From section 7.5 we are given that, speed = uV. = vLi+

( 4.006 840 442 ×10

−4

)(

V −1 cm 2 s −1 V −1 cm 2 s −1 173.958 423 9 V cm –1

)

vLi+ = 0.069 702 364 8 cm s −1 vLi= 7.0 ×10−2 cm s −1 + = vCl−

( 7.909 001 399 ×10

−4

)(

V −1 cm 2 s −1 173.958 423 9 V cm –1

)

vCl− = 0.137 583 7418 cm s −1 vCl− = 0.14 cm s −1

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7-61

Chapter 7: Solutions of Electrolytes

7.22.

Solutions

What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 × 10–19 C.

Solution: Given: vacuum: Q1 = −Q2 , e = 1.6 × 10–19 C Required: see above Recall that work is defined as the application of a force through a distance. This definition is given by Eq. 1.1, dw = F · dl. In this case, the force we are concerned with is an electrostatic force, and the distance in a vacuum we use as r. From Eq. 7.1, the electrostatic force is given by: F=

Q1Q2 4π ∈0 r 2

To determine the amount of work done, we will take the integral of F with respect to r. r2

w = ∫ Fdr r1

w=∫

r2

r1

Q1Q2 dr 4π ∈0 r 2

Since the particles have opposite charges, we will introduce a negative sign.

w =

∫

r2

r1

−

Q1Q2 dr 4π ∈0 r 2

QQ 1 1 w= − 1 2 − 4π ∈0 r2 r1 The permittivity of a vacuum has the value, = ∈0 8.854 ×10−12 C2 J −1 m −1 . This will be used when solving parts a, b and c.

7-62

Chapter 7: Solutions of Electrolytes

Solutions

a) r1 = 1.0 ×10−9 m, r2 = ∞m w= −

(

(1.6 ×10

–19

)

C

2

4π 8.854 ×10−12 C2 J −1 m −1

)

1 1 − −9 ∞ 1.0 ×10 m

w 2.300 862 1×10 –19 J = = w 2.3 ×10 –19 J b)

∞m r1 = 1.0 ×10−3 m, r2 = w= −

(

(1.6 ×10

–19

C

)

2

4π 8.854 ×10−12 C2 J −1 m −1

)

1 1 − −3 ∞ 1.0 ×10 m

)

1 1 − −9 0.10 m 1.0 ×10 m

= w 2.300 862 1×10 –25 J = w 2.3 ×10 –25 J c)

r1 = 1.0 ×10−9 m, r2 = 0.10 m w= −

(

(1.6 ×10

4π 8.854 ×10

–19

−12

C 2

)

2

C J

−1

m

−1

= w 2.300 862 1 ×10 –19 J = w 2.3 ×10 –19 J

Back to Problem 7.22

Back to Top 7-63

Chapter 7: Solutions of Electrolytes

7.23.

Solutions

According to Bjerrum’s theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is dN i N i exp(− zi zc e 2 /4π ∈0∈ rkBT ) 4π r 2 dr = where z i and z c are the charge numbers of the ion of type i and of the central ion and e, ∈0 , ∈ , and k B have their usual significance. Plot the exponential in this expression and also 4πr2 against r for a uni-univalent electrolyte in water at 25.0 °C (∈ = 78.3). Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN 1 /N 1 )dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion.

By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 °C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of k B T, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 °C. Solution:

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7-64

Chapter 7: Solutions of Electrolytes

7.24.

Solutions

The following are some conventional standard enthalpies of ions in aqueous solution at 25 °C: Ion

Δ f H˚/kJ mol–1

H+

0 +

–239.7

2+

–543.1

2+

Zn

–152.3

–

–167.4

–

–120.9

Na Ca Cl

Br

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl 2 , and ZnBr 2 , assuming complete dissociation. Solution: Given: standard enthalpies Required: enthalpies of formation In order to calculate the enthalpies of formation, we will simply sum up the standard enthalpies of the ions present in the solution.

7-65

Chapter 7: Solutions of Electrolytes

Solutions

∆ f H NaCl = ∆ f H Na° + ∆ f H Cl° ∆ f H NaCl = −239.7 kJ mol−1 − 167.4 kJ mol−1 ∆ f H NaCl = −407.1 kJ mol−1 ∆ f H CaCl2 =∆ f H Ca° + 2∆ f H Cl°

(

)

(

)

∆ f H CaCl2 = −543.1 kJ mol−1 − 2 167.4 kJ mol−1 ∆ f H CaCl2 = −877.9 kJ mol−1 ∆ f H ZnBr2 =∆ f H Zn° + 2∆ f H Br°

∆ f H ZnBr2 = −152.3 kJ mol−1 − 2 120.9 kJ mol−1 ∆ f H ZnBr2 = −394.1 kJ mol−1

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7-66

Chapter 7: Solutions of Electrolytes

7.25.

Solutions

One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is –1051.4 kJ mol–1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows: Ion

Δ hyd G˚ k J mol–1

H+

0 +

Na

2+

Mg

679.1 274.1

3+

–1346.4

–

–1407.1

–

–1393.3

Al Cl

Br Solution: Given: ∆ hyd G ° + ( absolute ) = −1051.4 kJ mol –1 H

Required: ∆ hyd G ° ( absolute ) for each ion In order to find the absolute Gibbs energies of hydration, we can either lower the cation’s standard Gibbs energies of hydration, or raise the anion’s absolute Gibbs energies of hydration by1051.4 kJ mol–1 (per charge).

7-67

Chapter 7: Solutions of Electrolytes

Solutions

∆ hyd G ° + ( absolute ) = 0 − 1051.4 kJ mol−1 H

∆ hyd G ° + ( absolute ) = −1051.4 kJ mol−1 H

∆ hyd G

679.1 kJ mol−1 − 1051.4 kJ mol−1 ( absolute ) =

° + Na

∆ hyd G ° + ( absolute ) = −372.3 kJ mol−1 Na

∆ hyd G

° 2+ Mg

∆ hyd G °

Mg

2+

274.1 kJ mol−1 − 2 (1051.4 kJ mol ) ( absolute ) =

( absolute ) =

−1828.7 kJ mol−1

(

∆ hyd G ° 3+ ( absolute ) = −1346.4 kJ mol−1 − 3 1051.4 kJ mol−1 Al

)

∆ hyd G ° 3+ ( absolute ) = −4500.6 kJ mol−1 Al

∆ hyd G ° − ( absolute ) = −1407.1 kJ mol−1 + 1051.4 kJ mol−1 Cl

∆ hyd G ° − ( absolute ) = −355.7 kJ mol−1 Cl

∆ hyd G ° − ( absolute ) = −1393.3 kJ mol−1 + 1051.4 kJ mol−1 Br

∆ hyd G ° − ( absolute ) = −341.9 kJ mol−1 Br

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7-68

Chapter 7: Solutions of Electrolytes

7.26.

Solutions

Calculate the ionic strengths of 0.1 M solutions of KNO 3 , K 2 SO 4 , ZnSO 4 , ZnCl 2 , and K 4 Fe(CN) 6 ; assume complete dissociation and neglect hydrolysis.

Solution: Given: c = 0.1 M Required: I The ionic strength of a solution is given by Eq. 7.103: I=

1 ∑ ci zi2 2 i

where z i is the valency of each ion present.

7-69

Chapter 7: Solutions of Electrolytes

Solutions

KNO3 → K + + NO3−

(

1 0.1 M ×12 + 0.1 M × 12 2 = 0.1 M

I= KNO3 I KNO3

)

K 2SO 4 → 2K + + SO 4 2− I= K 2SO 4 I K 2SO4

(

1 0.2 M ×12 + 0.1 M × 22 2 = 0.3 M

)

ZnSO 4 → Zn 2+ + SO 4 2− I= ZnSO 4 I ZnSO4

(

1 0.1 M × 22 + 0.1 M × 22 2 = 0.4 M

)

ZnCl2 → Zn 2+ + 2Cl− I= ZnCl2 I ZnCl2

(

1 0.1 M × 22 + 0.2 M ×12 2 = 0.3 M

K 4 Fe ( CN )6 → 4K + + Fe ( CN )6 I K 4 Fe= ( CN ) 6

I K 4 Fe( CN )

(

)

4−

1 0.4 M ×12 + 0.1 M × 44 2 = 1.0 M

)

6

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7-70

Chapter 7: Solutions of Electrolytes

Solutions

Calculate the mean activity coefficient γ ± for the Ba2+ and SO 24− ions in a saturated solution of BaSO 4 (K sp = 9.2 × 10–11 mol2 dm–6) in 0.2 M K 2 SO 4 , assuming the Debye-Hückel limiting law to apply.

7.27.

Solution: Given: BaSO 4 : K sp = 9.2×10–11 mol2 dm–6, cK 2SO4 = 0.2 M Required: γ ± When determining the mean activity coefficient, we use the Debye-Hückel limiting law given in Eq. 7.111:

log10 γ ± = − 0.51z+ z−

I/mol dm −3

We may then calculate the ionic strength from Eq. 7.103 in the following manner, I=

1 ∑ ci zi2 2 i

K 2SO 4 → 2K + + SO 4 2−

I K 2SO4

(

1 0.4 M ×12 + 0.2 M × 22 2 = 0.6 M

I= K 2SO 4

)

Finally we can solve for the mean activity coefficient which produces; log10 γ ± = − 0.51( 2 × 2 ) 0.6

γ ± = 10 −0.51( 2×2) 0.6 γ ± = 0.026 291 949 8 2.6 ×10−2 γ= ±

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7-71

Chapter 7: Solutions of Electrolytes

7.28.

Solutions

The solubility of AgCl in water at 25 °C is 1.274 × 10–5 mol dm–3. On the assumption t

hat the Debye-Hückel limiting law applies,

a. Calculate ΔG° for the process AgCl(s) → Ag+(aq) + Cl–(aq). b. Calculate the solubility of AgCl in an 0.005 M solution of K 2 SO 4 . Solution: Given: s = 1.274 ×10 –5 mol dm –3 , T = 25 °C Required: see above a. To calculate the Gibbs energy, we first need to determine the solubility product of AgCl in water. Eq. 7.121 shows that K s = [Ag + ][Cl− ]γ ±2 , and since [Ag + ] = [Cl− ] , we can write K s = s 2γ ±2 .

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength. log10 γ ± = − 0.51z+ z−

I /mol dm −3

− 0.51(1×1) 1.274 ×10 –5 log10 γ ± =

γ ± = 10 −0.51(1×1) 1.274×10 γ ± = 0.995 817 261 4

–5

The solubility product is then,

= Ks

(1.274 ×10

–5

M ) ( 0.995 817 261 4 ) 2

2

= K s 1.609 526 59 ×10 –10 M 2 Solving for Gibbs energy, using ∆G° = − RT ln K s , we obtain

7-72

Chapter 7: Solutions of Electrolytes

( −8.3145 J K

∆G° =

−1

)(

Solutions

) (

mol−1 298.15 K ln 1.609 526 59 ×10 –10 M 2

)

∆G° =55 900.511 31 J mol−1 ∆G° =55.90 kJ mol−1 b. To solve for the solubility in a solution of c = 0.005 M K 2 SO 4 , we need to calculate the ionic strength since we have a common ion present. We can calculate the ionic strength from Eq. 7.103 in the following manner, 1 ∑ ci zi2 2 i 1 0.01 M × 12 + 0.005 M × 22 = I 2 I = 0.015 M I=

(

)

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength.

log10 γ ± = −0.51 z+ z−

I / mol dm −3

−0.51(1× 1) 0.015 log10 γ ± = log10 γ ± = −0.062 461 988

γ ± = 10−0.062 461988 γ ± = 0.866 040 12 Solving for the solubility by rearranging Eq. 7.121,

7-73

Chapter 7: Solutions of Electrolytes

Solutions

K s = s 2γ ±2 s=

Ks

γ±

1.609 526 59 ×10−10 M 2 0.86604012 = s 1.464 91×10−5 M s=

= s 1.46 ×10−5 M

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7-74

Chapter 7: Solutions of Electrolytes

7.29.

Solutions

Employ Eq. 7.114 to make plots of log γ ± against I for a uni-univalent electrolyte in water at 25 °C, with B = 0.51 mol–1 dm3/2 and B′ = 0.33 × 1010 mol–1 dm3/2 m–1, and for the following values of the interionic distance a: a = 0, 0.1, 0.2, 0.4, and 0.8 nm

Solution:

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7-75

Chapter 7: Solutions of Electrolytes

7.30.

Solutions

Estimate the change in Gibbs energy ΔG when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution (∈ = 78) to the lipid environment of a cell membrane (∈ = 4) at 25 °C.

Solution: Given: n = 1 mol, rK + = 0.133 nm, ∈water = 78, ∈membrane = 4, T = 25 °C Required: ΔG Eq. 7.87 (given below) may be used to estimate the change in Gibbs energy. Ges° =

z 2e2 8π ∈0∈ r

Since we are given 1 mol of K+, we will multiply the expression above by L, Avogadro’s number.

7-76

Chapter 7: Solutions of Electrolytes

Ges° = ° es

G = Ges° =

Solutions

z 2e2 L 8π ∈0∈ r

( +1)

(

2

(1.602 ×10

−19

) ( 6.022 ×10 mol ) m ) ( 0.133 ×10 m ) ∈ 2

C

8π 8.854 ×10−12 C2 N −1

−1

23

−2

−9

5 222 197.4616 N m mol−1

∈

where 1 N m = 1 J ° es

G =

5 222 197.4616 J mol−1

∈

5 222 197.4616 J mol−1 78 ° Gwater = 6694.839 251 J mol−1 ° Gwater =

5 222 197.4616 J mol−1 4 ° Gmembrane = 130 549.3654 J mol−1 ° Gmembrane =

° ° Ges° Gmembrane ∆= − Gwater

= ∆Ge°s 130 549.3654 J mol−1 − 6694.839 251 J mol−1 ∆Ges° = 123 854.526 1 J mol−1 ∆Ges° = 124 kJ mol−1

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7-77

Chapter 7: Solutions of Electrolytes

Solutions

At 18 °C the electrolytic conductivity of a saturated solution of CaF 2 is 3.86 × 10–5 Ω–1 cm–1, and that of pure water is 1.5 × 10–6 Ω–1 1 cm–1. The molar ionic conductivities of Ca 2+ and F– are 51.1 Ω–1 cm2 mol–1 and 47.0 Ω–1 cm2 mol–1, respectively. Calculate the 2 solubility of CaF 2 in pure water at 18 °C and the solubility product.

7.31.

Solution: Given: T =18 °C, κ CaF2 =3.86 ×10 –5 Ω –1 cm –1 , κ H2O =1.5 ×10 –6 Ω –1 cm –1 ,

λ1 2

Ca

2+

= 51.1 Ω –1cm 2 mol –1 , λF− = 47.0 Ω –1cm 2 mol –1

Required: s, K s The expression for the molar conductivity is given by Eq. 7.9:

κ

Λ=

c

It is possible to use the concentration to determine the solubility. c=

κ Λ1 2

Λ1 2

Λ1 2

Λ1 2

CaF2

CaF2

= λ1 2

Ca 2+

+ λF-

= 51.1 Ω –1 cm 2 mol –1 + 47.0 Ω –1 cm 2 mol –1 CaF2

= 98.1 Ω –1 cm 2 mol –1 CaF2

The observed κ due to the salt is therefore,

7-78

Chapter 7: Solutions of Electrolytes

Solutions

κ = 3.86 ×10−5 Ω –1 cm –1 − 1.5 ×10−6 Ω –1 cm –1 κ = 3.71×10−5 Ω –1 cm –1 c=

3.71×10 –5 Ω –1 cm –1 98.1 Ω –1 cm 2 mol –1

= c 3.781855 25 ×10 –7 mol cm −3 = c 3.781855 25 ×10 –4 mol dm −3 1 CaF2 has the molar mass, 2

(

2

)

1 40.078 g mol−1 + 18.998 403 2 g mol−1 2

= M1 CaF2

M CaF2 = 39.037 403 2 g mol−1 The solubility can now be determined.

(

s = 39.037 403 2 g mol−1

) (3.781855 25 ×10

–4

mol dm −3

)

s = 0.014 763 380 8 g dm −3 = s 1.48 ×10 –2 g dm −3 The solubility product is given by, 2

K s = Ca 2+ F− 1 −3 −3 –4 –4 Ks = 3.781855 25 ×10 mol dm 3.781855 25 × 10 mol dm 2 −3 –11 = K s 2.704 485 84 ×10 mol dm

(

)(

)

2

= K s 2.70 ×10 –11 mol dm −3

Back to Problem 7.31

Back to Top 7-79

Chapter 7: Solutions of Electrolytes

7.32.

Solutions

What concentrations of the following have the same ionic strength as 0.1 M NaCl? CuSO 4 , Ni(NO 3 ) 2 , Assume complete dissociation and neglect hydrolysis.

Al 2 (SO 4 ) 3 ,

Na 3 PO 4

Solution: Given: cNaCl = 0.1 M Required: cCuSO4 , cNi( NO3 ) , cAl2 (SO4 ) , cNa3PO4 3

2

As we have previously seen, the ionic strength of a compound may be determined using Eq. 7.103. 1 ∑ ci zi2 2 i 1 0.1 M ×12 + 0.1 M × 12 = I NaCl 2 I NaCl = 0.1 M I=

(

)

For each salt, we set = I I= 0.1 M to solve for c . NaCl

7-80

Chapter 7: Solutions of Electrolytes

(

1 cCuSO4 × 22 + cCuSO4 × 22 2 0.1 M = 4cCuSO4 0.1 M =

Solutions

)

2.5 ×10−2 M cCuSO= 4

(

1 c × 22 + 2cNi( NO3 ) ×12 2 2 Ni( NO3 )2 0.1 M = 3cNi( NO3 )

0.1 M =

)

2

3.3 × 10−2 M cNi( NO3= ) 2

(

1 2cAl2 (SO4 ) × 32 + 3cAl2 (SO4 ) × 22 3 3 2 0.1 M = 15cAl2 (SO4 )

0.1 M =

)

3

6.7 ×10−3 M cAl2 (SO4= ) 3

(

1 3cNa3PO4 ×12 + cNa 3PO4 × 32 2 0.1 M = 6cNa3PO4

0.1 M =

)

cNa3PO= 1.7 ×10−2 M 4

Back to Problem 7.32

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7-81

Chapter 7: Solutions of Electrolytes

Solutions

The solubility product of PbF 2 at 25.0 °C is 4.0 × 10–9 mol3 dm–9. Assuming the Debye-Hückel limiting law to apply, calculate the solubility of PbF 2 in (a) pure water and (b) 0.01 M NaF.

7.33.

Solution: Given: K s = 4.0 ×10 –9 mol3 dm –9 , T = 25 °C Required: see above a) The dissolution of PbF 2 is written as: PbF2 → Pb 2+ + 2F− , hence the solubility product is given by: 2

K s = Pb 2+ F− . In order to solve for solubility, we must first neglect the effect of the activity coefficients and write, = Ks

s ][ 2 s ] [= 2

4s3 .

Solving for s, we obtain, 3 4 s= 4.0 ×10 –9 mol3 dm –9

= s 1.0 ×10 –3 mol dm –3 We will determine the activity coefficients of Pb2+ and F- by applying the Debye-Hückel limiting law. To solve, we must first calculate the ionic strength of PbF 2 from Eq. 7.103. 1 ∑ ci zi2 2 i 1 I= s × 22 + 2 s ×12 2 I = 3s I=

(

)

= I 3.0 ×10−3 mol dm −3

According to the Debye-Hückel limiting law, Eq. 7.111,

7-82

Chapter 7: Solutions of Electrolytes

log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

Solutions

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 3.0×10 γ ± = 0.879 290 334 3 −3

In order to find the true solubility, we must factor in the activity coefficients.

[γ= + s ][ 2γ − s ] 2

= Ks s3 =

s=

4s3

Ks

4γ +γ − 2

4.0 ×10 –9 mol3 dm –9 3

4 ( 0.879 290 334 3)

3

= s 1.137 280 8 ×10 –3 mol dm –3 = s 1.1×10 –3 mol dm –3

b) In 0.01 M NaF, the ionic strength is essentially 0.01 mol dm-3. Calculating the activity coefficients, we obtain,

γ ± =10−0.51z

+

z−

I /moldm −3

γ ± =10−0.51( 2×1) 0.01 γ ± = 0.790 678 628 If s is the solubility then,

7-83

Chapter 7: Solutions of Electrolytes

K s = Pb 2+ F− Pb 2+ = s

Solutions

2

F− = 0.01 mol dm –3 K s = [γ + s ] γ − 0.01 mol dm –3

(

2

)

K s = γ +γ − 2 0.0001 mol2 dm –6 s s=

(γ γ

+ −

s=

Ks 2

0.0001 mol2 dm –6

)

4.0 × 10 –9 mol 3 dm –9

( 0.790 678 628)

3

0.0001 mol2 dm –6

= s 8.092 076 715 ×10 –5 mol dm –3 = s 8.1×10 –5 mol dm –3

Back to Problem 7.33

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7-84

Chapter 7: Solutions of Electrolytes

Solutions

Calculate the solubility of silver acetate in water at 25 °C, assuming the DHLL to apply; the solubility product is 4.0 × 10–3 mol2 dm–6.

7.34.

Solution: Given: K s = 4.0 ×10 –3 mol2 dm –6 , T = 25 °C Required: s We will solve this problem in a similar manner as the previous question. We may start by neglecting the activity coefficients to find the first approximation of s. The dissolution of silver acetate is given by the following: AgC2 H 3O 2 → Ag + + CH 3COO − = Ks

s ][ s ] [=

s2

s = Ks = s

4.0 ×10 –3 mol2 dm –6

s = 0.063 245 553 2 mol dm –3 The ionic strength is thus,

(

)

1 s ×12 + s ×12 2 I= s= 0.063 245 553 2 mol dm –3 I=

According to the Debye-Hückel limiting law, Eq. 7.111, log10 γ ± = − 0.51z+ z−

I /mol dm −3

γ ± =10−0.51(1×1) 0.063 245 553 2 γ ± = 0.744 289 325 The second approximation of the solubility is therefore, 7-85

Chapter 7: Solutions of Electrolytes

= Ks s=

s=

[γ= + s ][γ − s ]

Solutions

γ +γ − s 2

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.744 289 325)

2

s = 0.084 974 419 3 mol dm –3 = s 8.5 ×10 –2 mol dm –3 We may continue to take third and fourth approximations by repeating the above steps.

γ ± =10−0.51(1×1) 0.084 974 419 3 γ ± = 0.710 122 220 4 s= s=

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.710 122 220 4 )

2

s = 0.089 062 912 5 mol dm –3

The third approximation is then; = s 8.9 ×10 –2 mol dm –3 .

γ ± =10−0.51(1×1) 0.089 062 912 5 γ ± = 0.704 366 363 8 s= s=

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.704 366 363 8)

2

s = 0.089 790 706 2 mol dm –3 7-86

Chapter 7: Solutions of Electrolytes

Solutions

The fourth approximation is then; = s 9.0 ×10 –2 mol dm –3 .

Back to Problem 7.34

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7-87

Chapter 7: Solutions of Electrolytes

7.35.

Solutions

Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 °C: ∈ = 78; ∂ ln ∈ /∂T = –0.0046 K–1. Suggest a qualitative explanation for the sign of the value you obtain.

Solution: Given: Problem 7.30: T = 25 °C, ∈= 78,

∂ln ∈ = −0.0046 ∂T

Required: ∆Ses° , explain the sign From Problem 7.30, we found the expression for the estimated Gibbs energy, Ges° =

5 222 197.4616 J mol−1

∈

For the transfer from water to lipid we can say that, 1 1 ° −1 ∆G= 5 222 197.4616 − es /J mol ∈ lipid ∈water

From Eq. 3.119:

∂G = −S ∂T P ∂G ° − es hence, ∆Ses° = ∂T P ∆Ses° / J K −1 mol−1 = −5 222 197.4616

∂ ∂T

1 1 − ∈lipid ∈water

Since ∈lipid is temperature independent, this leads to:

7-88

Chapter 7: Solutions of Electrolytes

Solutions

1 ∂ − ∂T ∈water 1 ∂∈ ∆Ses° / J K −1 mol−1 = −5 222 197.4616 2 ∈ water ∂T ∆Ses° / J K −1 mol−1 = −5 222 197.4616

since

1 ∂∈ 1 ∂ ln ∈ = 2 ∈ water ∂T ∈water ∂T

∆Ses° / J K −1 mol−1 = −5 222 197.4616

1

∈water

∂ ln ∈ ∂T

1 ∆Ses° / J K −1 mol−1 = −5 222 197.4616 ( −0.0046 ) 78 ° −1 −1 ∆Ses = 30.796 260 56 J K mol ∆Ses° = 31 J K −1 mol−1 The entropy increases due to the release of bound water molecules when the K+ ions pass into the lipid.

Back to Problem 7.35

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7-89

Chapter 7: Solutions of Electrolytes

7.36.

Solutions

Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl– in aqueous solution at 25 °C (∈ = 78), under the following conditions: a. The electrolyte is present at infinite dilution. b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70. The ionic radii are 95 pm for Na+ and 181 pm for Cl–.

Solution: Given: Eq. 7.86, = n 1 mol, T = 25 °C,= ∈ 78 Required: see above a) At infinite dilution, the work of charging an ion is given by Eq.7.86 which states, z 2e2 wrev = 8π ∈0∈ r For one mole of Na+, we multiply by Avogadro’s number, L and use the ionic radius of r = 95 pm. wNa + = wNa + =

z 2e2 L 8π ∈0∈ r

( +1)

(

2

(1.602 ×10

−19

C

8π 8.854 ×10−12 C2 N −1

) ( 6.022 ×10 mol ) m ) ( 78 ) ( 95 ×10 m ) 2

23

−2

−1

−12

wNa + = 9372.774 952 N m mol−1 wNa + = 9372.774 952 J mol−1 For one mole of Cl-, we will multiply by Avogadro’s number, L and use the ionic radius of r = 181 pm.

7-90

Chapter 7: Solutions of Electrolytes

wCl- =

( −1)

2

(

(1.602 ×10

−19

C

8π 8.854 ×10−12 C2 N −1

Solutions

) ( 6.022 ×10 mol ) m ) ( 78 ) (181×10 m ) 2

−1

23

−2

−12

wCl- = 4919.412 267 J mol−1 One mole of Na+Cl- at infinite dilution is thus,

wrev wNa + + wCl= wrev 9372.774 952 J mol−1 + 4919.412 267 J mol−1 = wrev = 14 292.187 22 J mol−1 wrev = 14 kJ mol−1 b) These values are reduced when the electrolyte is at a higher concentration. The work of charging the ionic atmosphere is negative and equal to kTlnγ i . Thus for one mol of Na+ ions, of activity γ + , the work of charging the atmosphere is RTlnγ + . Similarly, for the chloride ion, the work per mole is RTlnγ − . For one mole of Na+Clwrev RT ( lnγ + + ln γ − ) = wrev = RTln ( γ +γ − ) wrev = 2 RTlnγ ± where γ ± = 0.70

(

)(

)

wrev = 2 8.3145 J K −1 mol−1 298.15 K ln 0.70 wrev = −1768.371 67 J mol−1

The net work done is given by, = wrev 14 292.187 22 J mol−1 − 1768.371 67 J mol−1 wrev = 12 523.815 55 J mol−1 wrev = 13 kJ mol−1 7-91

Chapter 7: Solutions of Electrolytes

Back to Problem 7.36

Solutions

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7-92

Chapter 7: Solutions of Electrolytes

7.37.

Solutions

If the solubility product of barium sulfate is 9.2 × 10–11 mol2 dm–6, calculate the solubility of BaSO 4 in a solution that is 0.10 M in NaNO 3 and 0.20 M in Zn(NO 3 ) 2 ; assume the DHLL to apply.

Solution: Given: K s = 9.2 ×10 –11 mol2 dm –6 , cin NaNO3 = 0.10 M, cin Zn ( NO3 ) = 0.20 M 2

Required: s The expression for the solubility product is given by, K s = Ba 2+ SO 4 2− K s = [γ + s ][γ − s ] Ks = γ ±2s2 The ionic strength of the solution is calculated according to Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 I= (1 × 0.1 M + 12 × 0.1 M + 22 × 0.2 M + 12 × 0.4 M ) 2 I = 0.70 M I=

To find the activity coefficient, we will use the Debye-Hückel limiting law given by Eq. 7.111. log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×2) 0.70 γ ± = 0.019 643 259 1 If the solubility s is therefore,

7-93

Chapter 7: Solutions of Electrolytes

s= s=

Solutions

Ks

γ ±2 9.2 ×10 –11 mol2 dm –6

( 0.019 643 259 1)

2

= s 4.882 928 531×10 –4 M = s 4.9 × 10 –4 M

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7-94

Chapter 7: Solutions of Electrolytes

7.38.

Solutions

Silver chloride, AgCl, is found to have a solubility of 1.561 × 10–5 M in a solution that is 0.01 M in K 2 SO 4 . Assume the DHLL to apply and calculate the solubility in pure water.

Solution: Given: s = 1.561× 10 –5 M, c = 0.01 M Required: s The expression for the solubility product is given by, K s = Ag + Cl− K s = [γ + s ][γ − s ] Ks = γ ±2s2 The ionic strength of the solution is calculated according to Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 I= (1 × 0.02 M + 22 × 0.01 M ) 2 I = 0.03 M I=

To determine the activity coefficient, we will use the Debye-Hückel limiting law, Eq. 7.111. log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51(1×1) 0.03 γ ± = 0.815 953 498 4 If the solubility product is K s ,

7-95

Chapter 7: Solutions of Electrolytes

= Ks

( 0.815 953 498 4 )

2

(1.561×10

Solutions

–5

M

)

2

= K s 1.622 320 38 ×10 –10 M 2 Finally, the solubility in pure water is given by; s = Ks = s

1.622 320 38 ×10 –10 M 2

= s 1.273 703 411×10 –5 M = s 1.3 ×10 –5 M Back to Problem 7.38

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7-96

Chapter 7: Solutions of Electrolytes

7.39.

Solutions

The enthalpy of neutralization of a strong acid by a strong base, corresponding to the process H+(aq) + OH–(aq) → H 2 O is –55.90 kJ mol–1. The enthalpy of neutralization of HCN by NaOH is –12.13 kJ mol–1. Make an estimate of the enthalpy of dissociation of HCN.

Solution: Given: ∆H = −55.90 kJ mol−1 , ∆ neut H = −12.13 kJ mol−1 Required: ∆ diss H The enthalpy change for the neutralization of HCN by NaOH is less than the value of the 55.90 kJ mol-1 because the energy required for the dissociation of HCN, ∆ diss H is given by:

∆ neut H = ∆H − ∆ diss H , hence, ∆ diss H = ∆H − ∆ neut H = ∆ diss H 55.90 kJ mol−1 − 12.13 kJ mol−1 ∆ diss H = 43.77 kJ mol−1

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7-97

Chapter 7: Solutions of Electrolytes

7.40.

Solutions

Make use of the Debye-Hückel limiting law to estimate the activity coefficients of the ions in an aqueous 0.004 M solution of sodium sulfate at 298 K. Estimate also the mean activity coefficient.

Solution: Given: cNa 2SO4 0.004 M, T 298 K = = Required: γ Na + , γ SO 2− , γ ± 4

From Eq. 7.104 we can calculate the activity coefficients for each ion.

log10 γ i = − zi2 B I As we know, the ionic strength of the solution is calculated using Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 1 × 0.008 M + 22 × 0.004 M ) I= ( 2 I = 0.012 M I=

The activity coefficients are therefore,

γ Na = 10 −( +1) 0.51 2

0.012

+

γ Na = 0.879 290 334 3 +

γ Na = 0.879 +

γ SO

4

γ SO

4

γ SO

2−

= 10 −( −2)

2−

= 0.597 763 229 5

2−

2

0.51 0.012

= 0.598

4

The mean activity coefficient can be determined by using Eq. 7.111 which produces,

7-98

Chapter 7: Solutions of Electrolytes

log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

Solutions

I /mol dm −3

I /moldm −3

γ ± =10−0.51(1×2) 0.012 γ ± = 0.773151 491 9 γ ± = 0.773 Back to Problem 7.40

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7-99

Chapter 7: Solutions of Electrolytes

7.41.

Solutions

A 0.1 M solution of sodium palmitate, C 15 H 31 COONa, is separated from a 0.2 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl– ions but not to palmitate ions. Calculate the concentrations of Na+ and Cl– ions on the two sides of the membrane after equilibrium has become established. (For a calculation of the Nernst potential, see Problem 8.18.)

Solution: Given: = cNaP 0.1 = M, cNaCl 0.2 M Required: cNa + , cCl− on each side of the membrane at equilibrium Initial concentrations

Final concentrations

Palmitate Side Na + = 0.1 M

Other Side Na + = 0.2 M

P − = 0.1 M = Na + ( 0.1 M + x )

Cl− = 0.2 M = Na + ( 0.2 M − x )

P − = 0.1 M Cl− = x

Cl− =

( 0.2 M − x )

At equilibrium, the final concentration of NaCl on each side of the membrane will be the same. This allows us to solve for x.

( 0.2 M − x )

2

= ( 0.1 + x ) x

0.04 − 0.4 x + = x 2 0.1x + x 2 0.04 = 0.5 x x = 0.08

7-100

Chapter 7: Solutions of Electrolytes

Solutions

The final concentrations are thus, Final concentrations

Palmitate Side = Na + ( 0.1 M + 0.08 M )

Na + = 0.18 M Cl− = 0.08 M

Back to Problem 7.41

Other Side + Na = = Cl− + Na = = Cl−

( 0.2 M − x ) ( 0.2 M − 0.08 M )

+ = Cl− 0.12 M Na =

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7-101

Chapter 7: Solutions of Electrolytes

7.42.

Solutions

Consider the ionizations H + + H 3 N + CH 2 COO − H 3 N + CH 2 COOH H 2 NCH 2 COOH + H + Assume that the following acid dissociation constants apply to the ionizations:

− NH 3+ − NH 2 + H + ; K a = 1.5 ×10−10 M −COOH − COO − + H + ; K a =× 4.0 10−3 M Estimate a value for the equilibrium constant for the process H 3 N + CH 3COO − H 2 NCH 2 COOH

Solution: Given: K − NH+ = 1.5 ×10−10 M, K − COOH = 4.0 ×10−3 M 3

Required: K NH+ CH COO 3

3

The equilibrium constant K − NH+ is given by, 3

K − NH+ = 3

[ H 2 NCH 2COOH ] H + + H 3 N CH 2 COOH

And K − COOH is given by, K − COOH

H 3 N + CH 2 COO − H + = H 3 N + CH 2 COOH

We can rearrange the two expressions to obtain an expression for K NH+ CH COO , and solve for its value. 3

3

7-102

Chapter 7: Solutions of Electrolytes

K NH+ CH COO = 3

3

K NH+ CH COO = 3

3

K NH+ CH COO = 3

3

Solutions

K − NH+ 3

K − COOH

[ H 2 NCH 2COOH ]

H 3 N + CH 2 COO − 1.5 ×10−10 M 4.0 ×10−3 M

K NH+ CH COO = 3.75 ×10−8 3

3

K NH+ CH COO = 3.8 ×10−8 3

3

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7-103

Chapter 7: Solutions of Electrolytes

7.43.

Solutions

The pK values for the successive ionizations of phosphoric acid are given on p. 308. Which of the four species is predominant at the following values of the hydrogen or hydroxide concentration? a. [H+] = 0.1 M. b. [H+] = 2 × 10–3 M. c. [H+] = 5 × 10–5 M. d. [OH–] = 2 × 10–3 M. e. [OH–] = 1 M.

Solution: Given: p. 308 Required: see above The successive ionizations of phosphoric acid are given by the following expression. = pK1 2.1 = pK 2 7.2 − 3 4 2 4

H PO H PO

pK3 =12.3

HPO 24− PO34−

In order to determine the pH we will use, pH = − log10 H + and pH= 14 − pOH . a.

pH = − log10 ( 0.1)

H 3 PO 4 is predominant

pH = 1 b.

pH = − log10 ( 2 ×10−3 )

HPO 24− is predominant

pH = 2.698

c.

pH = − log10 ( 5 ×10−5 )

HPO 24− is predominant

pH = 4.301

d.

pOH = − log10 ( 2 ×10−3 )

HPO 24− is predominant

pOH = 2.698 pH= 14 − 2.698 pH = 11.304

7-104

Chapter 7: Solutions of Electrolytes

e.

pOH = − log10 1

Solutions

PO34− is predominant

pOH = 0 pH = 14 Back to Problem 7.43

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7-105

Chapter 7: Solutions of Electrolytes

7.44.

Solutions

Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl– ions but not to P– ions. The initial concentrations are as shown below.

[K+] = 0.05 M

[K+] = 0.15 M

[Cl–] = 0.05 M

[P–] = 0.15 M

Calculate the concentrations on each side of the membrane after equilibrium has become established. (See Problem 8.26 in Chapter 8 for the calculation of the Nernst potential for this system.) Solution: Given: see above Required: cK + , cCl− on each side of the membrane at equilibrium We can solve this problem in a similar manner as problem 7.41. Initial concentrations

Final concentrations

Left-hand Side K + = 0.05 M

Right-hand Side K + = 0.15 M

Cl− = 0.05 M K + 0.05 M − x =

P − = 0.15 M K + 0.15 M + x =

Cl− 0.05 M − x =

Cl− = x P − = 0.1 M

At equilibrium, the final concentration of KCl on each side of the membrane will be the same. This will allow us to solve for x.

( 0.05 M − x )

2

=

( 0.15 + x ) x

0.0025 − 0.1x += x 2 0.15x + x 2 0.0025 = 0.25 x x = 0.01

7-106

Chapter 7: Solutions of Electrolytes

Solutions

The final concentrations are thus, Final concentrations

Left-hand Side Right-hand Side + − K + 0.15 M + 0.01 M = K = Cl 0.05 M − 0.01 M = = K + = Cl− 0.04 M

Back to Problem 7.44

K + = 0.16 M Cl− = 0.01 M

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7-107

CHAPTER

8

Electrochemical Cells

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 8: Electrochemical Cells

Electrode Reactions and Electrode Potentials

Chapter 8 *problems with an asterisk are slightly more demanding Electrode Reactions and Electrode Potentials 8.1.

Write the electrode reactions, the overall reaction, and the expression for the emf for each of the following reversible cells. a. Pt, H 2 (1 bar)|HCl(aq)|Pt, Cl 2 (1 bar) b. Hg|Hg 2 Cl 2 (s)|HCl(aq)|Pt, H 2 (1 bar) c. Ag|AgCl(s)|KCl(aq)|Hg 2 Cl 2 (s)|Hg d. Pt, H 2 (1 bar)|HI(aq)|AuI(s)|Au e. Ag|AgCl(s)|KCl(c 1 )

KCl(c 2 )|AgCl(s)|Ag Solution

8.2.

At 25 °C and pH 7, a solution containing compound A and its reduced form AH 2 has a standard electrode potential of –0.60 V. A solution containing B and BH 2 has a standard potential of –0.16 V. If a cell were constructed with these systems as half-cells, a. Would AH 2 be oxidized by B or BH 2 oxidized by A under standard conditions? b. What would be the reversible emf of the cell? c. What would be the effect of pH on the equilibrium ratio [B][AH 2 ]/[A][BH 2 ]? Solution

8.3.

Calculate the standard electrode potential for the reaction Cr2+ + 2e– → Cr at 298 K. The necessary E° values are a. Cr3+ + 3e– → Cr

E° = –0.74 V

b. Cr3+ + e– → Cr2+

E° = –0.41 V Solution

8-2

Chapter 8: Electrochemical Cells

8.4.

Thermodynamics of Electrochemical Cells

Write the individual electrode reactions and the overall cell reaction for the following cell: Pt, H 2 |H+(1 m)

F2–, S2–, H+(aq)|Pt

where F2– represents the fumarate ion and S2– the succinate ion. Write the expression for the emf of the cell. Solution 8.5.

Design electrochemical cells in which each of the following reactions occurs: a. Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq) b. Ag+(aq) + Cl–(aq) → AgCl(s) c. HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) In each case, write the representation of the cell and the reactions at the two electrodes. Solution

Thermodynamics of Electrochemical Cells 8.6.

Calculate the equilibrium constant at 25 °C for the reaction 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I 2 (s) using the standard electrode potentials given in Table 8.1. Solution

8.7.

From data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction Sn + Fe2+ → Sn2+ + Fe Solution

8-3

Chapter 8: Electrochemical Cells

8.8.

Thermodynamics of Electrochemical Cells

The standard electrode potential at 25 °C for cytochrome c (Fe3+) + e– → cytochrome c (Fe2+) is 0.25 V. Calculate ∆G° for the process 1 H 2 (g) + cytochrome c (Fe3+) → H+ + cytochrome c (Fe2+) 2

Solution 8.9.

Using the values given in Table 8.1, calculate the standard Gibbs energy change ∆G° for the reaction H2 +

1 O2 → H2O 2 Solution

*8.10. From the data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction 2Cu+ → Cu2+ + Cu What will be produced if Cu 2 O is dissolved in dilute H 2 SO 4 ? Solution 8.11.

For the reaction 3H 2 (g, 1 atm) + Sb 2 O 3 (s, cubic) → 2Sb(s) + 3H 2 O(l), ∆G° = –83.7 kJ [Roberts and Fenwick, J. Amer. Chem. Soc., 50, 2146(1928)]. Calculate the potential developed by the cell Pt|H 2 (g, 1 atm)|H+|Sb 2 O 3 (s, cubic)|Sb(s) Which electrode will be positive? Solution

8-4

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

Nernst Equation and Nernst Potentials 8.12.

Calculate the emf for the following cell at 25 °C: Pt, H 2 (1 bar)|HCl(0.5 m)

HCl(1.0 m)|Pt, H 2 (1 bar) Solution

8.13.

The pyruvate-lactate system has an E°′ value of –0.185 V at 25 °C and pH 7.0. What will be the potential of this system if the oxidation has gone to 90% completion? Solution

8.14.

a. From the data in Table 8.1, calculate the standard electrode potential for the half-reaction Fe3+ + 3e– → Fe b. Calculate the emf at 25 °C of the cell Pt|Sn2+(0.1 m), Sn4+(0.01 m)

Fe3+(0.5 m)|Fe Solution

8.15.

The cell Pt|H 2 (1 bar), H+ KCl(saturated)|Hg 2 Cl 2 |Hg was used to measure the pH of a solution of 0.010 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential expected at 25 °C [K a = 1.81 × 10–5 for acetic acid]. Solution

8.16.

The voltage required to electrolyze certain solutions changes as the electrolysis proceeds because the concentrations in the solution are changing. In an experiment, 500 dm3 of a 0.0500 M solution of copper (II) bromide was electrolyzed until 2.872 g Cu was deposited. Calculate the theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end of the experiment. Solution

8-5

Chapter 8: Electrochemical Cells

8.17.

Nernst Equation and Nernst Potentials

Calculate the concentration of I3– in a standard solution of iodine in 0.5 M KI, making use of the following standard electrode potentials: I 2 + 2e– → 2I–

Eº = 0.5355 V

I3− + 2e– → 3I–

Eº = 0.5365 V

The molality of I– in the standard solution can be assumed to be 0.5 m. Solution 8.18.

Calculate the Nernst potential at 25 °C arising from the equilibrium established in Problem 7.41. Solution

8.19.

It might seem plausible to separate lead and gold by making use of the great difference between their standard electrode potentials (Table 8.1). In order to test this idea, one might electrolyze a solution containing 0.0100 M AuNO 3 and 0.0100 M Pb(NO 3 ) 2 in a well-stirred tank using platinum electrodes at low current density. As the potential difference is slowly increased from zero, which metal will be deposited first? What will be the concentration of this metal ion in solution when the second metal begins to be deposited? Do you think this is an acceptable method of separating the two metals? Solution

8.20.

Calculate the emf of the cell Pt, H 2 (1 bar)|HCl(0.1 m)

HCl(0.2 m)|Pt, H 2 (10 bar) Solution

*8.21. Suppose that the cell in Problem 8.20 is set up but that the two solutions are separated by a membrane that is permeable to H+ ions but impermeable to Cl– ions. What will be the emf of the cell at 25 °C? Solution

8-6

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

*8.22. A typical biological cell has a volume of 10–9 cm3, a surface area of 10–6 cm2, and a membrane thickness of 10–6 cm; the dielectric constant of the membrane may be taken as 3. Suppose that the concentration of K+ ions inside the cell is 0.155 M and that the Nernst potential across the cell wall is 0.085 V. a. Calculate the net charge on either side of the wall, and b. Calculate the fraction of the K+ ions in the cell that are required to produce this charge. Solution *8.23. Calculate the emf at 25 °C of the cell Pt, H 2 (1 bar)|H 2 SO 4 (0.001 m)|CrSO 4 (s)|Cr given the following standard electrode potential: CrSO 4 (s) + 2e– → Cr + SO 2– 4

Eº = –0.40 V

a. First make the calculation neglecting activity coefficient corrections. b. Then make the calculation using activity coefficients estimated on the basis of the Debye-Hückel limiting law. Solution *8.24. Write the individual electrode reactions and the overall reaction for Cu|CuCl 2 (aq)|AgCl(s)|Ag If the emf of the cell is 0.191 V when the concentration of CuCl 2 is 1.0 × 10–4 M and is –0.074 V when the concentration is 0.20 M, make an estimate of the mean activity coefficient in the latter solution. Solution

8-7

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

*8.25. a. Write both electrode reactions and the overall reaction for the cell Tl|TlCl(s)|CdCl 2 (0.01 m)|Cd b. Calculate E and E° for this cell at 25 °C from the following information: Tl+ + e– → Tl

Eº = –0.34 V

Cd + 2e → Cd

Eº = –0.40 V

2+

–

The solubility product for TlCl is 1.6 × 10–3 mol2 dm–6 at 25 °C. Solution 8.26.

Problem 7.44 involved calculating the concentrations on each side of a membrane after a Donnan equilibrium had become established. Which side of the membrane is positively charged? Calculate the Nernst potential across the membrane if the temperature is 37 °C. Solution

8.27.

The oxidation of lactate to pyruvate by the oxidized form of cytochrome c—represented as cytochrome c (Fe3+)—is an important biological reaction. The following are the relevant E°′ values, relating to pH 7 and 25 °C: Eº′/V

pyruvate + 2H + 2e → lactate –

+

–

–

–0.185

cytochrome c (Fe ) + e → cytochrome c (Fe ) 3+

–

2+

0.254

Calculate the equilibrium ratio [cytochrome c (Fe 2+ )]2 [pyruvate – ] [cytochrome c (Fe3+ )]2 [lactate – ] at pH 7 and 25 °C. Also calculate the ratio at pH 6. Solution

8-8

Chapter 8: Electrochemical Cells

8.28.

Nernst Equation and Nernst Potentials

Suppose that the cell Ag|AgCl(s)|HCl(0.10 m)

HCl(0.01 m)|AgCl(s)|Ag

is set up and that the membrane separating the two solutions is permeable only to H+ ions. What is the emf of the cell at 25 °C? Solution 8.29.

a. Consider the cell Pt, H 2 (1 bar)|HCl(m 1 )

HCl(m 2 )|Pt, H 2 (1 bar)

in which the solutions are separated by a partition that is permeable to both H+ and Cl–. The ratio of the speeds with which these ions pass through the membrane is the ratio of their transport numbers t + and t – . Derive an expression for the emf of this cell. b. If when m 1 = 0.01 m and m 2 = 0.01 m the emf is 0.0190 V, what are the transport numbers of the H+ and Cl– ions? Solution 8.30.

The metal M forms a soluble nitrate and a very slightly soluble chloride. The cell M|M+(0.1 m), HNO 3 (0.2 m)|H 2 (1 bar), Pt has a measured E = –0.40 V at 298.15 K. When sufficient solid KCl is added to make the solution of the cell 0.20 m in K+, the emf changes to –0.15 V at 298.15 K as MCl precipitates. Calculate the K sp of MCl, taking all activity coefficients to be unity. Solution

8-9

Chapter 8: Electrochemical Cells

8.31.

Temperature Dependence of Cell emfs

The substance nicotinamide adenine dinucleotide (NAD+) plays an important role in biological systems; under the action of certain enzymes it can react with a reducing agent and release a proton to the solution to form its reduced form NADH. With pyruvate the reduced form NADH undergoes the reaction NADH + pyruvate– + H+ NAD+ + lactate– The appropriate E°′ values, relating to 25 °C and pH 7, are pyruvate– + 2H+ +2e– → lactate–

Eº′ = –0.19 V

NAD + H + 2e → NADH

Eº′ = –0.34 V

+

+

–

Use these values to calculate ∆G°′ for the reaction, and also the equilibrium ratio

[lactate – ][NAD + ] [pyruvate – ][NADH] (a) at pH 7, and (b) at pH 8. Solution Temperature Dependence of Cell emfs 8.32.

a. Calculate the standard emf E° for the reaction fumarate2– + lactate– → succinate2– + pyruvate– on the basis of the following information: fumarate2– + 2H+ + 2e– → succinate2–

E°′ = 0.031 V

pyruvate + 2H + 2e → lactate

E°′ = –0.185 V

–

+

–

–

The E°′ values relate to pH 7. The temperature coefficient ∂E/∂T for this cell is 2.18 × 10–5 V K–1. b. Calculate ∆G°, ∆H°, and ∆S° at 25 °C. Solution 8-10

Chapter 8: Electrochemical Cells

8.33.

Temperature Dependence of Cell emfs

The Weston standard cell (see Figure 8.2b) is Cd amalgam|CdSO 4 ·

8 H 2 O(s)|Hg 2 SO 4 (s), Hg 3

(saturated solution) a. Write the cell reaction. b. At 25 °C, the emf is 1.018 32 V and ∂E °/∂T = –5.00 × 10–5 V K–1. Calculate ∆G°, ∆H°, and ∆S°. Solution 8.34.

Salstrom and Hildebrand [J. Amer. Chem. Soc., 52, 4650(1930)] reported the following data for the cell Ag(s)|AgBr(s)|HBr(aq)|Br 2 (g, 1 atm)|Pt

t/ºC

442.3

456.0

490.9

521.4

538.3

556.2

E/V

0.8031

0.7989

0.7887

0.7803

0.7751

0.7702

Find the temperature coefficient for this cell assuming a linear dependence of the cell potential with temperature. What is the entropy change for the cell reaction? Solution 8.35.

The reaction taking place in the cell Mg(s)|Mg2+(aq), Cl–(aq)|Cl 2 (g,1 atm)|Pt is found to have an entropy change of –337.3 J K–1 mol–1 under standard conditions. What is the temperature coefficient for the cell? Solution

8-11

Chapter 8: Electrochemical Cells

Temperature Dependence of Cell emfs

*8.36. a. Estimate the Gibbs energy of formation of the fumarate ion, using data in Problem 8.32 and the following values. ∆ f Gº (succinate, aq) = –690.44 kJ mol–1 ∆ f Gº (acetaldehyde, aq) = 139.08 kJ mol–1 ∆ f Gº (ethanol, aq) = –181.75 kJ mol–1 acetaldehyde + 2H+ + 2e– → ethanol Eº′ = –0.197 V b. If the ∂E °/∂T value for the process fumarate2– + ethanol → succinate2– + acetaldehyde is 1.45 × 10–4 V K–1, estimate the enthalpy of formation of the fumarate ion from the following values. ∆ f Hº (succinate, aq) = –908.68 kJ mol–1 ∆ f Hº (acetaldehyde, aq) = –210.66 kJ mol–1 ∆ f Hº (ethanol, aq) = –287.02 kJ mol–1 Solution *8.37. a. Calculate the emf at 298.15 K for the cell Tl|TlBr|HBr (unit activity)|H 2 (1 bar), Pt b. Calculate ∆H for the cell reaction in the following cell. Tl|Tl+ (unit activity), H+ (unit activity)|H 2 (1 bar), Pt For the half-cell Tl+ + e– → Tl E° = 0.34 V ∂E/∂T = –0.003 V/K and

K sp (TlBr) = 10–4 mol2 dm–6 Solution

8-12

Chapter 8: Electrochemical Cells

Applications of emf Measurements

Applications of emf Measurements 8.38.

Calculate the solubility product and the solubility of AgBr at 25 °C on the basis of the following standard electrode potentials: AgBr(s) + e– → Ag + Br–

E° = 0.0713 V

Ag + e → Ag

E° = 0.7996 V

+

–

Solution 8.39.

The emf of a cell Pt, H 2 (1 bar)|HCl(aq)|AgCl(s)|Ag was found to be 0.517 V at 25 °C. Calculate the pH of the HCl solution. Solution

8.40.

The emf of the cell Ag|AgI(s)I–(aq)

Ag+(ag)|Ag

is –0.9509 V at 25 °C. Calculate the solubility and the solubility product of AgI at that temperature. Solution 8.41.

An electrochemical cell M(s)|MCl(aq, 1.0 m)|AgCl(s)|Ag(s), where MCl is the chloride salt of the metal electrode M, yields a cell potential of 0.2053 V at 25 °C. What is the mean activity coefficient γ ± of the electrolyte MCl? E° for the M(s)|M+ electrode is 0.0254 V. Solution

8.42.

The following thermodynamic data apply to the complete oxidation of butane at 25 °C. C 4 H 10 (g) + (13/2)O 2 (g) → 4CO 2 (g) + 5H 2 O(l) ∆H ° = –2877 kJ mol –1 ∆S ° = –432.7 J K –1 mol –1 Suppose that a completely efficient fuel cell could be set up utilizing this reaction. Calculate (a) the maximum electrical work and (b) the maximum total work that could be obtained at 25 °C. 8-13

Chapter 8: Electrochemical Cells

Essay Questions

Solution *8.43. At 298 K the emf of the cell Cd, Hg|CdCl 2 (aq, 0.01 m), AgCl(s)|Ag is 0.7585 V. The standard emf of the cell is 0.5732 V. a. Calculate the mean activity coefficient for the Cd2+ and Cl– ions. b. Compare the value with that estimated from the Debye-Hückel limiting law, and comment on any difference. Solution

*8.44. The following emf values were obtained by H. S. Harned and Copson [J. Amer. Chem. Soc., 55, 2206(1933)] at 25 °C for the cell Pt,H 2 (1 bar)|LiOH(0.01 m), LiCl(m)|AgCl(s)|Ag at various molalities m of LiCl: m/mol kg–1 E/V

0.01

0.02

0.05

0.10

0.20

1.0498

1.0318

1.0076

1.9888

0.9696

Obtain from these data the ionic product of water. Solution

Essay Questions 8.45.

Explain how emf measurements can be used to obtain ∆G°, ∆H°, and ∆S° for a reaction.

8.46.

Suggest an additional example, giving details, for each of the electrochemical cells listed in Figure 8.8.

8-14

Chapter 8: Electrochemical Cells

Solutions

Solutions 8.1.

Write the electrode reactions, the overall reaction, and the expression for the emf for each of the following reversible cells. a. Pt, H 2 (1 bar)|HCl(aq)|Pt, Cl 2 (1 bar) b. Hg|Hg 2 Cl 2 (s)|HCl(aq)|Pt, H 2 (1 bar) c. Ag|AgCl(s)|KCl(aq)|Hg 2 Cl 2 (s)|Hg d. Pt, H 2 (1 bar)|HI(aq)|AuI(s)|Au e. Ag|AgCl(s)|KCl(c 1 )

KCl(c 2 )|AgCl(s)|Ag

Solution: Given: see above Required: electrode reactions, the overall reaction, and the expression for the emf We calculate the emf, or electromotive force, from Eq. 8.13 u

RT [Y] y [Z]z E = E° − ln where z is the number of electrons. zF [A]a [B]b

Electrode reactions

H 2 → 2H + + 2e − Cl2 + 2e − → 2Cl−

Overall reaction H 2 + Cl2 → 2H + + 2Cl− z=2

2Hg ( l ) + 2Cl− → Hg 2 Cl2 + 2e −

2Hg ( l ) + 2Cl− + 2H + → Hg 2 Cl2 + H 2

2H + + 2e − → H 2

z=2

EMF E= E ° −

(

2 2 RT ln H + Cl− 2F

)

u

RT 1 E= E° − ln 2 F H + 2 Cl− 2 2 2 u RT E= E° + ln H + Cl− 2F

(

u

)

8-15

Chapter 8: Electrochemical Cells

Solutions

Ag ( s ) + Cl− → AgCl ( s ) + e −

2Ag ( s ) + Hg 2 Cl2 ( s ) → 2AgCl ( s ) + 2Hg ( s )

Hg 2 Cl2 ( s ) + 2e − → 2Hg ( s ) + 2Cl−

z=2

1 H 2 ( g ) → H + + e− 2 AuI ( s ) + e − → Au ( s ) + I −

AuI ( s ) +

Ag ( s ) + Cl− ( c1 ) → AgCl ( s ) + e −

Cl− ( c1 ) → Cl− ( c2 )

AgCl ( s ) + e − → Ag ( s ) + Cl− ( c2 )

z =1

Back to Problem 8.1

z =1

1 H 2 ( g ) → Au ( s ) + H + + I − 2

No concentration dependence, therefore E= E ° E= E ° −

(

RT ln H + I − F

RT c1 E= E ° − ln F c2

)

u

u

Back to Top

8-16

Chapter 8: Electrochemical Cells

8.2.

Solutions

At 25 °C and pH 7, a solution containing compound A and its reduced form AH 2 has a standard electrode potential of –0.60 V. A solution containing B and BH 2 has a standard potential of –0.16 V. If a cell were constructed with these systems as half-cells, a. Would AH 2 be oxidized by B or BH 2 oxidized by A under standard conditions? b. What would be the reversible emf of the cell? c. What would be the effect of pH on the equilibrium ratio [B][AH 2 ]/[A][BH 2 ]?

Solution: Given: T = 25 °C, pH = 7, EA° = −0.60 V, EB° = −0.16 V Required: see above a. The reduction potential for each half reaction is: A + 2H + + 2e − → AH 2

E ° = −0.60 V

B + 2H + + 2e − → BH 2

E ° = −0.16 V

We reverse the first reaction since we require a positive potential for the overall reaction to be spontaneous in the forward direction. Therefore AH 2 is oxidized. AH 2 + B → A + BH 2 b. The reversible emf would be = E EB° − EA° E =−0.16 V − ( −0.60 V ) E = 0.44 V c. The equilibrium ratio is given by K =

[ B][ AH 2 ] . There is no dependence of [H O+ ] in the equilibrium expression, and the hydrogen3 [ A ][ BH 2 ]

containing entities cancel in the numerator and denominator. As a result, there is no effect of pH on the equilibrium ratio. Back to Problem 8.2

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8-17

Chapter 8: Electrochemical Cells

8.3.

Solutions

Calculate the standard electrode potential for the reaction Cr2+ + 2e– → Cr at 298 K. The necessary E° values are a. Cr3+ + 3e– → Cr E° = –0.74 V b. Cr3+ + e– → Cr2+

E° = –0.41 V

Solution: Given: see above Required: E° for Cr2+ To solve this problem, we follow Example 8.3. We first calculate the ∆G° values for these two reactions using Eq. 8.2 ∆G º = − zFE º 3+ Cr + 3e − → Cr ∆G1o =−3 × 96 485 J mol−1 × (−0.74V) = 2.22 V × 96 485 J mol−1 3+

−

Cr + e → Cr

2+

−1

∆G =−1× 96 485 J mol × (−0.41V) = 0.41 V × 96 485 J mol o 2

−1

(1) (2) The reaction Cr2+ + 2e– → Cr is obtained

by subtracting reaction (2) from reaction (1), and the ∆G° for this reaction is obtained from by subtracting ∆G2º from ∆G1º .

∆G° = ∆G1º − ∆G2º = ∆G° 2.22 V × 96 485 J mol –1 − 0.41 V × 96 485 J mol –1 = ∆G° 1.81 V × 96 485 J mol –1 = ∆G° 1.81 V × F We solve for E° by rearranging Eq. 8.2 and setting z = 2 . ∆G º Eº = − zF 1.81 V F Eº = −2 F E º = −0.905 V E º = −0.91 V Back to Problem 8.3

Back to Top

8-18

Chapter 8: Electrochemical Cells

8.4.

Solutions

Write the individual electrode reactions and the overall cell reaction for the following cell: Pt, H 2 |H+(1 m) F2–, S2–, H+(aq)|Pt 2– 2– where F represents the fumarate ion and S the succinate ion. Write the expression for the emf of the cell.

Solution: Given: see above Required: E We first write down the half reactions that are occurring in this cell. LHS electrode H 2 → 2H + (1 m ) + 2e − RHS electrode 2e − + 2H + ( aq ) + F2− → S2− The overall reaction is, 2H + ( aq ) + F2− + H 2 → 2H + (1 m ) + S2− The expression for the emf of the cell is determined from Eq. 8.13, RT [Y] y [Z]z E = E° − ln zF [A]a [B]b

u

+ 2 2− RT S H E= E° − ln 2 F F2 − H + 2 aq

u

2 2− RT S (1 m ) E= E° − ln 2 F F2− [ C]2 + Where [H ]=C is the concentration of [H+] on the LHS. We drop the superscript u since we have numerical values for the concentration.

Back to Problem 8.4

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8-19

Chapter 8: Electrochemical Cells

8.5.

Solutions

Design electrochemical cells in which each of the following reactions occurs: a. Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq) b. Ag+(aq) + Cl–(aq) → AgCl(s) c. HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) In each case, write the representation of the cell and the reactions at the two electrodes.

Solution: Given: see above Required: design the electrochemical cells a. In writing the representation of the cell, the oxidation reaction always occurs at the anode, which is placed at the left-hand position of the cell. In this case, Fe2+ is losing electrons, and therefore the oxidation process is: Fe 2+ → Fe3+ + e −

The cathode reaction is written on the right-hand side of the cell and is: Ce 4+ + e − → Ce3+

This is where reduction occurs. The overall reaction is the sum of these two reactions. The cell representation is: Fe3+(aq) | Fe2+(aq)

Ce4+(aq)| Ce3+(aq)

b. Upon examining the standard reduction potentials in Table 8.1, we see that the following half reactions can be combined to give the cited reaction. + Ag (aq) + e − → Ag Eo = 0.7996 V (1)

AgCl(s) + e − → Ag + Cl− (aq)

Eo = 0.22233 V

(2)

Reversal of equation (2), and then addition to equation (1) gives the overall desired equation: Ag + (aq) + Cl – (aq) → AgCl(s)

8-20

Chapter 8: Electrochemical Cells

Solutions

Equation (1) is the reduction reaction and is placed on the right-hand side of the cell. The anode reaction is placed on the left-hand side of the cell representation. Ag|AgCl(s)|Cl-(aq)

Ag+(aq)|Ag(s)

The voltage of this cell is the right-hand reduction potential minus the left-hand reduction potential. E= ° E1° − E2° = E ° 0.7996 V − 0.22233 V E ° =0.577 27 V c. HgO undergoes reduction to Hg and is the cathode. H 2 is oxidized and is the anode. The electrode potentials are obtained from Table 8.1 and the SRP Table. −0.8277 V Eo = 2H 2 O + 2e − → H 2 + 2OH − (3)

HgO + H 2 O + 2e − → Hg + 2OH −

Eo = 0.0977 V

(4)

Reversing the sense of equation (3) and adding to (4) gives, HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) The cell is represented by Pt, H 2 (g)|H 2 O(l),OH-(aq)

HgO (s),H 2 O|OH-(aq) |Hg(l)

The cell potential is E= ° E4° − E3°

E ° 0.0977 V − ( −0.8277 V ) = E ° =0.9254 V

Back to Problem 8.5

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8-21

Chapter 8: Electrochemical Cells

8.6.

Solutions

Calculate the equilibrium constant at 25 °C for the reaction 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I 2 (s) using the standard electrode potentials given in Table 8.1.

Solution: Given: Table 8.1 Required: K We can calculate the equilibrium constant from Eq. 8.7 Eo / V =

0.0257 ln K o z

z(E o / V ) K o = exp 0.0257 The half cell reactions are

Fe3+ + e − → Fe 2+

Eo = 0.771 V

I 2 + 2e − → 2I

Eo = 0.5355 V

And the overall emf is = E ° 0.771 V − 0.5355 V E ° =0.2355 V

We solve for K with z = 2 ,

8-22

Chapter 8: Electrochemical Cells

Solutions

2(0.2355) K o = exp 0.0257 K o = exp (18.32684825 ) K o = 91 043 525.2 = K o 9.10 ×107

Back to Problem 8.6

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8-23

Chapter 8: Electrochemical Cells

8.7.

Solutions

From data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction Sn + Fe2+ → Sn2+ + Fe

Solution: Given: Table 8.1 Required: K We follow the same procedure as we did in problem 8.6 to solve for the equilibrium constant. The half cell reactions are Fe 2+ + 2e − → Fe

Eo = −0.447 V

Sn 2+ + 2e − → Sn

Eo = −0.1375 V

The overall emf is E ° = −0.447 V − ( −0.1375 V ) E ° = −0.3095 V

Solving for K with z = 2 gives, 2(−0.3095 K o = exp 0.0257 = K o exp ( −24.08560311) = K o 3.46 541 679 ×10−11 = K o 3.47 ×10−11

Back to Problem 8.7

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8-24

Chapter 8: Electrochemical Cells

8.8.

Solutions

The standard electrode potential at 25 °C for cytochrome c (Fe3+) + e– → cytochrome c (Fe2+) is 0.25 V. Calculate ∆G° for the process 1 H 2 (g) + cytochrome c (Fe3+) → H+ + cytochrome c (Fe2+) 2

Solution: Given: E° = 0.25 V Required: ∆G° We calculate ∆G° for this reaction using Eq. 8.2, ∆G º = − zFE º , where z = 1

∆G o =−1× 96 485 mol−1 × 0.25 J ∆G o = −24 121.25 J mol−1 ∆G o = −24 kJ mol−1

Back to Problem 8.8

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8-25

Chapter 8: Electrochemical Cells

8.9.

Solutions

Using the values given in Table 8.1, calculate the standard Gibbs energy change ∆G° for the reaction H2 +

1 O2 → H2O 2

Solution: Given: Table 8.1 Required: ∆G° From Table 8.1, we write the following half reactions:

O 2 + 4H + + 4e – → 2H 2 O

E1° = 1.229 V

(1)

2H + + 2e – → H 2

E2° = 0

(2)

Subtracting (2) from ½ of (1) gives the desired equation, 1 H 2 + O2 H 2O 2 The overall emf is = E ° 1.229 V − 0 E ° =1.229 V

We calculate ∆G° for this reaction using Eq. 8.2, ∆G o = − zFE , where z=2 ∆G o =−2 × 96 485 mol−1 ×1.229 J ∆G o = −237 160.13 J mol−1 ∆G o = −237.2 kJ mol−1

Back to Problem 8.9

Back to Top 8-26

Chapter 8: Electrochemical Cells

Solutions

*8.10. From the data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction 2Cu+ → Cu2+ + Cu What will be produced if Cu 2 O is dissolved in dilute H 2 SO 4 ? Solution: Given: Table 8.1 Required: K We follow the same procedure as we did in problem 8.6 to solve for the equilibrium constant. From Table 8.1, we write the following half reactions:

Cu 2+ + 2e − → Cu

E1o = 0.3419 V (1)

Cu 2+ + e − → Cu +

E1o = 0.153 V (2)

To get the desired overall reaction, we subtract 2×(2) from (1): 2Cu + → Cu 2+ + Cu

E= ° E1° − E2° = E ° 0.3419 V − 0.153 V E ° =0.1889 V We solve for K from Eq. 8.7, with z = 2 , 2(0.1889 K o = exp 0.0257 K o = 2 422 690.131 = K o 2.42 ×106

If Cu 2 O is dissolved in dilute H 2 SO 4 , then half will form Cu2+ and half will form Cu. Back to Problem 8.10

Back to Top 8-27

Chapter 8: Electrochemical Cells

8.11.

Solutions

For the reaction 3H 2 (g, 1 atm) + Sb 2 O 3 (s, cubic) → 2Sb(s) + 3H 2 O(l), ∆G° = –83.7 kJ [Roberts and Fenwick, J. Amer. Chem. Soc., 50, 2146(1928)]. Calculate the potential developed by the cell Pt|H 2 (g, 1 atm)|H+|Sb 2 O 3 (s, cubic)|Sb(s) Which electrode will be positive?

Solution: Given: reaction above, ∆G° = –83.7 kJ Required: E°, positive electrode Note that the ∆G° given is for the reaction of 3 moles of H 2 to form 2 moles of Sb. The half cell reactions may be written as,

3H 2 → 6H + + 6e − Sb 2 O3 + 6H + + 6e − → 2Sb + 3H 2 O

We can solve for E° using Eq. 8.2,

∆G o = − zFE ∆G o ∆E o = − zF −83.7 ×103 J ∆E o = −6 × 96 485 o ∆E = 0.1 445 820 594 V ∆E o = 0.145 V

For this reaction to be spontaneous, the electron flow is from the hydrogen electrode (oxidation) to the antimony electrode (reduction).

Back to Problem 8.11

Back to Top 8-28

Chapter 8: Electrochemical Cells

8.12.

Solutions

Calculate the emf for the following cell at 25 °C: Pt, H 2 (1 bar)|HCl(0.5 m)

HCl(1.0 m)|Pt, H 2 (1 bar)

Solution: Given: T = 25 °C , cell above Required: E We calculate the emf for the cell from Eq. 8.27 where z = 1 ,

E=

RT m2 ln F m1

1.0 m 0.5 m E = 0.017 813 882 5 V

E = 0.0257 ln

E = 0.018 V

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8-29

Chapter 8: Electrochemical Cells

Solutions

The pyruvate-lactate system has an E°′ value of –0.185 V at 25 °C and pH 7.0. What will be the potential of this system if the oxidation has gone to 90% completion?

8.13.

Solution: Given: E°′ = −0.185 V, pH = 7.0, 90% completion Required: E The process is, pyruvate − + 2H + + 2e − lactate − And, the Nernst equation, given by Eq. 8.13 RT = E E − zF o

...[ Y ] y [ Z]z n [ A ]a [ B]b ...

u

For this process, this becomes

RT [lactate − ] E= E° − ln 2 F [pyruvate − ] Solving for E we get, 0.0257 10 ln 2 90 E = −0.156 765 664 2 V E= −0.185 V −

E = −0.157 V

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8-30

Chapter 8: Electrochemical Cells

8.14.

Solutions

a. From the data in Table 8.1, calculate the standard electrode potential for the half-reaction Fe3+ + 3e– → Fe b. Calculate the emf at 25 °C of the cell Pt|Sn2+(0.1 m), Sn4+(0.01 m)

Fe3+(0.5 m)|Fe

Solution: Given: Table 8.1 Required: see above a. To calculate the standard potential we follow example 8.3. We first calculate the ∆G° values for these two reactions using Eq. 8.2 ∆G o = − zFE

Fe3+ + e − → Fe 2+

∆G1o =−1× 96 485 J mol−1 × (0.771 V)=-0.771 V × 96 485 J mol−1

(1)

Fe 2+ + 2e − → Fe 2+

∆G o2 =−2 × 96 485 J mol−1 × (−0.447 V)=0.894 V × 96 485 J mol−1

(2)

Fe3+ + 3e − → Fe

The half-reaction given above is the sum of (1) and (2). The ∆G° value for the given reaction is

∆G o = ∆G1o + ∆G2o G o -0.771 V × 96 485 J mol−1 + 0.894 V × 96 485 J mol−1 ∆ = G o 0.123 V × 96 485 J mol−1 ∆= G o 0.123 V × F ∆=

We solve for E° by rearranging Eq. 8.2 and setting z = 3 .

8-31

Chapter 8: Electrochemical Cells

Solutions

∆G o − zF 0.123 V F Eo = −3 F Eo =

E o = −0.041 V b. The half cell reactions are: Fe3+ + 3e − → Fe

−0.041 V Eo =

−0.151 V Eo = Sn 2+ → Sn 4+ + 2e − The overall reaction is obtained from the sum of 2×(3) and 3×(4)

(3) (4)

2Fe3+ + 3Sn 2+ → 2Fe + 3Sn 4+

The electrical potential would be,

E= ° E3° + E4° E ° = −0.041 V −0.151 V E ° = −0.192 V From the Nernst equation, Eq. 8.13 we can calculate the emf using z = 6. u

RT [Y] y [Z]z ln E= E ° − zF [A]a [B]b 3 0.0257 ( 0.01) E= −0.192 − ln ( 0.1)3 ( 0.5 )2 6 E = −0.168 349 742 4 V E = −0.17 V

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8-32

Chapter 8: Electrochemical Cells

8.15.

Solutions

The cell Pt|H 2 (1 bar), H+ KCl(saturated)|Hg 2 Cl 2 |Hg was used to measure the pH of a solution of 0.010 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential expected at 25 °C [K a = 1.81 × 10–5 for acetic acid].

Solution: Given: cacetic acid = 0.010 M , csodium acetate = 0.0358 M , T = 25 °C, K a = 1.81× 10 –5 Required: E The equilibrium constant, K a is given by, H + CH 3COO − Ka = [CH3COOH ]

Rearranging for [H+], we obtain

K [ CH 3COOH ] H + = a CH 3COO −

(1.81×10 ) ( 0.010 –5

H + =

( 0.0358 M )

M

)

H + 5.055 865 922 ×10 –6 = From Table 8.1, the cell reactions are:

H 2 → 2H + + 2e –

E ° =0

Hg 2 Cl2 + 2e – → 2Hg + 2Cl –

E ° =0.2412 V

However, since the cathode reaction (the reduction reaction), is contained in the standard calomel (Hg 2 Cl 2(s) ) electrode, it is separated from the oxidation of H 2 , and its concentrations are constant throughout the measurement. This is a pH meter, so the only concentration that is relevant is [H+].

8-33

Chapter 8: Electrochemical Cells

Solutions

2 RT ln H + 2F E ° = 0 + 0.2412 V 2 0.0257 0.2412 V − ln ( 5.055 865 922 ×10 –6 ) E= 2 E = 0.554 610 508 5 V

E= E ° −

E = 0.55 V

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8-34

Chapter 8: Electrochemical Cells

8.16.

Solutions

The voltage required to electrolyze certain solutions changes as the electrolysis proceeds because the concentrations in the solution are changing. In an experiment, 500 dm3 of a 0.0500 M solution of copper (II) bromide was electrolyzed until 2.872 g Cu was deposited. Calculate the theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end of the experiment.

Solution: Given: V = 500 dm3, [Cu2+] = 0.0500 M, [Bi] = 0.100 M, m cu = 2.872 g Required: E i , E f The reactions taking place during the electrolysis are:

Cu 2+ + 2e − → Cu

Eo = 0.34190 V

2Br − → Br2 + 2e −

Eo = -1.80730 V

The overall reaction is: Cu 2+ + 2Br – → Cu + Br2

The electrical potential would be the sum of the two potentials, = Eo

0.34190 V − 1.08730 V

E o = −0.74540 V If the reverse reaction were taking place in a galvanic cell, the initial cell voltage would be: 2 RT ln Cu 2+ Br − 2F o E = 0.74540 V 0.0257 2 = Ei 0.74540 V − ln ( 0.0500 )( 0.100 ) 2 Ei = 0.8 430 715 966 V

= E

Eo −

Ei = 0.84 307 V

8-35

Chapter 8: Electrochemical Cells

Solutions

Therefore a minimum voltage of 0.84 307 V would have to be applied at the beginning in order for the electrolysis reaction to occur. At the end of the electrolysis, the concentrations are: m 1 + Cu 2= 0.0500 M − Cu × M Cu V 2+ = Cu 0.0500 M −

2.872 g

( 63.456 g mol ) ( 500 dm ) −1

3

Cu 2+ = 0.0500 M − 9.051 941 503 ×10−5 M 2+ Cu = 0.049 909 480 6 M = Br − 0.100 M − 2 × 9.051 941 503 ×10−5 M Br − = 0.099 818 961 2 M Therefore the final voltage required would be 0.0257 2 ln ( 0.0 499 094 806 M )( 0.0 998 189 612 ) 2 E f = 0.8 431 414 503 V

= Ef

0.74540 V −

E f = 0.84 314 V The E i and E f are close because a small amount of Cu2+ is plated out. Back to Problem 8.16

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8-36

Chapter 8: Electrochemical Cells

8.17.

Solutions

Calculate the concentration of I3– in a standard solution of iodine in 0.5 M KI, making use of the following standard electrode potentials: I 2 + 2e– → 2I–

Eº = 0.5355 V

I3− + 2e– → 3I–

Eº = 0.5365 V

The molality of I– in the standard solution can be assumed to be 0.5 m. Solution: Given: cKI = 0.5 M Required: cI– 3

The overall reaction of interest is obtained by reversing the second equation and adding it to the first. I 2 + I – → I3−

In this case, z = 2, and the standard electrode potential is = E º 0.5355 V − 0.5365 V E º = −0.0010 V

We can solve for the concentration of I3– using Eq. 8.7, where the equilibrium constant for this process is given by,

8-37

Chapter 8: Electrochemical Cells

Solutions

I3− Kc = − I 0.0257 Eo / V = ln K o z z ( Eo / V ) o K = exp 0.0257

z ( Eo / V ) I3− = exp 0.0257 I − o z(E /V ) cI− = cI− exp 3 0.0257

cI− =

2 −0.0010 ( 0.5 mol dm ) exp (0.0257 ) −3

cI− = 0.4 625 649 996 mol dm −3 3

3

cI− = 0.4 626 mol dm −3 3

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8-38

Chapter 8: Electrochemical Cells

8.18.

Solutions

Calculate the Nernst potential at 25 °C arising from the equilibrium established in Problem 7.41.

Solution: Given: Problem 7.41, T = 25 °C Required: ∆Φ ( ∆Φ = ∆E o for concentration cells) The Nernst potential is given by Eq. 8.19 as, ∆Φ =

RT c1 ln zF c2

From problem 7.41, we have the equilibrium concentrations of sodium and chloride as, Palmitate side

Other side

Na + 0.18 Na + 0.12 M = = M Cl− 0.08 Cl− 0.12 M = = M To calculate the Nernst potential, we only consider the Na+ equilibrium, therefore, 0.18 M 0.12 M ∆Φ = 0.010 420 453 3 V ∆Φ = 0.0257 ln

∆Φ = 10 mV

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8-39

Chapter 8: Electrochemical Cells

8.19.

Solutions

It might seem plausible to separate lead and gold by making use of the great difference between their standard electrode potentials (Table 8.1). In order to test this idea, one might electrolyze a solution containing 0.0100 M AuNO 3 and 0.0100 M Pb(NO 3 ) 2 in a well-stirred tank using platinum electrodes at low current density. As the potential difference is slowly increased from zero, which metal will be deposited first? What will be the concentration of this metal ion in solution when the second metal begins to be deposited? Do you think this is an acceptable method of separating the two metals?

Solution: Given:= Table 8.1, cAuNO3 0.0100 = M , cPb( NO3 ) 0.0100 M 2

Required: see above From Table 8.1 the two standard reduction potentials are given by,

Au + + e – → Au

E ° =1.692 V

Pb 2+ + 2e – → Pb

E ° = −0.1262 V

We can see that Au+ has a much higher reduction potential than Pb2+, therefore gold will be deposited first. As the Au+ concentration falls, the lead begins to be deposited. Therefore we have, 2Au(s) + Pb 2+ → 2Au + + Pb(s) With z = 2. The potential for this reaction is given by, E ° = −1.692 V − 0.1262 V E ° = −1.8182 V Following the procedure used in problem 8.17, we can calculate the concentration of Au+.

8-40

Chapter 8: Electrochemical Cells

Solutions

2

Au + Kc = Pb 2+

0.0257 ln K o z z ( Eo / V ) o K = exp 0.0257 Eo / V =

z ( Eo / V ) Au + = exp 0.0257 Pb 2+ 2

cAu + = cAu + = c= Au +

2 ( −1.8182 ) 0.0100 M exp 0.0257 1.88 335 766 ×10−32 M 1.88 ×10−32 M

The conclusion is that only an infinitesimal amount of gold will be left in the solution by the time the lead starts to deposit at the electrode. This is shown by the negligible concentration of gold. Therefore, this is an acceptable way to separate the two metals.

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8-41

Chapter 8: Electrochemical Cells

8.20.

Solutions

Calculate the emf of the cell Pt, H 2 (1 bar)|HCl(0.1 m)

HCl(0.2 m)|Pt, H 2 (10 bar)

Solution: Given: see above Required: E At the right-hand electrode we have the reaction: H + ( 0.2 m ) + e – →

1 H 2 (10 bar ) 2

And at the left-hand electrode we have, 1 H 2 (10 bar ) → H + ( 0.1 m ) + e – 2

The overall reaction, with z = 1, is: 1 1 H 2 (1 bar ) → H 2 (10 bar ) + H + ( 0.1 m ) 2 2

H + ( 0.2 m ) +

To calculate the cell emf, we use Eq. 8.7, = E°

0.0257 ln K °= , where z 1 z 1

E = 0.0257 ln

0.2 × (1 bar ) 2 1

0.1× (10 bar ) 2

E = −0.011 774 335 9 V E = −11.8 mV

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8-42

Chapter 8: Electrochemical Cells

Solutions

*8.21. Suppose that the cell in Problem 8.20 is set up but that the two solutions are separated by a membrane that is permeable to H+ ions but impermeable to Cl– ions. What will be the emf of the cell at 25 °C? Solution: Given: Problem 8.20, a membrane that is only permeable to H+ Required: E From problem 8.20, we have the half reactions at each electrode as: LHS : RHS :

1 H 2 (10 bar ) → H + ( 0.1 m ) + e – 2 1 H + ( 0.2 m ) + e – → H 2 (10 bar ) 2

Every H+ ion produced in the LHS solution will have to pass through the membrane to preserve the electrical neutrality (ion gradient).

H + ( 0.1 m ) → H + ( 0.2 m ) The net reaction is therefore, 1 1 H 2 (1 bar ) → H 2 (10 bar ) 2 2

The cell emf is calculated from Eq. 8.27, = E

RT m2 = ln , where z 1 F m1

We take the ratio of pressures as a measure of the ratio of molalities.

8-43

Chapter 8: Electrochemical Cells

Solutions

1

E = 0.0257 ln

(1 bar ) 2 1

(10 bar ) 2

E = −0.029 588 218 4 V E = −29.6 mV

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8-44

Chapter 8: Electrochemical Cells

Solutions

*8.22. A typical biological cell has a volume of 10–9 cm3, a surface area of 10–6 cm2, and a membrane thickness of 10–6 cm; the dielectric constant of the membrane may be taken as 3. Suppose that the concentration of K+ ions inside the cell is 0.155 M and that the Nernst potential across the cell wall is 0.085 V. a. Calculate the net charge on either side of the wall, and b. Calculate the fraction of the K+ ions in the cell that are required to produce this charge. Solution: 3 2 Given: = V 10 –9 cm= , A 10 –6 cm = , l 10 –6 cm, = ∈ 3,= cK + 0.155 M = , ∆Φ 0.085 V

Required: see above a. The net charge on either side of the wall can be calculated using Q = CV . The capacitance is given by Eq. 8.20 as, C=

∈0∈ A l

We solve for the net charge in the following expression, Q=

∈0∈ A l

V

(8.854 ×10 Q=

−12

C2 N −1 m −2

(10

–8

m)

) (3) (10

–10

m2

) ( 0.085 V )

= Q 2.257 77 ×10−14 C2 N −1 m −1 V = Q 2.257 77 ×10−14 F V where 1 F = 1 C V −1 = Q 2.3 ×10−14 C b. The fraction of K+ ions required to produce this charge is given by

8-45

Chapter 8: Electrochemical Cells

Solutions

Q 2.25 777 ×10−14 C = e 1.602 ×10−19 C Q = 140 900 ions e The number of ions inside the cell is, 0.155 ×10−12 × 6.022 ×1023= 9.3341×1010 Therefore the fraction of ions at the surface is, 140 934.4569 = 1.509 888 012 ×10−6 9.3341×1010 1.51×10−6

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8-46

Chapter 8: Electrochemical Cells

Solutions

*8.23. Calculate the emf at 25 °C of the cell Pt, H 2 (1 bar)|H 2 SO 4 (0.001 m)|CrSO 4 (s)|Cr given the following standard electrode potential: CrSO 4 (s) + 2e– → Cr + SO 2– 4

Eº = –0.40 V

a. First make the calculation neglecting activity coefficient corrections. b. Then make the calculation using activity coefficients estimated on the basis of the Debye-Hückel limiting law. Solution: Given: see above Required: see above a. At the left-hand electrode we have, 1 H2 → H+ + e– 2 and the right-hand electrode, we have, CrSO 4 (s) + 2e – → Cr(s) + SO 42– The overall reaction is given by, CrSO 4 (s) + H 2 → 2H + + Cr(s) + SO 42– With E º = −0.40 V and z = 2 The emf, neglecting the activity coefficients, is given by Eq. 8.13

8-47

Chapter 8: Electrochemical Cells

Solutions

u

RT [Y] y [Z]z ln E= E ° − zF [A]a [B]b u 2 RT ln H + SO 24− E= E ° − 2F 0.0257 2 ln ( 0.002 ) ( 0.001) E= −0.40 V − 2 E = −0.151 519 916 5 V

)

(

(

)

E = −0.152 V

b. To calculate the activity coefficients, we first calculate the ionic strength of the solution from Eq. 7.103, 1 I = ∑ ci zi2 2 i 1 2 1 × 0.002 + 22 × 0.001) I= ( 2 I = 0.003 M Now we rearrange the Debye-Hückel limiting law given by Eq. 7.111, log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 0.003 γ ± = 0.879 290 334 3 Substituting back into the expression for the emf obtained above we can solve for the true emf of the solution.

8-48

Chapter 8: Electrochemical Cells

Solutions

)

( (

u 2 RT ln H + γ ± 2 SO 24− γ ± 2F u 2 RT E= E ° − ln H + SO 24− γ ± 3 2F 0.0257 2 3 −0.40 V − E= ln ( 0.002 ) ( 0.001)( 0.879 290 334 3) 2 E = −0.146 560 839 3 V

E=

E° −

)

(

)

E = −0.147 V

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8-49

Chapter 8: Electrochemical Cells

Solutions

*8.24. Write the individual electrode reactions and the overall reaction for Cu|CuCl 2 (aq)|AgCl(s)|Ag If the emf of the cell is 0.191 V when the concentration of CuCl 2 is 1.0 × 10–4 M and is –0.074 V when the concentration is 0.20 M, make an estimate of the mean activity coefficient in the latter solution. Solution: Given: E1 = 0.191 V, c1 = 1.0 × 10 –4 M , E2 = 0.074 V, c2 = 0.20 M Required: γ ± At the left-hand electrode we have, Cu → Cu 2+ + 2e –

and the right-hand electrode, we have, AgCl(s) + e – → Ag + Cl – The overall reaction is given by, 2AgCl(s) + Cu(s) → 2Ag(s) + 2Cl – + Cu 2+ , where z = 2 To a good approximation, it can be assumed that the activity coefficients at 10-4 M are unity. (The DHLL gives γ ± = 0.988 ) Thus the standard electrical potential is obtained by rearranging Eq. 8.13 u

RT [Y] y [Z]z E= E ° − ln zF [A]a [B]b 2 u RT E °= E + ln Cu 2+ Cl− 2F 2 0.0257 = E ° 0.191 V + ln (10−4 )( 2 × 10−4 ) 2 E ° = −0.146 244 738 8 V

(

)

(

) 8-50

Chapter 8: Electrochemical Cells

Solutions

Suppose that at 0.20 M the activity coefficients are γ + and γ − , then we can solve for the mean activity coefficient from Eq. 8.40. E= E= E=

(

)

u RT ln aCu 2+ aCl− 2 2F 2 RT E° − ln γ + Cu 2+ γ − 2 Cl− 2F

E° −

(

E° −

(

2 RT ln Cu 2+ Cl− γ ±3 2F

−0.074 V = −0.146 244 738 8 V −

)

u

) (

)

0.0257 0.0257 2 ln ( 0.20 )( 0.40 ) − ln γ ±3 2 2

0.0257 ln γ ±3 2 3 ln γ ± = −2.180 139 285 0.028 014 789 9 = −

γ ± = 0.483 495 585 2 γ ± = 0.48

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8-51

Chapter 8: Electrochemical Cells

Solutions

*8.25. a. Write both electrode reactions and the overall reaction for the cell Tl|TlCl(s)|CdCl 2 (0.01 m)|Cd b. Calculate E and E° for this cell at 25 °C from the following information: Tl+ + e– → Tl

Eº = –0.34 V

Cd + 2e → Cd

Eº = –0.40 V

2+

–

The solubility product for TlCl is 1.6 × 10–3 mol2 dm–6 at 25 °C. Solution: Given: cell above, K sp = 1.6 ×10 –3 mol2 dm –6 , T = 25 °C Required: see above a. The left-hand or anode reaction is: Tl(s) + Cl− ( 0.02 m ) → TlCl(s) + e – The right-hand, or cathode reaction is: Cd 2+ ( 0.01 m ) + 2e – → Cd(s) The overall reaction is:

Cd 2+ ( 0.01 m ) + 2Tl(s) + 2Cl− ( 0.02 m ) → Cd(s) + 2TlCl(s) where z = 2 b. To use the electrical potentials given above, we rewrite the overall equation as (Cl- is a spectator ion): Cd 2+ ( 0.01 m ) + 2Tl(s) → Cd(s) + 2Tl+ ( in 0.01 m CdCl2 ) where z = 2 The standard electrical potential is given by, E ° = −0.40 V − ( −0.34 V ) E ° = −0.06 V 8-52

Chapter 8: Electrochemical Cells

Solutions

To solve for the emf, we use Eq. 8.13, + 2 RT Ti E= E ° − ln 2 F Cd 2+ where K sp = Ti + Cl−

u

K sp Ti + = Cl− E=

K sp 2 0.0257 E° − ln Cl− 2 Cd 2+ 2

u

–3 2 0.0257 (1.6 ×10 ) −0.06 V − E= ln ( 0.02 )2 ( 0.01) 2 E = −0.054 265 210 7 V

u

E = −0.054 V

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8-53

Chapter 8: Electrochemical Cells

8.26.

Solutions

Problem 7.44 involved calculating the concentrations on each side of a membrane after a Donnan equilibrium had become established. Which side of the membrane is positively charged? Calculate the Nernst potential across the membrane if the temperature is 37 °C.

Solution: Given: Problem 7.44, T = 37 °C Required: ∆Φ (used for concentration cells) We follow the same procedure as problem 8.18 to solve for the Nernst potential. From problem 7.44, we have the equilibrium concentrations of potassium and chloride as, right-hand side

left-hand side

+ + = = M K 0.04 K 0.16 M − − = = M Cl 0.04 Cl 0.01 M

The diffusible K+ ions are at a higher potential on the right-hand side of the membrane; there is thus a tendency for few of them to cross to the left-hand side and create a positive potential there. The same conclusion can be made by considering the diffusible Cl- ions; they are at a higher potential on the left-hand side, and few tend to cross to the right-hand side and create a negative potential. The Nernst potential is given by Eq. 8.19 as, ∆Φ =

∆Φ =

RT c1 ln zF c2

8.3145 J K −1 mol−1 × 310.15 K 1× 96 485 C mol−1

ln

0.16 M 0.04 M

∆Φ = 0.037 051 310 9 J C−1 where 1 J = 1 C V ∆Φ = 37 mV

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8-54

Chapter 8: Electrochemical Cells

8.27.

Solutions

The oxidation of lactate to pyruvate by the oxidized form of cytochrome c—represented as cytochrome c (Fe3+)—is an important biological reaction. The following are the relevant E°′ values, relating to pH 7 and 25 °C: E o '/V

pyruvate − + 2H + + 2e − → lactate − 3+

-0.185

−

2+

cytochrome c (Fe ) + e → cytochrome c (Fe )

0.254

Calculate the equilibrium ratio 2

cytochrome c (Fe 2+ ) pyruvate − H + 2 − 3+ cytochrome c (Fe ) lactate

2

at pH 7 and 25 °C. Also calculate the ratio at pH 6. Solution: Given: pH = 7, T = 25 °C Required: equilibrium ratio at pH 7 and pH 6 We begin by first writing the overall reaction as:

(

lactate – + 2cytochrome c Fe3+

)

(

)

→ 2cytochrome c Fe 2+ + pyruvate – + 2H +

Where z = 2 and E º ′ 0.254 V + 0.185 V = E º ′ =0.439 V

If K ′ is the equilibrium constant given at pH 7, then

K′ =

[cytochrome c (Fe 2+ )]2 [pyruvate – ][H + ]2 [cytochrome c (Fe3+ )]2 [lactate – ]

And the equilibrium ratio at pH 7 is just K ′ (unitless).

8-55

Chapter 8: Electrochemical Cells

Solutions

From Eq. 8.6 we can obtain K ′ ,

RT ln K o zF 0.0257 E o' = ln K ' 2 Eo =

2 K ' = exp E o' 0.0257 2 K ' = exp (0.439) 0.0257 K ' 6.870 472 098 × 1014 = K ' 6.87 × 1014 = At pH 6, the equilibrium ratio is K ′′ ′[H + ]2 = K true K=

(

[cytochrome c (Fe 2+ )]2 [pyruvate – ] [cytochrome c (Fe3+ )]2 [lactate – ]

)

(

2

′ 10−7 M = K true K= K ′′ 10−6 M = K ′′ 6.870 472 098 ×10

14

(10 × (10

−7 −6

)

2

) M) M

2

2

= K ′′ 6.870 472 098 ×1012 = K ′′ 6.87 ×1012

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8-56

Chapter 8: Electrochemical Cells

8.28.

Solutions

Suppose that the cell Ag|AgCl(s)|HCl(0.10 m)

HCl(0.01 m)|AgCl(s)|Ag

is set up and that the membrane separating the two solutions is permeable only to H+ ions. What is the emf of the cell at 25 °C? Solution: Given: see above, a membrane that is only permeable to H+ Required: E We can solve this problem in a similar manner as problem 8.21. The half reactions for each electrode are: LHS

Ag + Cl – → AgCl(s) + e –

RHS

Ag(s) + Cl – ( 0.01 m ) → AgCl(s) + e −

The electrical neutrality is maintained by the passage of H+ ions from right to left: H + ( 0.01 m ) → H + ( 0.10 m ) The net reaction is therefore,

H + ( 0.01 m ) + Cl− ( 0.01 m ) → H + ( 0.10 m ) + Cl− ( 0.10 m ) The cell emf is calculated from Eq. 8.13,

8-57

Chapter 8: Electrochemical Cells

Solutions

0.0257 ln K ° , where z = 1 and E °= 0 z + − H prod Clprod

E= E° − K° =

− H +react Clreact

E = −0.0257 ln

( 0.10 m ) ( 0.10 m ) ( 0.01 m ) ( 0.01 m )

E = −0.118 352 873 8 V E = −0.12 V

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8-58

Chapter 8: Electrochemical Cells

8.29.

Solutions

a. Consider the cell Pt, H 2 (1 bar)|HCl(m 1 )

HCl(m 2 )|Pt, H 2 (1 bar)

in which the solutions are separated by a partition that is permeable to both H+ and Cl–. The ratio of the speeds with which these ions pass through the membrane is the ratio of their transport numbers t + and t – . Derive an expression for the emf of this cell. b. If when m 1 = 0.01 m and m 2 = 0.01 m the emf is 0.0190 V, what are the transport numbers of the H+ and Cl– ions? Solution: Given: see above Required: see above a. The half reactions for each electrode are: 1 LHS : H 2 → H + ( m1 ) + e – 2 1 RHS : H + ( m2 ) + e – → H 2 2 To maintain electrical neutrality of the solutions, for every mole of H+ produced in the LHS solution, t + mol of H+ will cross the membrane from the left to the right, and t - mol of Cl- ions will pass fromright to left. In the LHS solution, there is therefore a net gain of,

t− mol of H + and of t− mol of Cl− . (1 − t+ ) mol = In the RHS solution, the net loss is t− mol of H + and of t− mol of Cl− (1 − t+ ) mol = The overall process is thus,

t− H + ( m2 ) + t− Cl− ( m2 ) → t− H + ( m1 ) + t− Cl− ( m1 ) The emf is given by Eq. 8.13,

8-59

Chapter 8: Electrochemical Cells

Solutions

0.0257 ln K ° , where z = 1 and E °= 0 z m1t− m1t− E = −0.0257 ln t− t− m2 m2

E= E° −

m2t− E = 0.0257 ln t− m1 = E 0.0257 × 2t− ln

2

m2 m1

b. To calculate the transport numbers we use the expression derived above= with m1 0.01 = m, m2 0.10 m m E 0.0257 × 2t− ln 2 = m1

0.0190 V 0.10 m ÷ ln 0.0257 × 2 0.01 m t− = 0.160 536 870 7

t− =

t− = 0.161 t+ = 1 − 0.160 536 870 7 t+ = 0.839 463129 3 t+ = 0.839

Back to Problem 8.29

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8-60

Chapter 8: Electrochemical Cells

8.30.

Solutions

The metal M forms a soluble nitrate and a very slightly soluble chloride. The cell M|M+(0.1 m), HNO 3 (0.2 m)|H 2 (1 bar), Pt has a measured E = –0.40 V at 298.15 K. When sufficient solid KCl is added to make the solution of the cell 0.20 m in K+, the emf changes to –0.15 V at 298.15 K as MCl precipitates. Calculate the K sp of MCl, taking all activity coefficients to be unity.

Solution: Given: E1 = −0.40 V, T = −0.15 V, γ ± = 298.15 K, cK + = 0.20 m, E2 = 1 Required: K sp The half reactions for each electrode are: LHS :

M → M + ( 0.1 m ) + e –

RHS :

H + ( 0.2 m ) + e – →

1 H 2 (1 bar ) 2

The overall reaction is given by, M + H + ( 0.2 m ) → M + ( 0.1 m ) +

1 H 2 (1 bar ) 2

with z = 1 and E o EHo + |H − EMo + |M = 2

. E= o

0 − EMo + |M

E o = − EMo + |M

Using Eq. 8.13,

8-61

Chapter 8: Electrochemical Cells

Solutions

u

RT [Y] y [Z]z E= E° − ln , where z = 1 zF [A]a [B]b + RT M γ M + E °= E + ln zF H + γ H+ −0.40 V + 0.0257 ln EM° + |M =

0.1 m 0.2 m

EM° + |M = −0.417 813 882 5 V Upon addition of KCl, almost all of the M+ precipitates, and 0.10 m Cl- is in excess. The value of M+ in solution is found from the K sp , K sp = M + Cl− K sp M + = Cl− Using the Nernst equation, we can solve for the solubility product, E Eo − =

K RT ln + sp − zF H Cl

K E − Eo = ln + sp − −0.0257 H Cl E − Eo K sp = H + Cl− exp −0.0257 −0.15 V-(-0.4 178 138 825 V) K sp = (0.20 m)(0.10 m) exp −0.0257 2 −7 = K sp 5.961 362 163 ×10 m 6.0 ×10−7 m 2 K= sp Back to Problem 8.30

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8-62

Chapter 8: Electrochemical Cells

8.31.

Solutions

The substance nicotinamide adenine dinucleotide (NAD+) plays an important role in biological systems; under the action of certain enzymes it can react with a reducing agent and release a proton to the solution to form its reduced form NADH. With pyruvate the reduced form NADH undergoes the reaction NADH + pyruvate– + H+ NAD+ + lactate– The appropriate E°′ values, relating to 25 °C and pH 7, are pyruvate– + 2H+ +2e– → lactate–

Eº′ = –0.19 V

NAD + H + 2e → NADH

Eº′ = –0.34 V

+

+

–

Use these values to calculate ∆G°′ for the reaction, and also the equilibrium ratio lactate − NAD − pyruvate − [ NADH ] H + (a) at pH 7, and (b) at pH 8. Solution: Given: see above Required: ∆G°′, equilibrium ratio a. The overall reaction that is obtained by reversing the second half reaction and summing the two together: pyruvate – + H + + NADH → lactate – + NAD + where z = 2 and,

E º ′ = −0.19 V − ( −0.34 V ) E º ′ =0.15 V To find ∆G°′ we use Eq. 8.2, ∆G º = − zFE º

8-63

Chapter 8: Electrochemical Cells

Solutions

∆G º′ =−2 × 96 485 C mol−1 × 0.15 V ∆G º′ = −28 945.5 C V mol−1 where 1 J = 1 C V ∆G º′ = −28 945.5 J mol−1 ∆G º′ = −29 kJ mol−1 To solve for the equilibrium ratio, we follow the thought process used in problem 8. 27. If K ′ is the equilibrium constant given at pH 7, then

K′ =

[lactate – ][NAD + ] [pyruvate – ][NADH][H + ]

And the equilibrium ratio at pH 7 is just K ′ (unitless) From Eq. 8.5, we can solve for K ′ . ∆G o ' = − RT ln K' ∆G o ' ) − RT −28945.5 J mol−1 K' = exp −1 −1 (−8.3145 J K mol )(298.15 K) K' = exp(

K' = 117 763.1471 dm3 mol−1 =' 1.2 ×105 K b. At pH 8, the equilibrium ratio is K ′′

8-64

Chapter 8: Electrochemical Cells

Solutions

lactate − NAD − ' H K true K= = pyruvate − [ NADH ] +

K true = K'(10−7 M )= K''(10−8 M )

(10 K''=117 763.1471× (10

−7 −8

M)

M)

K'' = 1177 631.471 '' 1.2 ×106 K=

Back to Problem 8.31

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8-65

Chapter 8: Electrochemical Cells

8.32.

Solutions

a. Calculate the standard emf E° for the reaction fumarate2– + lactate– → succinate2– + pyruvate– on the basis of the following information: fumarate2– + 2H+ + 2e– → succinate2–

E°′ = 0.031 V

pyruvate + 2H + 2e → lactate

E°′ = –0.185 V

–

+

–

–

The E°′ values relate to pH 7. The temperature coefficient ∂E/∂T for this cell is 2.18 × 10–5 V K–1. b. Calculate ∆G°, ∆H°, and ∆S° at 25 °C. Solution: Given: pH 7, ∂E /∂T= 2.18 × 10 –5 V K –1 Required: see above a. To find the standard emf, we first determine that the overall reaction is: fumarate 2– + lactate – → succinate 2– + pyruvate – where z = 2 Therefore, = E º ′ 0.031 V − ( −0.185 V ) E º=′ E=º 0.216 V (Note that this is also E° , the hydrogen ions having cancelled out.) b. To calculate ∆G°, we use E° from above and substitute into Eq. 8.2

8-66

Chapter 8: Electrochemical Cells

Solutions

∆G º = − zFE º ∆G º =−2 × 96 485 C mol−1 × 0.216 V ∆G º = −41 681.52 C V mol−1 where 1 J = 1 C V ∆G º = −41.7 kJ mol−1 The entropy change is obtained from Eq 8.23, ∂E ∆S = zF ∂T P ∆S = 2 × 96 485 C mol−1 × 2.18 × 10 –5 V K –1 ∆S =4.206 746 C V mol−1 K –1 where 1 J = 1 C V ∆S =4.21 J mol−1 K –1 To calculate enthalpy, we use the relationship between Gibbs energy and entropy we learned earlier as, ∆H =∆G + T ∆ S

(

∆H =− 41 681.52 J mol−1 + 298.15 K

) ( 4.206 746 J mol

−1

K –1

)

∆H =− 40 427.278 68 J mol−1 ∆H =− 40.4 kJ mol−1

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8-67

Chapter 8: Electrochemical Cells

8.33.

Solutions

The Weston standard cell (see Figure 8.2b) is Cd amalgam|CdSO 4 ·

8 H 2 O(s)|Hg 2 SO 4 (s), Hg 3

(saturated solution) a. Write the cell reaction. b. At 25 °C, the emf is 1.018 32 V and ∂E °/∂T = –5.00 × 10–5 V K–1. Calculate ∆G°, ∆H°, and ∆S°. Solution: Given: T= 25 °C, E= ° 1.018 32 V, ∂E /∂T= 5.00 × 10 –5 V K –1 Required: see above a. The half reactions for each electrode are: LHS : Cd(Hg) → Cd 2+ + 2e –

RHS :

Hg 22+ + 2e – → 2Hg

The overall reaction is given by, Cd(Hg) + Hg 22+ → Cd 2+ + 2Hg with z = 2 8 Since the solution is saturated with Hg 2SO 4 H 2 O , the overall reaction can be written as, 3

8 8 Cd(Hg) + Hg 2SO 4 (s) + H 2 O(l) → CdSO 4 H 2 O(s) + 2Hg(l) 3 3

b. We can calculate ∆G°, ∆H°, and ∆S° in a similar manner shown in problem 8.32. From Eq. 8.2,

8-68

Chapter 8: Electrochemical Cells

Solutions

∆G º = − zFE º ∆G º =−2 × 96 845 C mol−1 × 1.018 32 V ∆G º = −196 505.210 4 C V mol−1 where 1 J = 1 C V ∆G º = −196.5 kJ mol−1 The entropy change is obtained from Eq 8.23, ∂E ∆S = zF ∂T P ∆S = 2 × 96 845 C mol−1 × 5.00 × 10 –5 V K –1 ∆S = 9.6485 C V mol−1 K –1 where 1 J = 1 C V ∆S = 9.65 J mol−1 K –1 ∆H =∆G + T ∆ S

(

∆H =− 196 505.2104 J mol−1 + 298.15 K

) ( 9.6485 J mol

−1

K –1

)

∆H =− 199 381.9107 J mol−1 ∆H =− 199 kJ mol−1

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8-69

Chapter 8: Electrochemical Cells

8.34.

Solutions

Salstrom and Hildebrand [J. Amer. Chem. Soc., 52, 4650(1930)] reported the following data for the cell Ag(s)|AgBr(s)|HBr(aq)|Br 2 (g, 1 atm)|Pt

t/ºC

442.3

456.0

490.9

521.4

538.3

556.2

E/V

0.8031

0.7989

0.7887

0.7803

0.7751

0.7702

Find the temperature coefficient for this cell assuming a linear dependence of the cell potential with temperature. What is the entropy change for the cell reaction? Solution: Given: data above Required: ∆S

∂E To solve for ∆S , we need to find the temperature coefficient, from the data above. ∂T P Since we are only interested in the slope of the line best fit, we do not need to convert the temperature data to Kelvin. We can perform a linear regression analysis, using t as the independent variable and E as the dependent variable. The result is: = E 0.930 463 55 − 2.883 37 ×10−4t Differentiation with respect to t gives, ∂E −2.883 37 ×10−4 V °C−1 = ∂T P ∂E −2.883 37 ×10−4 V K −1 = ∂T P Substituting this value into Eq. 8.23 gives the entropy change,

8-70

Chapter 8: Electrochemical Cells

Solutions

∂E ∆S = zF ∂T P = ∆S 96 485 C mol−1 × −2.883 37 × 10−4 V K −1 ∆S =−27.820 195 45 C V K −1 mol−1 where 1 J = 1 C V ∆S =−27.82 J K −1 mol−1

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8-71

Chapter 8: Electrochemical Cells

8.35.

Solutions

The reaction taking place in the cell Mg(s)|Mg2+(aq), Cl–(aq)|Cl 2 (g,1 atm)|Pt is found to have an entropy change of –337.3 J K–1 mol–1 under standard conditions. What is the temperature coefficient for the cell?

Solution: Given: ∆S ° = −337.3J K –1 mol –1

∂E Required: ∂T P The overall reaction we are concerned with is, Mg(s) + Cl2 (g) → Mg 2+ (aq) + 2Cl− (aq), with z=2

Rearranging Eq. 8.23, we can solve for the temperature coefficient,

∂E ∆S = zF ∂T P ∆S ∂E = ∂T P zF −337.3J K –1 mol−1 ∂E = ∂T P 2 × 96 485 C mol−1 ∂E −1.747 940 1×10−3 J C−1 K −1 = ∂T P where 1 J = 1 C V ∂E −1.748 ×10−3 V K −1 = ∂ T P

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8-72

Chapter 8: Electrochemical Cells

Solutions

*8.36. a. Estimate the Gibbs energy of formation of the fumarate ion, using data in Problem 8.32 and the following values. ∆ f Gº (succinate, aq) = –690.44 kJ mol–1 ∆ f Gº (acetaldehyde, aq) = 139.08 kJ mol–1 ∆ f Gº (ethanol, aq) = –181.75 kJ mol–1 acetaldehyde + 2H+ + 2e– → ethanol Eº′ = –0.197 V b. If the ∂E °/∂T value for the process fumarate2– + ethanol → succinate2– + acetaldehyde is 2.18 × 10–5 V K–1, estimate the enthalpy of formation of the fumarate ion from the following values. ∆ f Hº (succinate, aq) = –908.68 kJ mol–1 ∆ f Hº (acetaldehyde, aq) = –210.66 kJ mol–1 ∆ f Hº (ethanol, aq) = –287.02 kJ mol–1 Solution: ∂E Given: problem 8.32, ∆ f Gº, = 2.18 ×10−5 V K −1 , ∆ f Hº ∂T P Required: see above a. The two reactions of interest are, fumarate 2– + 2H + + 2e – → succinate 2– +

acetaldehyde + 2H + 2e

–

→ ethanol

E °′ = 0.031 V

(1)

E º ′ = −0.197 V

(2)

Subtracting (2) from (1) gives the desired reaction: fumarate 2– + ethanol → succinate 2– + acetaldehyde with z = 2 And a standard electrical potential of, 8-73

Chapter 8: Electrochemical Cells

Solutions

= E °′ 0.031 V + 0.197 V E °′ = 0.228 V

The Gibbs energy for the reaction is calculated from Eq. 8.2 ∆G º = − zFE º ∆G º =−2 × 96 845 C mol−1 × 0.228 V ∆G º = −43 997.16 C V mol−1 where 1 J = 1 C V ∆G º = −43 997.16 J mol−1 To find the Gibbs energy of formation of fumarate, we use Eq. 3.91 ∆G o = ∑ ∆ f G o (products) − ∑ ∆ f G o (reactants) o o o o ∆G o = ∆ f Gsuccinate + ∆ f Gacetaldehyde − (∆ f Gethanol + ∆ f Gfumarate ) o o o o ∆ f Gfumarate = ∆ f Gsuccinate + ∆ f Gacetaldehyde − ∆ f Gethanol − ∆G o o ∆ f Gfumarate = −690.44 kJ mol−1 + 139.08 kJ mol−1 − (−181.74 kJ mol−1 ) − (−43.99716 kJ mol−1 ) o ∆ f Gfumarate = −325.61284 kJ mol−1 o ∆ f Gfumarate = −326 kJ mol−1

b. To find the enthalpy of formation of fumarate, we first need to find the enthalpy of the reaction. From Eq. 8.23, we calculate the entropy, and then use it to find the enthalpy.

8-74

Chapter 8: Electrochemical Cells

Solutions

∂E ∆S = zF ∂T P

∆S = 2 × 96 485 C mol−1 × ( 2.18 ×10−5 V K −1 ) ∆S =−4.222 442 C V mol−1 K −1 where 1 J=1 C V ∆S =−4.222 442 J mol−1 K −1 ∆H =∆G + T ∆S ∆H = −43 997.16 + (298.15)(−4.222 442) ∆H = −45 256.08 J mol−1 ∆H = −45.25 608 kJ mol−1 To find the enthalpy of formation of fumarate, we use Eq. 2.53

∆H ° = ∑ ∆ f H ° ( products ) − ∑ ∆ f H ° ( reactants )

(

° ° ° ° ∆H ° = ∆ f H succinate + ∆ f H acetaldehyde − ∆ f H ethanol + ∆ f H fumarate

)

o o o o ∆ f H fumarate = ∆ f H succinate + ∆ f H acetaldehyde − ∆ f H ethanol − ∆H o o ∆ f H fumarate =−908.68 kJ mol−1 − 210.66 kJ mol−1 − (−287.02 kJ mol−1 ) − (−45.25 608 kJ mol−1 ) o ∆ f H fumarate = −787.06 392 kJ mol−1 o ∆ f H fumarate = −787 kJ mol−1

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8-75

Chapter 8: Electrochemical Cells

Solutions

*8.37. a. Calculate the emf at 298.15 K for the cell Tl|TlBr|HBr (unit activity)|H 2 (1 bar), Pt b. Calculate ∆H for the cell reaction in the following cell. Tl|Tl+ (unit activity), H+ (unit activity)|H 2 (1 bar), Pt For the half-cell Tl+ + e– → Tl E° = 0.34 V ∂E/∂T = –0.003 V/K and

K sp (TlBr) = 10–4 mol2 dm–6 Solution: Given: T = 298.15 K, E ° = 0.34 V, ∂E /∂T = −0.003 V K −1 , K sp ( TlBr ) = 10 –4 mol2 dm –6 Required: see above, a. The half reactions for each electrode are: LHS : Tl → Tl+ ( in HBr; a = 1) + e – H + ( a = 1) + e – →

RHS :

1 H 2 (1 bar ) 2

The overall reaction is given by, Tl + H + ( a = 1) → M + ( in HBr; a =+ 1)

1 H 2 (1 bar ) with z = 1 and 2

= E o EHo + |H − ETlo + |Tl 2

E = 0 − ETlo + |Tl o

− ETlo + |Tl = −0.34 V Eo = The emf is given by Eq. 8.13, 8-76

Chapter 8: Electrochemical Cells

Tl+ E = E ° − 0.0257 ln + H

Solutions

u

K sp = Tl+ Br − K sp Tl+ = Br − K sp = E E o − 0.0257 ln Br − H +

u

10−4 mol−2 dm −6 −0.34 − 0.0257 ln E= 1× 1 E = −0.103 294 252 V

u

E = −0.10 V = E 0.0257 × 2t− ln

m2 m1

b. We can calculate the enthalpy using Eq. 8.25 and the data given for the half cell reaction. ∂E ∆H = − zF E − T ∂T

(

(

∆H =−1× 96 485 C mol−1 −0.34 V − ( 298.25 K ) −0.003 V K −1

))

∆H = 53 496.10825 C V mol−1 where 1J = 1 C V ∆H = 53 496.10825 J mol−1 ∆H = 535 kJ mol−1

Back to Problem 8.37

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Chapter 8: Electrochemical Cells

8.38.

Solutions

Calculate the solubility product and the solubility of AgBr at 25 °C on the basis of the following standard electrode potentials: AgBr(s) + e– → Ag + Br–

E° = 0.0713 V

Ag + e → Ag

E° = 0.7996 V

+

–

Solution: Given: T = 25 °C Required: K sp , s The desired reaction is obtained by subtracting the second reaction from the first. AgBr(s) → Ag + + Br – with z = 1 = E ° 0.0713 V − 0.7996 V E ° = −0.7283 V

We can calculate the solubility product from Eq. 8.13,

(

E o = 0.0257 ln Ag + Br −

)

u

K sp = Ag + Br − E o = 0.0257 ln K sp Eo K sp = exp 0.0257 −0.7283 K sp = exp 0.0257 = K sp 4.92 874 705 ×10−13 mol2 kg −2 = K sp 4.929 ×10−13 mol2 kg −2

The solubility is therefore,

8-78

Chapter 8: Electrochemical Cells

Solutions

= K sp = Ag + Br − s 2 s = K sp = s

4.928 747 05 ×10−13 mol2 kg −2

= s 7.020 503 58 ×10−7 mol kg −1 = s 7.021×10−7 mol kg −1

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8-79

Chapter 8: Electrochemical Cells

8.39.

Solutions

The emf of a cell Pt, H 2 (1 bar)|HCl(aq)|AgCl(s)|Ag was found to be 0.517 V at 25 °C. Calculate the pH of the HCl solution.

Solution: Given: E = 0.517 V, T = 25 °C Required: pH From Table 8.1, the standard emf of the AgCl|Ag electrode is 0.222 33 V and the cell reaction is: 1 AgCl(s) + H 2 → H + + Cl – + Ag(s) where z = 1, 2 To calculate the pH, we first need to find the concentration of H+ ions. Using Eq. 8.13,

(

)

u RT ln H + Cl− F u E − Eo ln H + Cl− = −0.0257 u E − Eo H + Cl− = exp −0.0257

= E Eo −

(

)

(

)

( H

+

Cl−

)

( H

+

Cl−

)

u

u

0.22 233 V − 0.517 V = exp −0.0257 = 95 392.83 548 mol2 dm −6

Since the concentrations of H+ and Cl- are the same,

8-80

Chapter 8: Electrochemical Cells

Solutions

H + = Cl− H + = 95 392.835 48 mol2 dm −6 H + = 308.857 306 mol dm −3

The pH is given by the logarithm of the hydrogen ion concentration pH = log H +

pH = log ( 308.857 306 mol dm −3 ) pH = 2.489 757 879 pH = 2.48

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8-81

Chapter 8: Electrochemical Cells

8.40.

Solutions

The emf of the cell Ag|AgI(s)I–(aq)

Ag+(aq)|Ag

is –0.9509 V at 25 °C. Calculate the solubility and the solubility product of AgI at that temperature. Solution: Given: E = 25 °C −0.9509 V, T = Required: s, K sp To solve this problem, we follow the example set in problem 8.38 The half reactions for each electrode are: Ag(s) → Ag + + e – AgI(s) + e – → Ag(s) + I − The reaction of interest is given by, AgI(s) → Ag + + I – with z = 1 We can calculate the solubility product from Eq. 8.13,

8-82

Chapter 8: Electrochemical Cells

(

E o = 0.0257 ln Ag + I −

)

Solutions

u

K sp = Ag + I − E o = 0.0257 ln K sp Eo K sp = exp 0.0257 −0.9509 K sp = exp 0.0257 = K sp 8.53 304 763 ×10−17 mol2 kg −2 = K sp 8.533 ×10−17 mol2 kg −2

The solubility is therefore, K sp = Ag + I − s 2 = s = K sp

= s

8.533 047 63 ×10−17 mol2 kg −2

= s 9.237 449 662 ×10−9 mol kg −1 = s 9.237 ×10−9 mol kg −1

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8-83

Chapter 8: Electrochemical Cells

8.41.

Solutions

An electrochemical cell M(s)|MCl(aq, 1.0 m)|AgCl(s)|Ag(s), where MCl is the chloride salt of the metal electrode M, yields a cell potential of 0.2053 V at 25 °C. What is the mean activity coefficient γ ± of the electrolyte MCl? E° for the M(s)|M+ electrode is 0.0254 V.

Solution: Given: E= 0.2053 V, T= 25 °C, E= ° 0.0254 V Required: γ ± The half reactions for each electrode are:

M → M+ + e– AgCl(s) + e – → Ag(s) + Cl− And the overall reaction is given by, AgCl(s) + M → Ag(s) + Cl− + M + = E ° 0.222 33 V − 0.0254 V E ° =0.196 93 V To solve for the activity coefficient, we use Eq. 8.43 E+

2 RT 2 RT ln mu = E ° − ln γ ± F F

Since the molality of MCl is exactly 1, this expression simplifies to, 2 RT 2 RT ln1 = E ° − ln γ ± F F 2 RT E = E° − ln γ ± F E+

Rearranging and solving for the mean activity coefficient, we obtain,

8-84

Chapter 8: Electrochemical Cells

Solutions

E = E ° − 2 × 0.0257 ln γ ±

γ± = e

E − E° − 2×0.0257 0.2053 − 0.196 93 − 2×0.0257

γ± = e γ ± = 0.849 726 737 1 γ ± = 0.850 Back to Problem 8.41

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8-85

Chapter 8: Electrochemical Cells

8.42.

Solutions

The following thermodynamic data apply to the complete oxidation of butane at 25 °C. C 4 H 10 (g) + (13/2)O 2 (g) → 4CO 2 (g) + 5H 2 O(l) ∆H ° = –2877 kJ mol –1 ∆S ° = –432.7 J K –1 mol –1 Suppose that a completely efficient fuel cell could be set up utilizing this reaction. Calculate (a) the maximum electrical work and (b) the maximum total work that could be obtained at 25 °C.

Solution: Given: ∆H ° = −2877 kJ mol –1 , ∆S ° = −432.7 J K –1 mol –1 , T = 25 °C Required: see above a. The maximum electrical work for the fuel cell is −∆G° . ∆G° =∆H ° − T ∆ S °

(

∆G° = − 2877 ×103 J mol –1 − 298.15 K

) ( −432.7 J K

–1

mol –1

)

∆G° = −2 747 990.495 J mol –1 ∆G° = −2748 kJ mol –1 electrical work = −∆G° electrical work = 2748 kJ mol –1 b. The maximum total work that can be obtained is −∆A

8-86

Chapter 8: Electrochemical Cells

Solutions

∆G o = ∆Ao − ∑ vRT

∆Ao = ∆G o + ∑ vRT

∑v = ∑v =

4 −1−

13 2

−3.5

∆Ao =−2747 990.495 J mol−1 + ( −3.5 ) ( 8.3145 J K −1 mol−1 ) ( 298.15 K ) ∆Ao = −2756 666.884 J mol−1 ∆Ao = −2758 kJ mol−1 total work = ∆Ao total work = 2758 kJ mol−1

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8-87

Chapter 8: Electrochemical Cells

Solutions

*8.43. At 298 K the emf of the cell Cd, Hg|CdCl 2 (aq, 0.01 m), AgCl(s)|Ag is 0.7585 V. The standard emf of the cell is 0.5732 V. a. Calculate the mean activity coefficient for the Cd2+ and Cl– ions. b. Compare the value with that estimated from the Debye-Hückel limiting law, and comment on any difference. Solution: Given: E 0.7585 = = V, E ° 0.5732 V Required: see above a. To calculate the mean activity coefficients, we follow a similar process as used in problem 8.24. The half reactions at each electrode are,

Cd(s) → Cd 2+ + 2e – AgCl(s) + e – → Ag + Cl – The overall reaction is given by, 2AgCl(s) + Cd(s) → 2Ag(s) + 2Cl− + Cd 2+ , where z = 2 Suppose that at 0. 01 m the activity coefficients are γ + and γ − , then we can solve for the mean activity coefficient from Eq. 8.40.

8-88

Chapter 8: Electrochemical Cells

(

Solutions

)

u RT ln aCd2+ aCl− 2 2F 2 u RT E Eo − = ln γ + Cd 2+ γ − 2 Cl− 2F u 2 RT E Eo − = ln Cd 2+ Cl− γ ±3 2F 0.0257 2 = 0.7585 V 0.5732 V − ln ( 0.01)( 0.02 ) γ ±3 2 −0.01 285ln ( 4 × 10−6 γ ±3 ) 0.1853 =

E Eo − =

)

( (

)

(

)

−14.42 023 346 ln ( 4 × 10−6 γ ±3 ) =

( 4 ×10 ) γ −6

3 ±

5.462 257 621×10−7 =

γ ± = 0.5 149 567 193 γ ± = 0.51 b. To calculate the activity coefficient according to the DHLL, we first calculate the ionic strength of the solution from Eq. 7.103, 1 I = ∑ ci zi2 2 i 1 2 1 × 0.02 + 22 × 0.01) I= ( 2 I = 0.03 m Now we rearrange the Debye-Hückel limiting law given by Eq. 7.111, log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 0.03 γ ± = 0.665 780 111 6 γ ± = 0.67

8-89

Chapter 8: Electrochemical Cells

Solutions

There is a considerable difference between the two methods.

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8-90

Chapter 8: Electrochemical Cells

Solutions

*8.44. The following emf values were obtained by H. S. Harned and Copson [J. Amer. Chem. Soc., 55, 2206(1933)] at 25 °C for the cell Pt,H 2 (1 bar)|LiOH(0.01 m), LiCl(m)|AgCl(s)|Ag at various molalities m of LiCl: m/mol kg–1 E/V

0.01

0.02

0.05

0.10

0.20

1.0498

1.0318

1.0076

1.9888

0.9696

Obtain from these data the ionic product of water. Solution: Given: emf values above Required: K w In this cell, we see that the LiOH is required for the hydrogen electrode, and the LiCl salt is used to complete the AgCl electrode. Both the Cl- ion and the H+ ion will behave according to their activities in solution. We begin solving the problem by first determining the emf of the cell: Ecell EAgCl − EH2 = ° Ecell = EAgCl −

RT RT ln aCl− − ln aH+ F F

Since K w = aH+ aOH− , we can substitute this into the above expression and develop a relationship involving K w . ° Ecell = EAgCl −

K RT RT ln aCl− − ln w F F aOH−

RT RT RT ln aCl− − ln K w + ln aOH− F F F RT aCl− RT ° EAgCl =− ln − ln K w F aOH− F

° Ecell = EAgCl −

Ecell

Rewriting this expression in terms of activity coefficients and molalities gives, 8-91

Chapter 8: Electrochemical Cells

Solutions

m −γ − RT RT ° Ecell = EAgCl − ln Cl Cl − ln K w F mOH− γ OH− F RT mCl− RT γ Cl− RT ° Ecell − EAgCl = − ln − ln − ln K w F mOH− F γ OH− F ° Ecell − EAgCl

0.0257

m − γ − = − ln Cl − ln Cl − ln K w mOH− γ OH−

° = 0.222 33 V . The molality of OH- is given as 0.01 m, and from Table 8.1, EAgCl

γ − m − Ecell − 0.222 33 + ln Cl = − ln Cl − ln K w γ OH− 0.0257 0.01 We plot the left-hand side of the equation against the ionic strength, which varies with concentration, and extrapolate to zero ionic strength. At zero ionic strength, the activity coefficients approach unity. Then the value of the curve is -lnK w . m − Ecell − 0.222 33 + ln Cl = − ln K w 0.0257 0.01 In the following data, I is based on m + 0.01 m OH − , where 0.01m is constant. m / mol kg −1

0.01

0.02

0.05

0.10

0.20

I / mol kg Ecell − 0.222 33 0.0257 m ln 0.01

0.02

0.03

0.06

0.11

0.21

32.3086

31.5079

30.566

29.834

29.087

0.000

0.693

1.609

2.303

2.996

Ecell − 0.222 33 m + ln 0.0257 0.01

32.209

32.301

32.175

32.137

32.083

−1

8-92

Chapter 8: Electrochemical Cells

Solutions

From the indicated plot shown, the value of -lnK w is 1.010×10-14.

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8-93

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7

Solutions of Electrolytes

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 7: Solutions of Electrolytes

Faraday’s Laws, Molar Conductivity, and Weak Electrolytes

Chapter 7 *problems with an asterisk are slightly more demanding Faraday’s Laws, Molar Conductivity, and Weak Electrolytes 7.1.

A constant current was passed through a solution of cupric sulfate, CuSO 4 , for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5). Solution

7.2.

After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9) was deposited from a solution of silver nitrate. Calculate the current. Solution

7.3.

Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C 6 H 5 OH) to monobromophenol using a current of 20 000 A? Solution

7.4.

The following are the molar conductivities Λ of chloroacetic acid in aqueous solution at 25 °C and at various concentrations c:

c −4

10 M Λ Ω cm 2 mol−1 −1

625

312.5

156.3

78.1

39.1

19.6

9.8

53.1

72.4

96.8

127.7

164.0

205.8

249.2

Plot Λ against c. If Λ° = 362 Ω–1 cm2 mol–1, are these values in accord with the Ostwald dilution law? What is the value of the dissociation constant? (See also Problem 7.11.) Solution 7.5.

The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 °C is 1.26 × 10–6 Ω–1 cm–1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 Ω–1 cm2 mol–1; Cl–, 76.4 Ω–1 cm2 mol–1. Solution 7-2

Chapter 7: Solutions of Electrolytes

*7.6.

Faraday’s Laws, Molar Conductivity, and Weak Electrolytes

The electrolytic conductivity of a 0.001 M solution of Na 2 SO 4 is 2.6 × 10–4 Ω–1 cm–1. If the solution is saturated with CaSO 4 , the conductivity becomes 7.0 × 10–4 Ω–1 cm–1. Calculate the solubility product for CaSO 4 using the following molar conductivities at 1 2+ −1 −1 2 these concentrations: λ(Na+) = 50.1 Ω–1 cm2 mol–1; λ Ca = 59.5 Ω cm mol . 2 Solution

7.7.

The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 °C. The equivalent conductance for 0.100 M KCl at 25 °C is 128.96 S cm2 mol–1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte? Solution

*7.8.

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: λ(K+) = 73.5 Ω–1 cm2 mol–1; λ(Cl–) = 76.4 Ω–1 cm2 mol–1; λ (NH +4 ) = 73.4 Ω–1 cm2 mol–1; λ(OH–) = 198.6 Ω–1 cm2 mol–1. Solution

7.9.

The conductivity of a 0.0312 M solution of a weak base is 1.53 × 10–4 S cm–1. If the sum of the limiting ionic conductances for BH+ and OH– is 237.0 S cm2 mol–1, what is the value of the base constant K b ? Solution

7.10.

The equivalent conductance of KBr solutions as a function of concentration at 25 °C is given in the following table. By a linear regression analysis of suitable variables, find the value of Λ° for KBr.

c/10–3 M

0.25

0.36

0.50

0.75

1.00

1.60

2.00

5.00

10.00

Λ/S cm2 mol–1

150.16

149.87

149.55

149.12

148.78

148.02

147.64

145.47

143.15 Solution

7-3

Chapter 7: Solutions of Electrolytes

7.11.

Debye-Hückel Theory and Transport of Electrolytes

Equation 7.20 is one form of Ostwald’s dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of Λ at various concentrations to be tested by means of a straight-line plot). Explain how Λ° and K can be obtained from the plot. Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities κ for the dissociation of tetramethyl tin chloride, (CH 3 ) 4 SnCl, in ethyl alcohol solution at 25.0 °C and at various concentrations c:

c/10–4 mol dm–3

1.566

2.600

6.219

10.441

κ/10–6 Ω–1 cm–1

1.788

2.418

4.009

5.336

By the use of the linear plot you have devised, determine Λ° and K. Solution 7.12.

A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 °C? (Assume that the cell constant is 1.0 cm–1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The Λ° for acetic acid is 390.7 S cm2 mol–1 and K a = 1.81 × 10–5 mol dm–3 at 25 °C. Ignore the conductance of water.) Solution

7.13.

How far can the conductivity of water at 25 °C be lowered in theory by removing impurities? The Λ° (in S cm2 mol–1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. K w = 1.008 × 10–14. Compare your answer to the experimental value of 5.8 × 10–8 S cm–1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894). Solution

Debye-Hückel Theory and Transport of Electrolytes 7.14.

The radius of the ionic atmosphere (1/κ) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 °C (∈ = 78). Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of ∈ = 38 at a concentration of 0.10 M. Solution

7-4

Chapter 7: Solutions of Electrolytes

7.15.

Debye-Hückel Theory and Transport of Electrolytes

1 The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate K 2SO 4 are 149.9, 2 –1 2 –1 126.5, and 153.3 Ω cm mol , respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution 7.16.

The molar conductivity at 18 °C of a 0.0100 M aqueous solution of ammonia is 9.6 Ω–1 cm2 mol–1. For NH 4 Cl, Λ° = 129.8 Ω–1 cm2 mol–1 and the molar ionic conductivities of OH– and Cl– are 174.0 and 65.6 Ω–1 cm2 mol–1, respectively. Calculate Λ° for NH 3 and the degree of ionization in 0.01 M solution. Solution

7.17.

A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g. a. Calculate the transport numbers of the Li+ and Cl– ions. b. If Λ° (LiCl) is 115.0 Ω–1 cm2 mol–1, what are the molar ionic conductivities and the ionic mobilities? Solution

7.18.

A solution of cadmium iodide, CdI2 , having a molality of 7.545 × 10–3 mol kg–1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I–. Solution

7.19.

The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t– = 0.179 and the molar conductivity is 426.16 Ω–1 cm2 mol–1. Calculate the mobilities of the hydrogen and chloride ions. Solution

7.20.

If a potential gradient of 100 V cm–1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl– ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291. Solution

7-5

Chapter 7: Solutions of Electrolytes

Debye-Hückel Theory and Transport of Electrolytes

*7.21. A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl– ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3. Solution 7.22.

What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 × 10–19 C. Solution

*7.23. According to Bjerrum’s theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is = dN i N i exp(− zi zc e 2 /4π ∈0∈ rkBT ) 4π r 2 dr

where z i and z c are the charge numbers of the ion of type i and of the central ion and e, ∈0 , ∈ , and k B have their usual significance. Plot the exponential in this expression and also 4πr2 against r for a uni-univalent electrolyte in water at 25.0 °C (∈ = 78.3). Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN 1 /N 1 )dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion. By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 °C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of k B T, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 °C. Solution

7-6

Chapter 7: Solutions of Electrolytes

Thermodynamics of Ions

Thermodynamics of Ions 7.24.

The following are some conventional standard enthalpies of ions in aqueous solution at 25 °C: Ion

Δ f H˚/kJ mol–1

H+

0 +

–239.7

2+

–543.1

2+

Zn

–152.3

–

–167.4

–

–120.9

Na Ca Cl

Br

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl 2 , and ZnBr 2 , assuming complete dissociation. Solution 7.25.

One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is –1051.4 kJ mol–1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows: Ion

Δ hyd G˚ k J mol–1

H+

0

Na+

679.1

Mg2+

274.1

Al3+

–1346.4

Cl–

–1407.1

Br–

–1393.3 Solution

7-7

Chapter 7: Solutions of Electrolytes

7.26.

Thermodynamics of Ions

Calculate the ionic strengths of 0.1 M solutions of KNO 3 , K 2 SO 4 , ZnSO 4 , ZnCl 2 , and K 4 Fe(CN) 6 ; assume complete dissociation and neglect hydrolysis. Solution

7.27.

Calculate the mean activity coefficient γ ± for the Ba2+ and SO 24− ions in a saturated solution of BaSO 4 (K sp = 9.2 × 10–11 mol2 dm–6) in 0.2 M K 2 SO 4 , assuming the Debye-Hückel limiting law to apply. Solution

7.28.

The solubility of AgCl in water at 25 °C is 1.274 × 10–5 mol dm–3. On the assumption t

hat the Debye-Hückel limiting law applies,

a. Calculate ΔG° for the process AgCl(s) → Ag+(aq) + Cl–(aq). b. Calculate the solubility of AgCl in an 0.005 M solution of K 2 SO 4 . Solution 7.29.

Employ Eq. 7.114 to make plots of log γ ± against I for a uni-univalent electrolyte in water at 25 °C, with B = 0.51 mol–1 dm3/2 and B′ = 0.33 × 1010 mol–1 dm3/2 m–1, and for the following values of the interionic distance a: a = 0, 0.1, 0.2, 0.4, and 0.8 nm Solution

7.30.

Estimate the change in Gibbs energy ΔG when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution (∈ = 78) to the lipid environment of a cell membrane (∈ = 4) at 25 °C. Solution

7.31.

At 18 °C the electrolytic conductivity of a saturated solution of CaF2 is 3.86 × 10–5 Ω–1 cm–1, and that of pure water is 1.5 × 10–6 Ω–1 cm–1. 1 The molar ionic conductivities of Ca 2+ and F– are 51.1 Ω–1 cm2 mol–1 and 47.0 Ω–1 cm2 mol–1, respectively. Calculate the solubility of 2 CaF2 in pure water at 18 °C and the solubility product. Solution

7-8

Chapter 7: Solutions of Electrolytes

7.32.

Thermodynamics of Ions

What concentrations of the following have the same ionic strength as 0.1 M NaCl? CuSO 4 , Ni(NO 3 ) 2 , Assume complete dissociation and neglect hydrolysis.

Al 2 (SO 4 ) 3 ,

Na 3 PO 4 Solution

7.33.

The solubility product of PbF 2 at 25.0 °C is 4.0 × 10–9 mol3 dm–9. Assuming the Debye-Hückel limiting law to apply, calculate the solubility of PbF 2 in (a) pure water and (b) 0.01 M NaF. Solution

7.34.

Calculate the solubility of silver acetate in water at 25 °C, assuming the DHLL to apply; the solubility product is 4.0 × 10–3 mol2 dm–6. Solution

*7.35. Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 °C: ∈ = 78; ∂ ln ∈ /∂T = –0.0046 K–1. Suggest a qualitative explanation for the sign of the value you obtain. Solution *7.36. Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl– in aqueous solution at 25 °C (∈ = 78), under the following conditions: a. The electrolyte is present at infinite dilution. b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70. The ionic radii are 95 pm for Na+ and 181 pm for Cl–. Solution 7.37.

If the solubility product of barium sulfate is 9.2 × 10–11 mol2 dm–6, calculate the solubility of BaSO 4 in a solution that is 0.10 M in NaNO 3 and 0.20 M in Zn(NO 3 ) 2 ; assume the DHLL to apply. Solution

7-9

Chapter 7: Solutions of Electrolytes

7.38.

Ionic Equilibria

Silver chloride, AgCl, is found to have a solubility of 1.561 × 10–5 M in a solution that is 0.01 M in K 2 SO 4 . Assume the DHLL to apply and calculate the solubility in pure water. Solution

7.39.

The enthalpy of neutralization of a strong acid by a strong base, corresponding to the process H+(aq) + OH–(aq) → H 2 O is –55.90 kJ mol–1. The enthalpy of neutralization of HCN by NaOH is –12.13 kJ mol–1. Make an estimate of the enthalpy of dissociation of HCN. Solution

7.40.

Make use of the Debye-Hückel limiting law to estimate the activity coefficients of the ions in an aqueous 0.004 M solution of sodium sulfate at 298 K. Estimate also the mean activity coefficient. Solution

Ionic Equilibria 7.41.

A 0.1 M solution of sodium palmitate, C 15 H 31 COONa, is separated from a 0.2 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl– ions but not to palmitate ions. Calculate the concentrations of Na+ and Cl– ions on the two sides of the membrane after equilibrium has become established. (For a calculation of the Nernst potential, see Problem 8.18.) Solution

7.42.

Consider the ionizations H + + H 3 N + CH 2 COO − H 3 N + CH 2 COOH H 2 NCH 2 COOH + H +

Assume that the following acid dissociation constants apply to the ionizations: 1.5 ×10−10 M − NH 3+ − NH 2 + H + ; K a = −COOH − COO − + H + ; K a =× 4.0 10−3 M Estimate a value for the equilibrium constant for the process H 3 N + CH 3COO − H 2 NCH 2 COOH

Solution 7-10

Chapter 7: Solutions of Electrolytes

7.43.

Essay Questions

The pK values for the successive ionizations of phosphoric acid are given on p. 308. Which of the four species is predominant at the following values of the hydrogen or hydroxide concentration? a. [H+] = 0.1 M. b. [H+] = 2 × 10–3 M. c. [H+] = 5 × 10–5 M. d. [OH–] = 2 × 10–3 M. e. [OH–] = 1 M. Solution

7.44.

Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl– ions but not to P– ions. The initial concentrations are as shown below.

[K+] = 0.05 M

[K+] = 0.15 M

[Cl–] = 0.05 M

[P–] = 0.15 M

Calculate the concentrations on each side of the membrane after equilibrium has become established. (See Problem 8.26 in Chapter 8 for the calculation of the Nernst potential for this system.) Solution Essay Questions 7.45.

State Faraday’s two laws of electrolysis and discuss their significance in connection with the electrical nature of matter.

7.46.

Discuss the main ideas that lie behind the Debye-Hückel theory, as applied to the conductivities of solutions of strong electrolytes.

7.47.

Outline two important methods for determining transport numbers of ions.

7.48.

Explain why Li+ has a lower ionic conductivity than Na+ and why the value for H+ is so much higher than the values for both of these ions.

7.49.

Describe briefly the type of hydration found with the following ions in aqueous solution: Li+, Br–, H+, OH–.

7.50.

What modifications to the Debye-Hückel limiting law are required to explain the influence of ionic strength on solubilities?

7-11

Chapter 7: Solutions of Electrolytes

Solutions

Solutions 7.1.

A constant current was passed through a solution of cupric sulfate, CuSO 4 , for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5 g mol-1).

Solution: Given:= t 1= h 3600 s, = m 0.040 g, M = 63.5 g mol−1 Required: I To solve this problem we must use Eq. 7.6 and apply Faraday’s Laws of Electrolysis. Eq. 7.6 is given by, Q = It

Rearranging to solve for I we obtain, I=

Q t

where Q, is the quantity of electricity. Q is proportional to the mass of the element produced at the electrode. Faraday’s constant, given by the symbol F, relates the amount of substance deposited to the quantity of electricity, Q, passed through the solution. The charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1. Hence, m Q = zF M

Using the expression above, we can solve for the current through the solution. I=

zF m t M

The charge on copper in CuSO 4 is Cu2+, therefore z = 2

7-12

Chapter 7: Solutions of Electrolytes

2 × 96 485 C mol−1 0.040 g 63.5 g mol−1 3600 s −1 I = 0.033 765 529 3 C s I=

Solutions

where 1 C s −1 = 1 A I = 33.8 mA Back to Problem 7.1

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7-13

Chapter 7: Solutions of Electrolytes

7.2.

Solutions

After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9 g mol-1) was deposited from a solution of silver nitrate. Calculate the current.

Solution: Given: = t 45 = min 2700 = s, m 7.19 = mg 0.00719= g, M 107.9 g mol−1 Required: I This problem can be solved in a similar manner as problem 7.1, using the expression for current as, I =

zF m . t M

The charge on silver in AgNO 3 is Ag+, therefore z = 1 1× 96 485 C mol−1 0.00719 g 107.9 g mol−1 2700 s −1 I = 0.002 381 241 7 C s

I=

where 1 C s −1 = 1 A I = 2.4 mA

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7-14

Chapter 7: Solutions of Electrolytes

7.3.

Solutions

Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C 6 H 5 OH) to monobromophenol using a current of 20 000 A?

Solution: Given: m 500.00 = = kg, I 20 000 A Required: t To solve this problem, we must first outline the chemical reactions that are taking place.

(1) ( 2) ( 3)

2Br − → Br2 (g) + 2e − 2K + + 2e − → 2K(s) C6 H 5OH + Br2 (g) → C6 H 4 (Br)OH + HBr

Two moles of electrons are involved in the generation of each mole of bromine gas, which reacts with one mole of phenol, therefore z = 2. Each batch consists of 500.00 kg of phenol therefore converting to the number of moles and we can determine the time required to convert all 500.00 kg of phenol into monobromophenol. m Using Eq. 7.6 and Q = zF from problem 6.1, we can solve for t, M

7-15

Chapter 7: Solutions of Electrolytes

Q I zF m t= I M

Solutions

t=

(

) (

)

6 12.011 g mol−1 + 6 1.007 94 g mol−1 + 15.9994 g mol−1 M C6 H5OH = M C6 H5OH = 94.11304 g mol−1 3 2 × 96 485 C mol−1 500.00 ×10 g t= −1 20 000 A 94.113 04 g mol where 1 C s −1 = 1 A 1h t 51 260.165 44 s × = 3600 s t = 14.238 934 84 h

t = 14.239 h

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7-16

Chapter 7: Solutions of Electrolytes

7.4.

Solutions

The following are the molar conductivities Λ of chloroacetic acid in aqueous solution at 25 °C and at various concentrations c:

c −4

10 M Λ Ω cm 2 mol−1 −1

625

312.5

156.3

78.1

39.1

19.6

9.8

53.1

72.4

96.8

127.7

164.0

205.8

249.2

Plot Λ against c. If Λ° = 362 Ω–1 cm2 mol–1, are these values in accord with the Ostwald Dilution Law? What is the value of the dissociation constant? (See also Problem 7.11.) Solution: Given: c, Λ, = Λ° 362 Ω –1 cm 2 mol –1 Required: plot of Λ against c, are these values in accord with the Ostwald Dilution Law, K Using the data above, we can create the following plot of Λ against c.

7-17

Chapter 7: Solutions of Electrolytes

Solutions

The Ostwald dilution law is given by Eq. 7.20 as: c ( Λ/ Λ ° ) 2 K= 1 − ( Λ/ Λ ° ) To determine if the data given above follows the Ostwald Dilution Law, we see if we can calculate a fixed value for K, the dissociation constant of the solution. The results are plotted in the table below. c

Λ

K

10-4 M

Ω–1 cm2 mol–1

M

625

53.1

0.001575951

312.5

72.4

0.0015625

156.3

96.8

0.001525554

78.1

127.7

0.001501592

39.1

164

0.001467205

19.6

205.8

0.001468105

9.8

249.2

0.001490406

Since the values of K are reasonably constant, we can say that data given above follows the Ostwald Dilution Law. The value of the dissociation constant, K, can be calculated from the average of the K values obtained above. K average = 0.001513045 M K average = 1.5 ×10−3 M

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7-18

Chapter 7: Solutions of Electrolytes

Solutions

The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 °C is 1.26 × 10–6 Ω–1 cm–1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 Ω–1 cm2 mol–1; Cl–, 76.4 Ω–1 cm2 mol–1.

7.5.

Solution: Given: ∆κ= 1.26 ×10 –6 Ω –1 cm –1 , λAg+= 61.9 Ω –1 cm 2 mol –1 , λCl-= 76.4 Ω –1 cm 2 mol –1 Required: solubility The expression for molar conductivity is given by Eq. 7.9.

κ

Λ=

c

In order to solve this problem we can use the concentration as a measure of solubility. solubility =

∆κ Λ AgCl

Λ AgCl = λAg+ + λClsolubility = solubility =

∆κ λAg+ + λCl1.26 ×10 –6 Ω –1 cm –1 61.9 Ω –1 cm 2 mol –1 + 76.4 Ω –1 cm 2 mol –1

= solubility 9.110 629 067 ×10 –9 mol cm −3 = solubility 9.110 629 067 ×10 –6 mol dm −3 = 9.11×10 –6 mol dm −3 solubility

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7-19

Chapter 7: Solutions of Electrolytes

7.6.

Solutions

The electrolytic conductivity of a 0.001 M solution of Na 2 SO 4 is 2.6 × 10–4 Ω–1 cm–1. If the solution is saturated with CaSO 4 , the conductivity becomes 7.0 × 10–4 Ω–1 cm–1. Calculate the solubility product for CaSO 4 using the following molar conductivities at these concentrations:

1 2+ −1 2 −1 λ (Na+) = 50.1 Ω–1 cm2 mol–1; λ Ca = 59.5 Ω cm mol . 2 Solution: Given: cNa 2SO4 =0.001 M , κ 1 2

Na 2SO 4

=2.6 ×10 –4 Ω –1 cm −1 , κ 1 2

1 2

=7.0 × 10 –4 Ω –1 cm −1 CaSO 4

λ ( Na + ) = 50.1 Ω –1 cm 2 mol –1 , λ Ca 2+ = 59.5 Ω −1 cm 2 mol−1 Required: K s for CaSO 4 To determine the base dissociation constant for CaSO 4 , we must first realize which chemical reactions are taking place.

(1) ( 2)

Na 2SO 4 2Na + + SO 4 2− CaSO 4 Ca 2+ + SO 4 2−

The solubility product is therefore given by: K s = Ca 2+ SO 4 2− , To determine the concentrations of the species in the solution, we will determine the electrolytic and molar conductivities. We are given that the initial electrolytic conductivity of the Na 2 SO 4 solution is 2.6 × 10–4 Ω–1 cm–1 which is raised to 7.0 × 10–4 Ω–1 cm–1 upon saturation with CaSO 4 .This means the increase in electrolytic conductivity is: ∆κ =7.0 ×10 –4 Ω –1 cm −1 − 2.6 ×10 –4 Ω –1 cm −1 ∆κ = 4.4 ×10 –4 Ω –1 cm −1 The molar conductivity of the ½CaSO 4 solution can be calculated using Eq. 7.9.

7-20

Chapter 7: Solutions of Electrolytes

Solutions

κ

Λ= Λ1 2

c

CaSO4

∆κ = 2c

where c is the concentration of CaSO 4 and 2c is the concentration of ½ CaSO 4 . The molar conductivity of the ½Na 2 SO 4 solution is:

κ1 Λ1 2

Λ1 2

Λ1 2

Na 2SO4

Na 2SO4

Na 2SO4

2 = 2c

2.6 ×10 –4 Ω –1 cm −1 = 2 0.001×10 –3 mol cm −3

(

)

= 130 Ω –1 cm 2 mol−1 Na 2SO 4

The molar conductivity of the ½Na 2 SO 4 is the same as, Λ1 2

Na 2SO4

=λNa + + λ1 2

SO42−

Therefore we can solve for λ1 2

SO42−

to calculate the molar conductivity of the ½CaSO 4 .

7-21

Chapter 7: Solutions of Electrolytes

λ1 2

λ1 2

= Λ1

SO42−

2

Na 2SO4

Solutions

− λNa +

=130 Ω –1 cm 2 mol –1 − 50.1 Ω –1 cm 2 mol –1

SO42−

λ1 = 79.9 Ω –1 cm 2 mol –1 2

Λ1 2

Λ1 2

Λ1 2

SO42−

CaSO4

=λCa 2+ + λ1 2

SO42−

= 59.5 Ω −1 cm 2 mol−1 + 79.9 Ω –1 cm 2 mol –1 CaSO4

= 139.4 Ω −1 cm 2 mol−1 CaSO4

The concentration of ½CaSO 4 is therefore given by:

c=

∆κ 2Λ 1 2

c=

(

CaSO4

4.4 ×10 –4 Ω –1 cm −1

2 139.4 Ω –1 cm 2 mol−1

)

c 1.578192 253 ×10 –6 mol cm −3 = = c 1.578192 253 ×10 –3 mol dm −3 Solving for K s , Ca 2+ = c SO 4 2− = 1.0 ×10−3 mol dm −3 + c The concentration of SO 4 2- is influenced by the contributions of ½CaSO 4 and ½Na 2 SO 4 .

7-22

Chapter 7: Solutions of Electrolytes

(

Ks = c 1.0 ×10−3 mol dm −3 + c

Solutions

)

= K s 4.068 883 038 ×10−6 mol2 dm −6 = K s 4.07 ×10−6 mol2 dm −6

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7-23

Chapter 7: Solutions of Electrolytes

7.7.

Solutions

The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 °C. The equivalent conductance for 0.100 M KCl at 25 °C is 128.96 S cm2 mol–1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte?

Solution: Given: G= 0.01178 S, T= 25 °C, Λ= 128.96 S cm 2 mol –1 at cKCl= 0.100 M = G 0.00824 = S, celectrolyte 0.0500 M Required: l/A, Λ electrolyte In order to determine the cell constant, we must first calculate the value for the electrolytic conductance. As we know, this can be obtained using Eq. 7.9.

κ

Λ=

c κ = cΛ

κ =

( 0.100 ×10

−3

)(

mol cm −3 128.96 S cm 2 mol –1

)

κ = 0.012896 S cm −1 We can now use Eq. 7.8 to solve for the cell constant, l/A.

7-24

Chapter 7: Solutions of Electrolytes

G (conductance) = κ

Solutions

A l

l κ = A G l 0.012896 S cm −1 = A 0.01178 S l =1.094 736 842 cm −1 A l =1.09 cm −1 A The equivalent conductance of the electrolyte in this same cell can be found using the cell constant calculated above and solve for Λ from Eq. 7.9.

κ

Λ=

c

κ =G

l A

l G A Λ electrolyte = celectrolyte

( 0.00824 S) (1.094 736 842 cm −1 )

Λ electrolyte = 0.0500 ×10−3 mol cm −3 Λ electrolyte = 180.412 631 6 S cm 2 mol−1 Λ electrolyte = 180 S cm 2 mol−1

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7-25

Chapter 7: Solutions of Electrolytes

7.8.

Solutions

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: λ(K+) = 73.5 Ω–1 cm2 mol–1; λ(Cl–) = 76.4 Ω–1 cm2 mol–1; λ (NH +4 ) = 73.4 Ω–1 cm2 mol–1; λ(OH–) = 198.6 Ω–1 cm2 mol–1.

Solution: Given: c= 0.01 M= , R 189 Ω, c= 0.01 M= , R 2460 Ω KCl NH3

λ (K+ ) = 73.5 Ω –1 cm 2 mol –1 , λ ( Cl – ) = 76.4 Ω –1 cm 2 mol –1 , λ ( NH 4+ ) = 73.4 Ω –1 cm 2 mol –1 , λ ( OH – ) = 198.6 Ω –1 cm 2 mol –1

Required: K b In order to determine the base dissociation constant for ammonia, we must first outline which chemical reactions are taking place.

(1) ( 2)

NH 4 OH NH 3 + H 2 O Kb + − NH 4 OH NH 4 + OH

The base dissociation constant is therefore given by: NH 4 + OH − Kb = [ NH 4OH ]

To solve for the concentrations of each species, we may begin by calculating the value for the electrolytic conductance of the standard KCl

κ

Λ=

c κ = cΛ Λ KCl = λK + + λClsolution in the cell. Λ KCl = 73.5 Ω –1 cm 2 mol –1 + 76.4 Ω –1 cm 2 mol –1 Λ KCl= 149.9 Ω –1 cm 2 mol –1

κ= KCl

( 0.01×10

−3

)(

mol cm −3 149.9 Ω –1 cm 2 mol –1

)

= κ KCl 0.001 499 Ω –1 cm −1

7-26

Chapter 7: Solutions of Electrolytes

Solutions

Recall that the resistance is inversely proportional to the conductance. The electrolytic conductivity of the ammonia solution is therefore,

κ NH = κ KCl × 3

RKCl RNH3

κ NH= 0.001 499 Ω –1 cm −1 × 3

189 Ω 2460 Ω

= κ NH3 1.151 670 732 ×10−4 Ω –1 cm −1 The molar conductivity of NH 4 + + OH − is: Λ NH4OH = λNH + + λOH4

Λ NH4OH =73.4 Ω –1 cm 2 mol –1 + 198.6 Ω –1 cm 2 mol –1 Λ NH4OH =272 Ω –1 cm 2 mol –1

Using Eq. 7.9, we can calculate the concentrations of NH 4 + and OH − . Λ= c= c=

κ c

κ NH

3

Λ NH4OH 1.151 670 732 ×10−4 Ω –1 cm –1 272 Ω –1 cm 2 mol –1

= c 4.234 083 572 ×10 –7 mol cm −3 = c 4.234 083 572 ×10 –4 mol dm −3 Knowing= that; c = NH 4 + OH − , we can solve for K b . NH 4 OH C initial C equilibrium

0.01 0.01 − c

NH 4 + 0 c

+

OH − 0 c

mol dm-3 mol dm-3 7-27

Chapter 7: Solutions of Electrolytes

Solutions

NH 4 + OH − c2 = Kb = 0.01 − c [ NH 4OH ] Kb =

( 4.234 083 572 ×10

–4

mol dm −3

)

2

0.01 − 4.234 083 572 ×10 –4 mol dm −3

= K b 1.872 008 786 ×10 –5 mol dm −3 K= 1.9 ×10 –5 mol dm −3 b Back to Problem 7.8

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7-28

Chapter 7: Solutions of Electrolytes

Solutions

The conductivity of a 0.0312 M solution of a weak base is 1.53 × 10–4 S cm–1. If the sum of the limiting ionic conductances for BH+ and OH– is 237.0 S cm2 mol–1, what is the value of the base constant K b ?

7.9.

Solution: Given:= c 0.0312 M, = κ 1.53 × 10 –4 S cm –1 ,= Λ° 237.0 S cm 2 mol –1 Required: K b In order to determine the base dissociation constant for the solution, we must write out the chemical reaction that is taking place. Kb + − B + H 2 O BH + OH

Since we are told we have a weak base, it is possible to apply Ostwald’s Dilution Law and introduce the degree of dissociation, α, given by Eq. 7.11.

α=

Λ Λ°

B

c (1 − α )

+

H 2O

BH + αc

+

OH − αc

The base dissociation constant is therefore given by Eq. 7.18. Kb =

cα 2 1− α

And the degree of dissociation is determined by calculating the molar conductivity of B + H 2 O using Eq. 7.9. Λ=

κ c

1.53 ×10 –4 S cm –1 Λ= 0.0312 ×10 –3 mol cm −3 Λ =4.903 846 154 S cm 2 mol –1

7-29

Chapter 7: Solutions of Electrolytes

Solutions

The degree of dissociation is therefore,

α=

4.903 846 154 S cm 2 mol –1

237.0 S cm 2 mol –1 α = 0.020 691 334

Solving for K b , we obtain the following:

( 0.0312 mol dm ) ( 0.020 691 334 ) = −3

Kb

2

1 − 0.020 691 334

= K b 1.363 992 486 ×10−5 mol dm −3 = K b 1.36 ×10−5 mol dm −3

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7-30

Chapter 7: Solutions of Electrolytes

7.10.

Solutions

The equivalent conductance of KBr solutions as a function of concentration at 25 °C is given in the following table. By a linear regression analysis of suitable variables, find the value of Λ° for KBr.

c/10–3 M

0.25

0.36

0.50

0.75

1.00

1.60

2.00

5.00

10.00

Λ/S cm2 mol–1

150.16

149.87

149.55

149.12

148.78

148.02

147.64

145.47

143.15

Solution: Given: T = 25 °C , data given above Required: Λ°KBr The relationship between Λ and c is given by the Debye-Hückel-Onsager Equation, Eq. 7.53. Λ = Λ ° − ( P + QΛ ° ) c

In order to solve for Λ°KBr , we can plot

c ×10−3

c c against Λ, where = c

(

M

)

(

Λ S cm 2 mol –1

0.01581139

150.16

0.01897367

149.87

0.02236068

149.55

0.02738613

149.12

0.03162278

148.78

0.04

148.02

0.04472136

147.64

0.07071068

145.47

)

7-31

Chapter 7: Solutions of Electrolytes

Solutions

0.1

143.15

From the linear regression, the y- intercept will be the value of Λ°KBr

Λ°KBr = 151.41268 S cm 2 mol –1 Λ°KBr = 151.41S cm 2 mol –1

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7-32

Chapter 7: Solutions of Electrolytes

7.11.

Solutions

Equation 7.20 is one form of Ostwald’s dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of Λ at various concentrations to be tested by means of a straight-line plot). Explain how Λ° and K can be obtained from the plot. Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities κ for the dissociation of tetramethyl tin chloride, (CH 3 ) 4 SnCl, in ethyl alcohol solution at 25.0 °C and at various concentrations c:

c/10–4 mol dm–3

1.566

2.600

6.219

10.441

κ/10–6 Ω–1 cm–1

1.788

2.418

4.009

5.336

By the use of the linear plot you have devised, determine Λ° and K.

Solution: Given: T = 25 °C , data above Required: Λ°, K Ostwald’s dilution law, given by Eq. 7.20 can be linearized in the following manner:

7-33

Chapter 7: Solutions of Electrolytes

Solutions

c ( Λ /Λ ° ) 2 K= 1 − ( Λ /Λ ° )

(

)

K 1 − ( Λ /Λ ° ) = c ( Λ /Λ ° ) c=

K − K ( Λ /Λ ° )

( Λ /Λ ) ( K − K ( Λ /Λ ) ) Λ ° 2

°

c=

2

°2

Λ2 K Λ°2 − K Λ°Λ c= Λ2 K Λ°2 cΛ = − K Λ° Λ

From here, we can plot cΛ against

1 and determine Λ° and K. Λ

κ

We can calculate Λ from Eq. 7.9 at each concentration given. Λ = . This leads to a table of values with the following: c c 10–4 mol dm–3

κ

Λ

10–6 Ω–1 cm–1 Ω–1 cm2 mol-1

cΛ

1/Λ

Ω-1 cm-1

Ω cm mol

-2

1.566

1.788

11.417625

1.788×10-6

0.087584

2.6

2.418

9.3

2.418×10-6

0.107527

6.219

4.009

6.446374

4.009×10-6

0.155126

10.441

5.336

5.1106216

5.336×10-6

0.195671

Now we obtain the following graph: 7-34

Chapter 7: Solutions of Electrolytes

Solutions

From the linear regression, the y- intercept will be KΛ° and the slope will be KΛ°2

7-35

Chapter 7: Solutions of Electrolytes

Solutions

= K Λ° 0.00111 Ω −1 cm −1 = K Λ°2 0.03294 Ω −2 cm mol−1 K Λ°2 K Λ° 0.03294 Ω −2 cm mol−1 Λ° = 0.00111 Ω −1 cm −1 Λ° =

= Λ° 29.675 675 68 Ω −1 cm 2 mol−1 Λ° = 30 Ω −1 cm 2 mol−1 K= K=

0.00111 Ω −1 cm −1 Λ° 0.00111 Ω −1 cm −1 29.675 675 68 Ω −1 cm 2 mol−1

= K 3.740 437 158 ×10−5 mol cm −3 = K 3.7 ×10−2 mol dm −3

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7-36

Chapter 7: Solutions of Electrolytes

7.12.

Solutions

A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 °C? (Assume that the cell constant is 1.0 cm–1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The Λ° for acetic acid is 390.7 S cm2 mol–1 and K a = 1.81 × 10–5 mol dm–3 at 25 °C. Ignore the conductance of water.)

Solution: A Given: cmax = 1500 ppm, = 1.0 cm −1 , ρacetic acid ≈ ρ water , Λ° = 390.7 S cm 2 mol –1 , K = 1.81×10 –5 mol dm –3 a l

T = 25 °C

Required: G In order to solve this problem, we must first convert the concentration from parts per million to SI units. 1500 g acid 106 g solution 1.500 g acid c= 1000 g solution

c 1500 ppm = =

(

) (

) (

M acetic acid = 2 12.011 g mol−1 + 4 1.007 94 g mol−1 + 2 15.9994 g mol−1

)

M acetic acid = 60.052 56 g mol−1 c=

1.500 g acid 60.052 56 g mol

−1

×

1 1 kg solution

c = 0.024 978119 2 mol kg −1 Since the solution has the same density as water, 1.00 kg of solution has a volume of 1.0 dm3. Therefore we can assume the solution has concentration, c = 0.0249781192 M. The conductance of a solution is given by Eq. 7.8.

7-37

Chapter 7: Solutions of Electrolytes

G (conductance) = κ

Solutions

A l

Since acetic acid is a weak acid, we may we use the Ostwald’s Dilution Law, Eq. 7.20, to solve for Λ . K=

c ( Λ /Λ ° ) 2 1 − ( Λ /Λ ° )

K − K ( Λ /Λ ° )= c ( Λ /Λ ° )

2

K − K ( Λ /Λ ° ) − c ( Λ /Λ ° ) =0 2

To solve for Λ /Λ ° , we use the quadratic equation. x=

−b ± b 2 − 4ac 2a

K ± K 2 − 4 ( −c ) K Λ /Λ = 2 ( −c ) °

K ± K 2 + 4cK Λ /Λ ° = −2c 1.81×10 –5 M ± ° Λ /Λ =

(1.81×10

–5

M ) + 4 ( 0.0249781192 M ) (1.81× 10 –5 M ) 2

−2 ( 0.0249781192 M )

= Λ /Λ ° 0.026 939 569 1 and − 0.026 903 369 1 We will disregard the negative value and take Λ /Λ ° =0.026 939 569 1 to solve for Λ .

( 0.026 939 569 1) ( Λ° ) Λ =( 0.026 939 569 1) ( 390.7 S cm 2 mol –1 )

= Λ

Λ =10.525 289 64 S cm 2 mol –1

7-38

Chapter 7: Solutions of Electrolytes

Solutions

Using Eq. 7.9, we can substitute for the value of κ and determine the conductance of the solution.

κ

Λ=

c κ = cΛ A G =κ l G=

( cΛ )

(

A l

)(

G = 0.024 978119 2 mol dm −3 10.525 289 64 S cm 2 mol –1

) (1.0 cm ) −1

G = 0.262 901 938 9 dm −3 S cm G 0.262 901 938 9 ×10−3 cm −3 S cm = G 2.629 019 389 ×10−4 S cm −2 = G 2.63 ×10−4 S cm −2 =

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7-39

Chapter 7: Solutions of Electrolytes

7.13.

Solutions

How far can the conductivity of water at 25 °C be lowered in theory by removing impurities? The Λ° (in S cm2 mol–1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. K w = 1.008 × 10–14. Compare your answer to the experimental value of 5.8 × 10–8 S cm–1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894).

Solution: 2 2 Given: Λ °KOH 274.4 S cm = = = mol –1 , Λ °HCl 426.04 S cm mol –1 , Λ °KCl 149.86 S cm 2 mol –1

T= 25 °C, K w = 1.008 ×10 –14 , κ exp = 5.8 ×10 –8 S cm –1 Required: κ The dissociations of each salt in water are given by:

(1) ( 2) ( 3)

KOH K + + OH − HCl H + + Cl− KCl K + + Cl−

By rearranging we find that, Λ °H2O = Λ °KOH + Λ °HCl − Λ °KCl = Λ °H2O 274.4 S cm 2 mol –1 + 426.04 S cm 2 mol –1 − 149.86 S cm 2 mol –1 550.58 S cm 2 mol –1 Λ °H2O =

In pure water, the only species that conduct electricity are H + and OH − ions. According to K w = H + OH − , each have a concentration of ;

K w =× 1.008 10−14 mol dm −3 = 1.003 999 203 2 ×10−7 mol dm −3 . Since this concentration is very low, we can assume that Λ H2O ≈ Λ °H2O .

7-40

Chapter 7: Solutions of Electrolytes

Solutions

κ

Λ= = κ

c

(1.003 999 203 2 ×10

−10

)(

mol cm −3 550.58 S cm 2 mol−1

)

= κ 5.527 779 329 ×10−8 S cm −1 = κ 5.528 ×10−8 S cm −1 Compared to the experimental value of 5.8 ×10 –8 S cm –1 , the conductivity determined through this process produces a very similar result. Back to Problem 7.13

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7-41

Chapter 7: Solutions of Electrolytes

7.14.

Solutions

The radius of the ionic atmosphere (1/κ) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 °C (∈ = 78). Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of ∈ = 38 at a concentration of 0.10 M.

Solution: 1 Given: = 0.964 nm,= c 0.10 M, T = 25 °C,= ∈ 78

κ

Required: see above a) Eq. 7.50 indicates that the thickness of the ionic atmosphere is inversely proportional to the square root of the concentration. 1/ 2

1 ∈0∈ kBT = κ e 2 ∑ ci zi2 L i 1 1 ∝ c κ Therefore the radius in water, where the electrolyte has c = 0.0001 M, can be obtained from the ratio of proportions.

7-42

Chapter 7: Solutions of Electrolytes

Solutions

1 1 κ 1 = κ 2 c2 c1 1 c1 1 κ 1 = c2 κ 2 c1 1 1 = c2 κ 1 κ 2 0.1 M 1 ( 0.964 nm ) = 0.0001 M κ 2 1 = 30.484 356 64 nm κ 2 1 = 30.5 nm κ 2 b) Similarly, we see from Eq. 7.50 that the thickness of the ionic atmosphere is proportional to the square root of the permittivity 1 ∝ ∈ κ The radius in water where∈ = 38 , can be obtained from the ratio of proportions.

7-43

Chapter 7: Solutions of Electrolytes

Solutions

∈2 1 1 = ∈1 κ 1 κ 2 38 1 ( 0.964 nm ) = 78 κ 2 1 = 0.672 855 072 6 nm κ 2 1 = 0.673 nm κ 2

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7-44

Chapter 7: Solutions of Electrolytes

7.15.

Solutions

1 The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate K 2SO 4 are 149.9, 2 –1 2 –1 126.5, and 153.3 Ω cm mol , respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution: Given: c = 0.001 M Λ KCl = 149.9 Ω –1 cm 2 mol –1 , Λ NaCl = 126.5 Ω –1 cm 2 mol –1 , Λ 1 2

Required: Λ 1 2

= 153.3 Ω –1 cm 2 mol –1 K 2SO 4

Na 2SO 4

1 The molar conductivity of Na 2SO 4 is given by the combination of the molar conductivities of each salt. We must also subtract the molar 2 conductivity of KCl since we are considering the solution containing only sodium and sulfate ions. Λ1 2

Λ1 2

Na 2SO 4

= Λ NaCl + Λ 1 2

K 2SO 4

− Λ KCl

126.5 –1 cm 2 mol –1 + 153.3 Ω –1 cm 2 mol –1 − 149.9 Ω –1 cm 2 mol –1 =Ω Na 2SO 4

Λ1 2

=129.9 Ω –1 cm 2 mol –1 Na 2SO 4

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7-45

Chapter 7: Solutions of Electrolytes

7.16.

Solutions

The molar conductivity at 18 °C of a 0.0100 M aqueous solution of ammonia is 9.6 Ω–1 cm2 mol–1. For NH 4 Cl, Λ° = 129.8 Ω–1 cm2 mol–1 and the molar ionic conductivities of OH– and Cl– are 174.0 and 65.6 Ω–1 cm2 mol–1, respectively. Calculate Λ° for NH 3 and the degree of ionization in 0.01 M solution.

Solution: Given: T =18 °C, cNH3 =0.0100 M, Λ NH3 =9.6 Ω –1 cm 2 mol –1 , Λ °NH4Cl = 129.8 Ω –1cm 2 mol –1 , ° λOH = 174.0 Ω –1cm 2 mol –1 , λCl° = 65.6 Ω –1cm 2 mol –1 , c = 0.01 M −

−

Required: Λ °NH3 , α In solution, ammonia reacts in following manner,

NH 3 + H 2 O NH 4 OH NH 4 OH NH 4 + + OH − As a result, we can obtain Λ °NH3 from the molar conductivity of NH 4 OH. ° ° Λ °NH4OH = Λ °NH4Cl + λOH − −λ − Cl

Λ °NH4OH = 129.8 Ω –1cm 2 mol –1 + 174.0 Ω –1cm 2 mol –1 − 65.6 Ω –1cm 2 mol –1 Λ °NH4OH = 238.2 Ω –1cm 2 mol –1 Λ °NH4OH =238 Ω –1cm 2 mol –1 The degree of dissociation is defined by Eq. 7.11 which states,

7-46

Chapter 7: Solutions of Electrolytes

α= α=

Solutions

Λ Λ° 9.6 Ω –1 cm 2 mol –1

238.2 Ω –1 cm 2 mol –1 α = 0.040 302 267 = α 4.0 ×10−2

Back to Problem 7.16

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7-47

Chapter 7: Solutions of Electrolytes

7.17.

Solutions

A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g. a. Calculate the transport numbers of the Li+ and Cl– ions. b. If Λ° (LiCl) is 115.0 Ω–1 cm2 mol–1, what are the molar ionic conductivities and the ionic mobilities?

Solution: Given: I= 0.79 A, t= 2 h, ∆m= 0.793 g, Λ °LiCl= 115.0 Ω –1 cm 2 mol –1 Required: see above a) To solve this problem, we can use the Hittorf method. This method gives the transport numbers according to Eq. 7.75 and Eq.7.76. amount lost from cathode compartment amount lost from anode compartment = t− = t+ And amount deposited amount deposited We will use the number of moles to measure the amounts of the Li+ and Cl– ions. To determine the total amount deposited, we use Eq. 7.6. Q = It s Q = ( 0.79 A ) 2 h × 3600 h Q = 5688 A s Q = 5688 C In problem 7.1 we found that Q = zFn since the charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1. Solving for n, we can determine the total amount deposited. n=

Q zF

amount deposited =

5688 C

(1) ( 96 485 C mol−1 )

amount deposited = 0.058 952 168 7 mol The amount lost of LiCl lost in the anode compartment is given by n Cl . 7-48

Chapter 7: Solutions of Electrolytes

nLiCl =

Solutions

mLiCl M LiCl

= M LiCl 6.941 g mol−1 + 35.4527 g mol−1 M LiCl = 42.3927 g mol−1 nLiCl =

0.793 g 42.3927 g mol−1

nLiCl = 0.018 705 609 6 mol 1 The anode reaction that is occurring is Cl− → Cl2 + e − , therefore 0.058 952 168 7 mol of Cl- are removed by electrolysis. The amount lost 2 from the anode compartment is given by, n= ntotal − nLiCl net

= nnet 0.058 952 168 7 mol − 0.018 705 609 6 mol nnet = 0.040 246 559 2 mol Solving for t Cl - we obtain, 0.040 246 559 2 mol 0.058 952 168 7 mol = 0.682 698 534

= t+ tCl= − tCl−

tCl− = 0.68

The second transport number is given by1 − t+ , t− = tLi+ = 1 − t+ tLi+ = 1 − 0.682 698 534 tLi+ = 0.317 301 466 tLi+ = 0.32

7-49

Chapter 7: Solutions of Electrolytes

Solutions

b) In order to determine the molar ionic conductivities we will use Eq. 7.79 which states,

λ+°

= t+

λ°

− = and t− Λ° Λ°

To solve, we rearrange and use Λ °LiCl= 115.0 Ω –1 cm 2 mol –1 .

λCl° = Λ °tCl −

= λCl° −

−

(115.0 Ω

–1

cm 2 mol –1 ) ( 0.682 698 534 )

= λCl° − 78.510 331 41 Ω –1 cm 2 mol –1

λCl°= 79 Ω –1 cm 2 mol –1 −

λLi° = Λ °tLi +

= λLi° +

+

(115.0 Ω

–1

cm 2 mol –1 ) ( 0.317 301 466 )

= λLi° + 36.489 668 59 Ω –1 cm 2 mol –1

λLi°= 36 Ω –1 cm 2 mol –1 +

The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64.

7-50

Chapter 7: Solutions of Electrolytes

κ+

° λ= +

= Fu+ c+

u+ = uCl−

Solutions

λ+°

F 78.510 331 41 Ω –1 cm 2 mol –1 = 96 485 C

uCl− 8.137 050 465 ×10−4 Ω –1 cm 2 mol –1 C−1 = where 1 Ω –1 1 A= V −1 and 1 A 1 C s −1 = therefore, 1 Ω –1 = 1 C s −1 V −1

(

)

uCl− 8.137 050 465 ×10−4 C s −1 V −1 cm 2 mol –1 C –1 = uCl− 8.137 050 465 ×10−4 cm 2 mol –1 V −1 s −1 = uCl= 8.1×10−4 cm 2 mol –1 V −1 s −1 − uLi+ =

36.489 668 59 Ω –1 cm 2 mol –1 96 485 C

uLi+ 3.781 900 667 ×10−4 cm 2 mol –1 V −1 s −1 = 3.8 ×10−4 cm 2 mol –1 V −1 s −1 uLi= +

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7-51

Chapter 7: Solutions of Electrolytes

7.18.

Solutions

A solution of cadmium iodide, CdI2 , having a molality of 7.545 × 10–3 mol kg–1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I–.

Solution: Given: molality = 7.545 ×10 –3 mol kg –1 , mCd2+ = 0.03462 g, manode = 152.64 g, mCdI2 in anode = 0.3718 g

Required: tCd2+ , tI− When working with a Hittorf cell, we must use Eq. 7.75 and Eq. 7.76 to solve for tCd2+ and tI− . amount lost from anode compartment amount lost from cathode compartment = t− = t+ And amount deposited amount deposited

The number of coulombs of charge will be used as a measure of each amount. The anode compartment initially contained the following number of moles: = ni molality × manode kg mol kg –1 152.64 g ×10 –3 g ni = 0.001151 669 mol

= ni

( 7.545 ×10

–3

)

The anode compartment finally contained,

7-52

Chapter 7: Solutions of Electrolytes

nf =

Solutions

mCdI2 in anode M CdI2

(

M CdI2 112.411 g mol−1 + 2 126.904 47 g mol−1 =

)

M CdI2 = 366.219 94 g mol−1 nf =

0.3718 g 366.219 94 g mol−1

n f = 0.001 015 236 9 mol The number of moles lost from the anode compartment is therefore,

n= ni − n f = n 0.001151 669 mol − 0.001 015 236 9 mol = n 1.364 318 907 ×10−4 mol The total amount of Cd2+ deposited is calculated by, nCd2+ =

mCd2+ M Cd2+

M Cd2+ = 112.411 g mol−1 nCd2+ =

0.03462 g 112.411 g mol−1

= nCd2+ 3.079 769 773 ×10−4 mol

Now it is possible to determine the transport number at the anode.

1.364 318 907 ×10−4 mol 3.079 769 773 ×10−4 mol t− = 0.442 993 797 4 t− =

t− = 0.4430 7-53

Chapter 7: Solutions of Electrolytes

Solutions

The second transport number is given by1 − t − .

t + = 1 − 0.442 993 797 4 t + = 0.557 006 202 6 t + = 0.5570

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7-54

Chapter 7: Solutions of Electrolytes

Solutions

The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t– = 0.179 and the molar conductivity is 426.16 Ω–1 cm2 mol–1. Calculate the mobilities of the hydrogen and chloride ions.

7.19.

Solution: ° Given: = t+ 0.821,= t− 0.179, Λ= 426.16 Ω –1 cm 2 mol –1 HCl

Required: u+ , u− The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64. ° λ= +

κ+

= Fu+ c+

The molar ionic conductivities are given by Eq. 7.79. = t+

λ+°

λ°

− = and t− ° Λ Λ°

By rearranging and substituting equations 7.64 and 7.79, we can obtain an expression for the ionic mobility. u=

λ°

F Λ °t u= F Now it is possible to solve for u+ and u− .

7-55

Chapter 7: Solutions of Electrolytes

u+ =

( 426.16 Ω

–1

Solutions

cm 2 mol –1 ) ( 0.821)

96 485 C

u+ 0.003 626 236 Ω –1 cm 2 mol –1 C−1 = where 1 Ω –1 1 A= V −1 and 1 A 1 C s −1 = therefore, 1 Ω –1 = 1 C s −1 V −1

(

)

u+ = 0.003 626 236 C s −1 V −1 cm 2 mol –1 C−1 = u+ 3.63 ×10−3 V −1 cm 2 mol –1 s −1 u−

( 426.16 Ω =

cm 2 mol –1 ) ( 0.179 )

–1

96 485 C

= u− 7.906 165 725 ×10−4 V −1 cm 2 mol –1 s −1 = u− 7.91×10−4 V −1 cm 2 mol –1 s −1

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7-56

Chapter 7: Solutions of Electrolytes

Solutions

If a potential gradient of 100 V cm–1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl– ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291.

7.20.

Solution: Given: V 100 = = V cm –1 , cNaCl 0.01 M, Table 7.3 Required: vNa + , vCl− The ionic mobility is given in terms of the molar ionic conductivity by Eq. 7.64.

κ+

° λ= +

= Fu+ c+

u+ =

λ+° F

From Table 7.3 we are given that, ° λNa

+

uNa + =

= 50.08 S cm 2 mol−1 and λCl° − 76.31S cm 2 mol−1 50.08 S cm 2 mol−1 96 485 C mol−1

uNa + 5.190 444 11×10−4 S cm 2 C −1 = where 1 S =1 Ω –1 and 1 Ω –1 =1 A V −1 and 1 A =1 C s −1 therefore, 1S = 1 C s −1 V −1

(

)

uNa + 5.190 444 11×10−4 C s −1 V −1 cm 2 C−1 = −4

−1

= uNa + 5.190 444 11×10 V cm s uCl− =

2

−1

76.31S cm 2 mol−1 96 485 C mol−1

= uCl− 7.909 001 399 ×10−4 V −1 cm 2 s −1

7-57

Chapter 7: Solutions of Electrolytes

Solutions

From section 7.5 we know that, speed = uV. The velocities in a gradient of 100 V cm–1 are thus, vNa + =

(5.190 444 11×10

−4

)(

V −1 cm 2 s −1 100 V cm –1

)

vNa + 5.190 444 11×10−2 cm s −1 = 5.19 ×10−2 cm s −1 v= Na + vCl− =

( 7.909 001 399 ×10

−4

)(

V −1 cm 2 s −1 100 V cm –1

)

= vCl− 7.909 001 399 ×10−2 cm s −1 v= 7.91×10−2 cm s −1 Cl−

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7-58

Chapter 7: Solutions of Electrolytes

7.21.

Solutions

A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl– ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3.

Solution: Given: = cLiCl 0.01 = M, A 5= cm 2 , I 1 A , Table 7.3 Required: vLi+ , vCl− In order to determine the speeds of the ions, we must find the potential gradient. Once we have this information, we can proceed in a similar manner as was done in problem 7.20. The potential gradient can be calculated using Ohm’s Law (Eq.7.7) in conjunction with Eq. 7.9. R=

V κ and Λ = I c

The specific conductivity of a 0.01 M solution is calculated according to:

κ = cΛ We determine the molar conductivity of LiCl using the data for the ionic conductivities of Li+ and Cl- found in Table 7.3. Λ °LiCl = λLi° + + λCl° − = Λ °LiCl 38.66 S cm 2 mol−1 + 76.31S cm 2 mol−1 Λ °LiCl = 114.97 S cm 2 mol−1

Hence, = κ

( 0.01×10

−3

)(

mol cm −3 114.97 S cm 2 mol−1

)

= κ 1.1497 ×10−3 S cm −1

Let us consider the fact that the resistance is inversely proportional to κ, and we must factor in the 5 cm2 of surface area.

7-59

Chapter 7: Solutions of Electrolytes

R=

Solutions

1 1.1497 ×10−3 S cm −1 × 5 cm 2

= R 173.958 423 9 Ω cm −1 The potential gradient required to produce a current of 1 A is therefore, = V

(1 A ) (173.958 423 9 Ω cm −1 )

where 1 Ω =1 V A –1 V = 173.958 423 9 V cm −1 The ionic mobilities can be calculated using Eq. 7.64. ° λ= +

u+ =

κ+

= Fu+ c+

λ+° F

From Table 7.3 we know that,

7-60

Chapter 7: Solutions of Electrolytes

Solutions

uLi+ =

38.66 S cm 2 mol−1 96 485 C mol−1

uLi+ 4.006 840 442 ×10−4 S cm 2 C−1 = where 1 S =1 Ω –1 and 1 Ω –1 =1 A V −1 and 1 A =1 C s −1

λ

° Li +

therefore, 1S = 1 C s −1 V −1

= 38.66 S cm mol and λ 76.31S cm mol uLi+ 4.006 840 442 ×10−4 C s −1 V −1 cm 2 C−1 = 2

−1

° Cl−

2

−1

(

)

uLi+ 4.006 840 442 ×10−4 V −1 cm 2 s −1 = uCl− =

76.31S cm 2 mol−1 96 485 C mol−1

= uCl− 7.909 001 399 ×10−4 V −1 cm 2 s −1

From section 7.5 we are given that, speed = uV. = vLi+

( 4.006 840 442 ×10

−4

)(

V −1 cm 2 s −1 V −1 cm 2 s −1 173.958 423 9 V cm –1

)

vLi+ = 0.069 702 364 8 cm s −1 vLi= 7.0 ×10−2 cm s −1 + = vCl−

( 7.909 001 399 ×10

−4

)(

V −1 cm 2 s −1 173.958 423 9 V cm –1

)

vCl− = 0.137 583 7418 cm s −1 vCl− = 0.14 cm s −1

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7-61

Chapter 7: Solutions of Electrolytes

7.22.

Solutions

What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 × 10–19 C.

Solution: Given: vacuum: Q1 = −Q2 , e = 1.6 × 10–19 C Required: see above Recall that work is defined as the application of a force through a distance. This definition is given by Eq. 1.1, dw = F · dl. In this case, the force we are concerned with is an electrostatic force, and the distance in a vacuum we use as r. From Eq. 7.1, the electrostatic force is given by: F=

Q1Q2 4π ∈0 r 2

To determine the amount of work done, we will take the integral of F with respect to r. r2

w = ∫ Fdr r1

w=∫

r2

r1

Q1Q2 dr 4π ∈0 r 2

Since the particles have opposite charges, we will introduce a negative sign.

w =

∫

r2

r1

−

Q1Q2 dr 4π ∈0 r 2

QQ 1 1 w= − 1 2 − 4π ∈0 r2 r1 The permittivity of a vacuum has the value, = ∈0 8.854 ×10−12 C2 J −1 m −1 . This will be used when solving parts a, b and c.

7-62

Chapter 7: Solutions of Electrolytes

Solutions

a) r1 = 1.0 ×10−9 m, r2 = ∞m w= −

(

(1.6 ×10

–19

)

C

2

4π 8.854 ×10−12 C2 J −1 m −1

)

1 1 − −9 ∞ 1.0 ×10 m

w 2.300 862 1×10 –19 J = = w 2.3 ×10 –19 J b)

∞m r1 = 1.0 ×10−3 m, r2 = w= −

(

(1.6 ×10

–19

C

)

2

4π 8.854 ×10−12 C2 J −1 m −1

)

1 1 − −3 ∞ 1.0 ×10 m

)

1 1 − −9 0.10 m 1.0 ×10 m

= w 2.300 862 1×10 –25 J = w 2.3 ×10 –25 J c)

r1 = 1.0 ×10−9 m, r2 = 0.10 m w= −

(

(1.6 ×10

4π 8.854 ×10

–19

−12

C 2

)

2

C J

−1

m

−1

= w 2.300 862 1 ×10 –19 J = w 2.3 ×10 –19 J

Back to Problem 7.22

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Chapter 7: Solutions of Electrolytes

7.23.

Solutions

According to Bjerrum’s theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is dN i N i exp(− zi zc e 2 /4π ∈0∈ rkBT ) 4π r 2 dr = where z i and z c are the charge numbers of the ion of type i and of the central ion and e, ∈0 , ∈ , and k B have their usual significance. Plot the exponential in this expression and also 4πr2 against r for a uni-univalent electrolyte in water at 25.0 °C (∈ = 78.3). Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN 1 /N 1 )dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion.

By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 °C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of k B T, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 °C. Solution:

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7-64

Chapter 7: Solutions of Electrolytes

7.24.

Solutions

The following are some conventional standard enthalpies of ions in aqueous solution at 25 °C: Ion

Δ f H˚/kJ mol–1

H+

0 +

–239.7

2+

–543.1

2+

Zn

–152.3

–

–167.4

–

–120.9

Na Ca Cl

Br

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl 2 , and ZnBr 2 , assuming complete dissociation. Solution: Given: standard enthalpies Required: enthalpies of formation In order to calculate the enthalpies of formation, we will simply sum up the standard enthalpies of the ions present in the solution.

7-65

Chapter 7: Solutions of Electrolytes

Solutions

∆ f H NaCl = ∆ f H Na° + ∆ f H Cl° ∆ f H NaCl = −239.7 kJ mol−1 − 167.4 kJ mol−1 ∆ f H NaCl = −407.1 kJ mol−1 ∆ f H CaCl2 =∆ f H Ca° + 2∆ f H Cl°

(

)

(

)

∆ f H CaCl2 = −543.1 kJ mol−1 − 2 167.4 kJ mol−1 ∆ f H CaCl2 = −877.9 kJ mol−1 ∆ f H ZnBr2 =∆ f H Zn° + 2∆ f H Br°

∆ f H ZnBr2 = −152.3 kJ mol−1 − 2 120.9 kJ mol−1 ∆ f H ZnBr2 = −394.1 kJ mol−1

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7-66

Chapter 7: Solutions of Electrolytes

7.25.

Solutions

One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is –1051.4 kJ mol–1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows: Ion

Δ hyd G˚ k J mol–1

H+

0 +

Na

2+

Mg

679.1 274.1

3+

–1346.4

–

–1407.1

–

–1393.3

Al Cl

Br Solution: Given: ∆ hyd G ° + ( absolute ) = −1051.4 kJ mol –1 H

Required: ∆ hyd G ° ( absolute ) for each ion In order to find the absolute Gibbs energies of hydration, we can either lower the cation’s standard Gibbs energies of hydration, or raise the anion’s absolute Gibbs energies of hydration by1051.4 kJ mol–1 (per charge).

7-67

Chapter 7: Solutions of Electrolytes

Solutions

∆ hyd G ° + ( absolute ) = 0 − 1051.4 kJ mol−1 H

∆ hyd G ° + ( absolute ) = −1051.4 kJ mol−1 H

∆ hyd G

679.1 kJ mol−1 − 1051.4 kJ mol−1 ( absolute ) =

° + Na

∆ hyd G ° + ( absolute ) = −372.3 kJ mol−1 Na

∆ hyd G

° 2+ Mg

∆ hyd G °

Mg

2+

274.1 kJ mol−1 − 2 (1051.4 kJ mol ) ( absolute ) =

( absolute ) =

−1828.7 kJ mol−1

(

∆ hyd G ° 3+ ( absolute ) = −1346.4 kJ mol−1 − 3 1051.4 kJ mol−1 Al

)

∆ hyd G ° 3+ ( absolute ) = −4500.6 kJ mol−1 Al

∆ hyd G ° − ( absolute ) = −1407.1 kJ mol−1 + 1051.4 kJ mol−1 Cl

∆ hyd G ° − ( absolute ) = −355.7 kJ mol−1 Cl

∆ hyd G ° − ( absolute ) = −1393.3 kJ mol−1 + 1051.4 kJ mol−1 Br

∆ hyd G ° − ( absolute ) = −341.9 kJ mol−1 Br

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7-68

Chapter 7: Solutions of Electrolytes

7.26.

Solutions

Calculate the ionic strengths of 0.1 M solutions of KNO 3 , K 2 SO 4 , ZnSO 4 , ZnCl 2 , and K 4 Fe(CN) 6 ; assume complete dissociation and neglect hydrolysis.

Solution: Given: c = 0.1 M Required: I The ionic strength of a solution is given by Eq. 7.103: I=

1 ∑ ci zi2 2 i

where z i is the valency of each ion present.

7-69

Chapter 7: Solutions of Electrolytes

Solutions

KNO3 → K + + NO3−

(

1 0.1 M ×12 + 0.1 M × 12 2 = 0.1 M

I= KNO3 I KNO3

)

K 2SO 4 → 2K + + SO 4 2− I= K 2SO 4 I K 2SO4

(

1 0.2 M ×12 + 0.1 M × 22 2 = 0.3 M

)

ZnSO 4 → Zn 2+ + SO 4 2− I= ZnSO 4 I ZnSO4

(

1 0.1 M × 22 + 0.1 M × 22 2 = 0.4 M

)

ZnCl2 → Zn 2+ + 2Cl− I= ZnCl2 I ZnCl2

(

1 0.1 M × 22 + 0.2 M ×12 2 = 0.3 M

K 4 Fe ( CN )6 → 4K + + Fe ( CN )6 I K 4 Fe= ( CN ) 6

I K 4 Fe( CN )

(

)

4−

1 0.4 M ×12 + 0.1 M × 44 2 = 1.0 M

)

6

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7-70

Chapter 7: Solutions of Electrolytes

Solutions

Calculate the mean activity coefficient γ ± for the Ba2+ and SO 24− ions in a saturated solution of BaSO 4 (K sp = 9.2 × 10–11 mol2 dm–6) in 0.2 M K 2 SO 4 , assuming the Debye-Hückel limiting law to apply.

7.27.

Solution: Given: BaSO 4 : K sp = 9.2×10–11 mol2 dm–6, cK 2SO4 = 0.2 M Required: γ ± When determining the mean activity coefficient, we use the Debye-Hückel limiting law given in Eq. 7.111:

log10 γ ± = − 0.51z+ z−

I/mol dm −3

We may then calculate the ionic strength from Eq. 7.103 in the following manner, I=

1 ∑ ci zi2 2 i

K 2SO 4 → 2K + + SO 4 2−

I K 2SO4

(

1 0.4 M ×12 + 0.2 M × 22 2 = 0.6 M

I= K 2SO 4

)

Finally we can solve for the mean activity coefficient which produces; log10 γ ± = − 0.51( 2 × 2 ) 0.6

γ ± = 10 −0.51( 2×2) 0.6 γ ± = 0.026 291 949 8 2.6 ×10−2 γ= ±

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7-71

Chapter 7: Solutions of Electrolytes

7.28.

Solutions

The solubility of AgCl in water at 25 °C is 1.274 × 10–5 mol dm–3. On the assumption t

hat the Debye-Hückel limiting law applies,

a. Calculate ΔG° for the process AgCl(s) → Ag+(aq) + Cl–(aq). b. Calculate the solubility of AgCl in an 0.005 M solution of K 2 SO 4 . Solution: Given: s = 1.274 ×10 –5 mol dm –3 , T = 25 °C Required: see above a. To calculate the Gibbs energy, we first need to determine the solubility product of AgCl in water. Eq. 7.121 shows that K s = [Ag + ][Cl− ]γ ±2 , and since [Ag + ] = [Cl− ] , we can write K s = s 2γ ±2 .

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength. log10 γ ± = − 0.51z+ z−

I /mol dm −3

− 0.51(1×1) 1.274 ×10 –5 log10 γ ± =

γ ± = 10 −0.51(1×1) 1.274×10 γ ± = 0.995 817 261 4

–5

The solubility product is then,

= Ks

(1.274 ×10

–5

M ) ( 0.995 817 261 4 ) 2

2

= K s 1.609 526 59 ×10 –10 M 2 Solving for Gibbs energy, using ∆G° = − RT ln K s , we obtain

7-72

Chapter 7: Solutions of Electrolytes

( −8.3145 J K

∆G° =

−1

)(

Solutions

) (

mol−1 298.15 K ln 1.609 526 59 ×10 –10 M 2

)

∆G° =55 900.511 31 J mol−1 ∆G° =55.90 kJ mol−1 b. To solve for the solubility in a solution of c = 0.005 M K 2 SO 4 , we need to calculate the ionic strength since we have a common ion present. We can calculate the ionic strength from Eq. 7.103 in the following manner, 1 ∑ ci zi2 2 i 1 0.01 M × 12 + 0.005 M × 22 = I 2 I = 0.015 M I=

(

)

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength.

log10 γ ± = −0.51 z+ z−

I / mol dm −3

−0.51(1× 1) 0.015 log10 γ ± = log10 γ ± = −0.062 461 988

γ ± = 10−0.062 461988 γ ± = 0.866 040 12 Solving for the solubility by rearranging Eq. 7.121,

7-73

Chapter 7: Solutions of Electrolytes

Solutions

K s = s 2γ ±2 s=

Ks

γ±

1.609 526 59 ×10−10 M 2 0.86604012 = s 1.464 91×10−5 M s=

= s 1.46 ×10−5 M

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7-74

Chapter 7: Solutions of Electrolytes

7.29.

Solutions

Employ Eq. 7.114 to make plots of log γ ± against I for a uni-univalent electrolyte in water at 25 °C, with B = 0.51 mol–1 dm3/2 and B′ = 0.33 × 1010 mol–1 dm3/2 m–1, and for the following values of the interionic distance a: a = 0, 0.1, 0.2, 0.4, and 0.8 nm

Solution:

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7-75

Chapter 7: Solutions of Electrolytes

7.30.

Solutions

Estimate the change in Gibbs energy ΔG when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution (∈ = 78) to the lipid environment of a cell membrane (∈ = 4) at 25 °C.

Solution: Given: n = 1 mol, rK + = 0.133 nm, ∈water = 78, ∈membrane = 4, T = 25 °C Required: ΔG Eq. 7.87 (given below) may be used to estimate the change in Gibbs energy. Ges° =

z 2e2 8π ∈0∈ r

Since we are given 1 mol of K+, we will multiply the expression above by L, Avogadro’s number.

7-76

Chapter 7: Solutions of Electrolytes

Ges° = ° es

G = Ges° =

Solutions

z 2e2 L 8π ∈0∈ r

( +1)

(

2

(1.602 ×10

−19

) ( 6.022 ×10 mol ) m ) ( 0.133 ×10 m ) ∈ 2

C

8π 8.854 ×10−12 C2 N −1

−1

23

−2

−9

5 222 197.4616 N m mol−1

∈

where 1 N m = 1 J ° es

G =

5 222 197.4616 J mol−1

∈

5 222 197.4616 J mol−1 78 ° Gwater = 6694.839 251 J mol−1 ° Gwater =

5 222 197.4616 J mol−1 4 ° Gmembrane = 130 549.3654 J mol−1 ° Gmembrane =

° ° Ges° Gmembrane ∆= − Gwater

= ∆Ge°s 130 549.3654 J mol−1 − 6694.839 251 J mol−1 ∆Ges° = 123 854.526 1 J mol−1 ∆Ges° = 124 kJ mol−1

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7-77

Chapter 7: Solutions of Electrolytes

Solutions

At 18 °C the electrolytic conductivity of a saturated solution of CaF 2 is 3.86 × 10–5 Ω–1 cm–1, and that of pure water is 1.5 × 10–6 Ω–1 1 cm–1. The molar ionic conductivities of Ca 2+ and F– are 51.1 Ω–1 cm2 mol–1 and 47.0 Ω–1 cm2 mol–1, respectively. Calculate the 2 solubility of CaF 2 in pure water at 18 °C and the solubility product.

7.31.

Solution: Given: T =18 °C, κ CaF2 =3.86 ×10 –5 Ω –1 cm –1 , κ H2O =1.5 ×10 –6 Ω –1 cm –1 ,

λ1 2

Ca

2+

= 51.1 Ω –1cm 2 mol –1 , λF− = 47.0 Ω –1cm 2 mol –1

Required: s, K s The expression for the molar conductivity is given by Eq. 7.9:

κ

Λ=

c

It is possible to use the concentration to determine the solubility. c=

κ Λ1 2

Λ1 2

Λ1 2

Λ1 2

CaF2

CaF2

= λ1 2

Ca 2+

+ λF-

= 51.1 Ω –1 cm 2 mol –1 + 47.0 Ω –1 cm 2 mol –1 CaF2

= 98.1 Ω –1 cm 2 mol –1 CaF2

The observed κ due to the salt is therefore,

7-78

Chapter 7: Solutions of Electrolytes

Solutions

κ = 3.86 ×10−5 Ω –1 cm –1 − 1.5 ×10−6 Ω –1 cm –1 κ = 3.71×10−5 Ω –1 cm –1 c=

3.71×10 –5 Ω –1 cm –1 98.1 Ω –1 cm 2 mol –1

= c 3.781855 25 ×10 –7 mol cm −3 = c 3.781855 25 ×10 –4 mol dm −3 1 CaF2 has the molar mass, 2

(

2

)

1 40.078 g mol−1 + 18.998 403 2 g mol−1 2

= M1 CaF2

M CaF2 = 39.037 403 2 g mol−1 The solubility can now be determined.

(

s = 39.037 403 2 g mol−1

) (3.781855 25 ×10

–4

mol dm −3

)

s = 0.014 763 380 8 g dm −3 = s 1.48 ×10 –2 g dm −3 The solubility product is given by, 2

K s = Ca 2+ F− 1 −3 −3 –4 –4 Ks = 3.781855 25 ×10 mol dm 3.781855 25 × 10 mol dm 2 −3 –11 = K s 2.704 485 84 ×10 mol dm

(

)(

)

2

= K s 2.70 ×10 –11 mol dm −3

Back to Problem 7.31

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Chapter 7: Solutions of Electrolytes

7.32.

Solutions

What concentrations of the following have the same ionic strength as 0.1 M NaCl? CuSO 4 , Ni(NO 3 ) 2 , Assume complete dissociation and neglect hydrolysis.

Al 2 (SO 4 ) 3 ,

Na 3 PO 4

Solution: Given: cNaCl = 0.1 M Required: cCuSO4 , cNi( NO3 ) , cAl2 (SO4 ) , cNa3PO4 3

2

As we have previously seen, the ionic strength of a compound may be determined using Eq. 7.103. 1 ∑ ci zi2 2 i 1 0.1 M ×12 + 0.1 M × 12 = I NaCl 2 I NaCl = 0.1 M I=

(

)

For each salt, we set = I I= 0.1 M to solve for c . NaCl

7-80

Chapter 7: Solutions of Electrolytes

(

1 cCuSO4 × 22 + cCuSO4 × 22 2 0.1 M = 4cCuSO4 0.1 M =

Solutions

)

2.5 ×10−2 M cCuSO= 4

(

1 c × 22 + 2cNi( NO3 ) ×12 2 2 Ni( NO3 )2 0.1 M = 3cNi( NO3 )

0.1 M =

)

2

3.3 × 10−2 M cNi( NO3= ) 2

(

1 2cAl2 (SO4 ) × 32 + 3cAl2 (SO4 ) × 22 3 3 2 0.1 M = 15cAl2 (SO4 )

0.1 M =

)

3

6.7 ×10−3 M cAl2 (SO4= ) 3

(

1 3cNa3PO4 ×12 + cNa 3PO4 × 32 2 0.1 M = 6cNa3PO4

0.1 M =

)

cNa3PO= 1.7 ×10−2 M 4

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7-81

Chapter 7: Solutions of Electrolytes

Solutions

The solubility product of PbF 2 at 25.0 °C is 4.0 × 10–9 mol3 dm–9. Assuming the Debye-Hückel limiting law to apply, calculate the solubility of PbF 2 in (a) pure water and (b) 0.01 M NaF.

7.33.

Solution: Given: K s = 4.0 ×10 –9 mol3 dm –9 , T = 25 °C Required: see above a) The dissolution of PbF 2 is written as: PbF2 → Pb 2+ + 2F− , hence the solubility product is given by: 2

K s = Pb 2+ F− . In order to solve for solubility, we must first neglect the effect of the activity coefficients and write, = Ks

s ][ 2 s ] [= 2

4s3 .

Solving for s, we obtain, 3 4 s= 4.0 ×10 –9 mol3 dm –9

= s 1.0 ×10 –3 mol dm –3 We will determine the activity coefficients of Pb2+ and F- by applying the Debye-Hückel limiting law. To solve, we must first calculate the ionic strength of PbF 2 from Eq. 7.103. 1 ∑ ci zi2 2 i 1 I= s × 22 + 2 s ×12 2 I = 3s I=

(

)

= I 3.0 ×10−3 mol dm −3

According to the Debye-Hückel limiting law, Eq. 7.111,

7-82

Chapter 7: Solutions of Electrolytes

log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

Solutions

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 3.0×10 γ ± = 0.879 290 334 3 −3

In order to find the true solubility, we must factor in the activity coefficients.

[γ= + s ][ 2γ − s ] 2

= Ks s3 =

s=

4s3

Ks

4γ +γ − 2

4.0 ×10 –9 mol3 dm –9 3

4 ( 0.879 290 334 3)

3

= s 1.137 280 8 ×10 –3 mol dm –3 = s 1.1×10 –3 mol dm –3

b) In 0.01 M NaF, the ionic strength is essentially 0.01 mol dm-3. Calculating the activity coefficients, we obtain,

γ ± =10−0.51z

+

z−

I /moldm −3

γ ± =10−0.51( 2×1) 0.01 γ ± = 0.790 678 628 If s is the solubility then,

7-83

Chapter 7: Solutions of Electrolytes

K s = Pb 2+ F− Pb 2+ = s

Solutions

2

F− = 0.01 mol dm –3 K s = [γ + s ] γ − 0.01 mol dm –3

(

2

)

K s = γ +γ − 2 0.0001 mol2 dm –6 s s=

(γ γ

+ −

s=

Ks 2

0.0001 mol2 dm –6

)

4.0 × 10 –9 mol 3 dm –9

( 0.790 678 628)

3

0.0001 mol2 dm –6

= s 8.092 076 715 ×10 –5 mol dm –3 = s 8.1×10 –5 mol dm –3

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7-84

Chapter 7: Solutions of Electrolytes

Solutions

Calculate the solubility of silver acetate in water at 25 °C, assuming the DHLL to apply; the solubility product is 4.0 × 10–3 mol2 dm–6.

7.34.

Solution: Given: K s = 4.0 ×10 –3 mol2 dm –6 , T = 25 °C Required: s We will solve this problem in a similar manner as the previous question. We may start by neglecting the activity coefficients to find the first approximation of s. The dissolution of silver acetate is given by the following: AgC2 H 3O 2 → Ag + + CH 3COO − = Ks

s ][ s ] [=

s2

s = Ks = s

4.0 ×10 –3 mol2 dm –6

s = 0.063 245 553 2 mol dm –3 The ionic strength is thus,

(

)

1 s ×12 + s ×12 2 I= s= 0.063 245 553 2 mol dm –3 I=

According to the Debye-Hückel limiting law, Eq. 7.111, log10 γ ± = − 0.51z+ z−

I /mol dm −3

γ ± =10−0.51(1×1) 0.063 245 553 2 γ ± = 0.744 289 325 The second approximation of the solubility is therefore, 7-85

Chapter 7: Solutions of Electrolytes

= Ks s=

s=

[γ= + s ][γ − s ]

Solutions

γ +γ − s 2

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.744 289 325)

2

s = 0.084 974 419 3 mol dm –3 = s 8.5 ×10 –2 mol dm –3 We may continue to take third and fourth approximations by repeating the above steps.

γ ± =10−0.51(1×1) 0.084 974 419 3 γ ± = 0.710 122 220 4 s= s=

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.710 122 220 4 )

2

s = 0.089 062 912 5 mol dm –3

The third approximation is then; = s 8.9 ×10 –2 mol dm –3 .

γ ± =10−0.51(1×1) 0.089 062 912 5 γ ± = 0.704 366 363 8 s= s=

Ks

γ +γ − 4.0 ×10 –3 mol2 dm –6

( 0.704 366 363 8)

2

s = 0.089 790 706 2 mol dm –3 7-86

Chapter 7: Solutions of Electrolytes

Solutions

The fourth approximation is then; = s 9.0 ×10 –2 mol dm –3 .

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7-87

Chapter 7: Solutions of Electrolytes

7.35.

Solutions

Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 °C: ∈ = 78; ∂ ln ∈ /∂T = –0.0046 K–1. Suggest a qualitative explanation for the sign of the value you obtain.

Solution: Given: Problem 7.30: T = 25 °C, ∈= 78,

∂ln ∈ = −0.0046 ∂T

Required: ∆Ses° , explain the sign From Problem 7.30, we found the expression for the estimated Gibbs energy, Ges° =

5 222 197.4616 J mol−1

∈

For the transfer from water to lipid we can say that, 1 1 ° −1 ∆G= 5 222 197.4616 − es /J mol ∈ lipid ∈water

From Eq. 3.119:

∂G = −S ∂T P ∂G ° − es hence, ∆Ses° = ∂T P ∆Ses° / J K −1 mol−1 = −5 222 197.4616

∂ ∂T

1 1 − ∈lipid ∈water

Since ∈lipid is temperature independent, this leads to:

7-88

Chapter 7: Solutions of Electrolytes

Solutions

1 ∂ − ∂T ∈water 1 ∂∈ ∆Ses° / J K −1 mol−1 = −5 222 197.4616 2 ∈ water ∂T ∆Ses° / J K −1 mol−1 = −5 222 197.4616

since

1 ∂∈ 1 ∂ ln ∈ = 2 ∈ water ∂T ∈water ∂T

∆Ses° / J K −1 mol−1 = −5 222 197.4616

1

∈water

∂ ln ∈ ∂T

1 ∆Ses° / J K −1 mol−1 = −5 222 197.4616 ( −0.0046 ) 78 ° −1 −1 ∆Ses = 30.796 260 56 J K mol ∆Ses° = 31 J K −1 mol−1 The entropy increases due to the release of bound water molecules when the K+ ions pass into the lipid.

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7-89

Chapter 7: Solutions of Electrolytes

7.36.

Solutions

Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl– in aqueous solution at 25 °C (∈ = 78), under the following conditions: a. The electrolyte is present at infinite dilution. b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70. The ionic radii are 95 pm for Na+ and 181 pm for Cl–.

Solution: Given: Eq. 7.86, = n 1 mol, T = 25 °C,= ∈ 78 Required: see above a) At infinite dilution, the work of charging an ion is given by Eq.7.86 which states, z 2e2 wrev = 8π ∈0∈ r For one mole of Na+, we multiply by Avogadro’s number, L and use the ionic radius of r = 95 pm. wNa + = wNa + =

z 2e2 L 8π ∈0∈ r

( +1)

(

2

(1.602 ×10

−19

C

8π 8.854 ×10−12 C2 N −1

) ( 6.022 ×10 mol ) m ) ( 78 ) ( 95 ×10 m ) 2

23

−2

−1

−12

wNa + = 9372.774 952 N m mol−1 wNa + = 9372.774 952 J mol−1 For one mole of Cl-, we will multiply by Avogadro’s number, L and use the ionic radius of r = 181 pm.

7-90

Chapter 7: Solutions of Electrolytes

wCl- =

( −1)

2

(

(1.602 ×10

−19

C

8π 8.854 ×10−12 C2 N −1

Solutions

) ( 6.022 ×10 mol ) m ) ( 78 ) (181×10 m ) 2

−1

23

−2

−12

wCl- = 4919.412 267 J mol−1 One mole of Na+Cl- at infinite dilution is thus,

wrev wNa + + wCl= wrev 9372.774 952 J mol−1 + 4919.412 267 J mol−1 = wrev = 14 292.187 22 J mol−1 wrev = 14 kJ mol−1 b) These values are reduced when the electrolyte is at a higher concentration. The work of charging the ionic atmosphere is negative and equal to kTlnγ i . Thus for one mol of Na+ ions, of activity γ + , the work of charging the atmosphere is RTlnγ + . Similarly, for the chloride ion, the work per mole is RTlnγ − . For one mole of Na+Clwrev RT ( lnγ + + ln γ − ) = wrev = RTln ( γ +γ − ) wrev = 2 RTlnγ ± where γ ± = 0.70

(

)(

)

wrev = 2 8.3145 J K −1 mol−1 298.15 K ln 0.70 wrev = −1768.371 67 J mol−1

The net work done is given by, = wrev 14 292.187 22 J mol−1 − 1768.371 67 J mol−1 wrev = 12 523.815 55 J mol−1 wrev = 13 kJ mol−1 7-91

Chapter 7: Solutions of Electrolytes

Back to Problem 7.36

Solutions

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7-92

Chapter 7: Solutions of Electrolytes

7.37.

Solutions

If the solubility product of barium sulfate is 9.2 × 10–11 mol2 dm–6, calculate the solubility of BaSO 4 in a solution that is 0.10 M in NaNO 3 and 0.20 M in Zn(NO 3 ) 2 ; assume the DHLL to apply.

Solution: Given: K s = 9.2 ×10 –11 mol2 dm –6 , cin NaNO3 = 0.10 M, cin Zn ( NO3 ) = 0.20 M 2

Required: s The expression for the solubility product is given by, K s = Ba 2+ SO 4 2− K s = [γ + s ][γ − s ] Ks = γ ±2s2 The ionic strength of the solution is calculated according to Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 I= (1 × 0.1 M + 12 × 0.1 M + 22 × 0.2 M + 12 × 0.4 M ) 2 I = 0.70 M I=

To find the activity coefficient, we will use the Debye-Hückel limiting law given by Eq. 7.111. log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×2) 0.70 γ ± = 0.019 643 259 1 If the solubility s is therefore,

7-93

Chapter 7: Solutions of Electrolytes

s= s=

Solutions

Ks

γ ±2 9.2 ×10 –11 mol2 dm –6

( 0.019 643 259 1)

2

= s 4.882 928 531×10 –4 M = s 4.9 × 10 –4 M

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7-94

Chapter 7: Solutions of Electrolytes

7.38.

Solutions

Silver chloride, AgCl, is found to have a solubility of 1.561 × 10–5 M in a solution that is 0.01 M in K 2 SO 4 . Assume the DHLL to apply and calculate the solubility in pure water.

Solution: Given: s = 1.561× 10 –5 M, c = 0.01 M Required: s The expression for the solubility product is given by, K s = Ag + Cl− K s = [γ + s ][γ − s ] Ks = γ ±2s2 The ionic strength of the solution is calculated according to Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 I= (1 × 0.02 M + 22 × 0.01 M ) 2 I = 0.03 M I=

To determine the activity coefficient, we will use the Debye-Hückel limiting law, Eq. 7.111. log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51(1×1) 0.03 γ ± = 0.815 953 498 4 If the solubility product is K s ,

7-95

Chapter 7: Solutions of Electrolytes

= Ks

( 0.815 953 498 4 )

2

(1.561×10

Solutions

–5

M

)

2

= K s 1.622 320 38 ×10 –10 M 2 Finally, the solubility in pure water is given by; s = Ks = s

1.622 320 38 ×10 –10 M 2

= s 1.273 703 411×10 –5 M = s 1.3 ×10 –5 M Back to Problem 7.38

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7-96

Chapter 7: Solutions of Electrolytes

7.39.

Solutions

The enthalpy of neutralization of a strong acid by a strong base, corresponding to the process H+(aq) + OH–(aq) → H 2 O is –55.90 kJ mol–1. The enthalpy of neutralization of HCN by NaOH is –12.13 kJ mol–1. Make an estimate of the enthalpy of dissociation of HCN.

Solution: Given: ∆H = −55.90 kJ mol−1 , ∆ neut H = −12.13 kJ mol−1 Required: ∆ diss H The enthalpy change for the neutralization of HCN by NaOH is less than the value of the 55.90 kJ mol-1 because the energy required for the dissociation of HCN, ∆ diss H is given by:

∆ neut H = ∆H − ∆ diss H , hence, ∆ diss H = ∆H − ∆ neut H = ∆ diss H 55.90 kJ mol−1 − 12.13 kJ mol−1 ∆ diss H = 43.77 kJ mol−1

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7-97

Chapter 7: Solutions of Electrolytes

7.40.

Solutions

Make use of the Debye-Hückel limiting law to estimate the activity coefficients of the ions in an aqueous 0.004 M solution of sodium sulfate at 298 K. Estimate also the mean activity coefficient.

Solution: Given: cNa 2SO4 0.004 M, T 298 K = = Required: γ Na + , γ SO 2− , γ ± 4

From Eq. 7.104 we can calculate the activity coefficients for each ion.

log10 γ i = − zi2 B I As we know, the ionic strength of the solution is calculated using Eq.7.103.

1 ci zi2 ∑ 2 i 1 2 1 × 0.008 M + 22 × 0.004 M ) I= ( 2 I = 0.012 M I=

The activity coefficients are therefore,

γ Na = 10 −( +1) 0.51 2

0.012

+

γ Na = 0.879 290 334 3 +

γ Na = 0.879 +

γ SO

4

γ SO

4

γ SO

2−

= 10 −( −2)

2−

= 0.597 763 229 5

2−

2

0.51 0.012

= 0.598

4

The mean activity coefficient can be determined by using Eq. 7.111 which produces,

7-98

Chapter 7: Solutions of Electrolytes

log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

Solutions

I /mol dm −3

I /moldm −3

γ ± =10−0.51(1×2) 0.012 γ ± = 0.773151 491 9 γ ± = 0.773 Back to Problem 7.40

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7-99

Chapter 7: Solutions of Electrolytes

7.41.

Solutions

A 0.1 M solution of sodium palmitate, C 15 H 31 COONa, is separated from a 0.2 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl– ions but not to palmitate ions. Calculate the concentrations of Na+ and Cl– ions on the two sides of the membrane after equilibrium has become established. (For a calculation of the Nernst potential, see Problem 8.18.)

Solution: Given: = cNaP 0.1 = M, cNaCl 0.2 M Required: cNa + , cCl− on each side of the membrane at equilibrium Initial concentrations

Final concentrations

Palmitate Side Na + = 0.1 M

Other Side Na + = 0.2 M

P − = 0.1 M = Na + ( 0.1 M + x )

Cl− = 0.2 M = Na + ( 0.2 M − x )

P − = 0.1 M Cl− = x

Cl− =

( 0.2 M − x )

At equilibrium, the final concentration of NaCl on each side of the membrane will be the same. This allows us to solve for x.

( 0.2 M − x )

2

= ( 0.1 + x ) x

0.04 − 0.4 x + = x 2 0.1x + x 2 0.04 = 0.5 x x = 0.08

7-100

Chapter 7: Solutions of Electrolytes

Solutions

The final concentrations are thus, Final concentrations

Palmitate Side = Na + ( 0.1 M + 0.08 M )

Na + = 0.18 M Cl− = 0.08 M

Back to Problem 7.41

Other Side + Na = = Cl− + Na = = Cl−

( 0.2 M − x ) ( 0.2 M − 0.08 M )

+ = Cl− 0.12 M Na =

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7-101

Chapter 7: Solutions of Electrolytes

7.42.

Solutions

Consider the ionizations H + + H 3 N + CH 2 COO − H 3 N + CH 2 COOH H 2 NCH 2 COOH + H + Assume that the following acid dissociation constants apply to the ionizations:

− NH 3+ − NH 2 + H + ; K a = 1.5 ×10−10 M −COOH − COO − + H + ; K a =× 4.0 10−3 M Estimate a value for the equilibrium constant for the process H 3 N + CH 3COO − H 2 NCH 2 COOH

Solution: Given: K − NH+ = 1.5 ×10−10 M, K − COOH = 4.0 ×10−3 M 3

Required: K NH+ CH COO 3

3

The equilibrium constant K − NH+ is given by, 3

K − NH+ = 3

[ H 2 NCH 2COOH ] H + + H 3 N CH 2 COOH

And K − COOH is given by, K − COOH

H 3 N + CH 2 COO − H + = H 3 N + CH 2 COOH

We can rearrange the two expressions to obtain an expression for K NH+ CH COO , and solve for its value. 3

3

7-102

Chapter 7: Solutions of Electrolytes

K NH+ CH COO = 3

3

K NH+ CH COO = 3

3

K NH+ CH COO = 3

3

Solutions

K − NH+ 3

K − COOH

[ H 2 NCH 2COOH ]

H 3 N + CH 2 COO − 1.5 ×10−10 M 4.0 ×10−3 M

K NH+ CH COO = 3.75 ×10−8 3

3

K NH+ CH COO = 3.8 ×10−8 3

3

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7-103

Chapter 7: Solutions of Electrolytes

7.43.

Solutions

The pK values for the successive ionizations of phosphoric acid are given on p. 308. Which of the four species is predominant at the following values of the hydrogen or hydroxide concentration? a. [H+] = 0.1 M. b. [H+] = 2 × 10–3 M. c. [H+] = 5 × 10–5 M. d. [OH–] = 2 × 10–3 M. e. [OH–] = 1 M.

Solution: Given: p. 308 Required: see above The successive ionizations of phosphoric acid are given by the following expression. = pK1 2.1 = pK 2 7.2 − 3 4 2 4

H PO H PO

pK3 =12.3

HPO 24− PO34−

In order to determine the pH we will use, pH = − log10 H + and pH= 14 − pOH . a.

pH = − log10 ( 0.1)

H 3 PO 4 is predominant

pH = 1 b.

pH = − log10 ( 2 ×10−3 )

HPO 24− is predominant

pH = 2.698

c.

pH = − log10 ( 5 ×10−5 )

HPO 24− is predominant

pH = 4.301

d.

pOH = − log10 ( 2 ×10−3 )

HPO 24− is predominant

pOH = 2.698 pH= 14 − 2.698 pH = 11.304

7-104

Chapter 7: Solutions of Electrolytes

e.

pOH = − log10 1

Solutions

PO34− is predominant

pOH = 0 pH = 14 Back to Problem 7.43

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7-105

Chapter 7: Solutions of Electrolytes

7.44.

Solutions

Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl– ions but not to P– ions. The initial concentrations are as shown below.

[K+] = 0.05 M

[K+] = 0.15 M

[Cl–] = 0.05 M

[P–] = 0.15 M

Calculate the concentrations on each side of the membrane after equilibrium has become established. (See Problem 8.26 in Chapter 8 for the calculation of the Nernst potential for this system.) Solution: Given: see above Required: cK + , cCl− on each side of the membrane at equilibrium We can solve this problem in a similar manner as problem 7.41. Initial concentrations

Final concentrations

Left-hand Side K + = 0.05 M

Right-hand Side K + = 0.15 M

Cl− = 0.05 M K + 0.05 M − x =

P − = 0.15 M K + 0.15 M + x =

Cl− 0.05 M − x =

Cl− = x P − = 0.1 M

At equilibrium, the final concentration of KCl on each side of the membrane will be the same. This will allow us to solve for x.

( 0.05 M − x )

2

=

( 0.15 + x ) x

0.0025 − 0.1x += x 2 0.15x + x 2 0.0025 = 0.25 x x = 0.01

7-106

Chapter 7: Solutions of Electrolytes

Solutions

The final concentrations are thus, Final concentrations

Left-hand Side Right-hand Side + − K + 0.15 M + 0.01 M = K = Cl 0.05 M − 0.01 M = = K + = Cl− 0.04 M

Back to Problem 7.44

K + = 0.16 M Cl− = 0.01 M

Back to Top

7-107

CHAPTER

8

Electrochemical Cells

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 8: Electrochemical Cells

Electrode Reactions and Electrode Potentials

Chapter 8 *problems with an asterisk are slightly more demanding Electrode Reactions and Electrode Potentials 8.1.

Write the electrode reactions, the overall reaction, and the expression for the emf for each of the following reversible cells. a. Pt, H 2 (1 bar)|HCl(aq)|Pt, Cl 2 (1 bar) b. Hg|Hg 2 Cl 2 (s)|HCl(aq)|Pt, H 2 (1 bar) c. Ag|AgCl(s)|KCl(aq)|Hg 2 Cl 2 (s)|Hg d. Pt, H 2 (1 bar)|HI(aq)|AuI(s)|Au e. Ag|AgCl(s)|KCl(c 1 )

KCl(c 2 )|AgCl(s)|Ag Solution

8.2.

At 25 °C and pH 7, a solution containing compound A and its reduced form AH 2 has a standard electrode potential of –0.60 V. A solution containing B and BH 2 has a standard potential of –0.16 V. If a cell were constructed with these systems as half-cells, a. Would AH 2 be oxidized by B or BH 2 oxidized by A under standard conditions? b. What would be the reversible emf of the cell? c. What would be the effect of pH on the equilibrium ratio [B][AH 2 ]/[A][BH 2 ]? Solution

8.3.

Calculate the standard electrode potential for the reaction Cr2+ + 2e– → Cr at 298 K. The necessary E° values are a. Cr3+ + 3e– → Cr

E° = –0.74 V

b. Cr3+ + e– → Cr2+

E° = –0.41 V Solution

8-2

Chapter 8: Electrochemical Cells

8.4.

Thermodynamics of Electrochemical Cells

Write the individual electrode reactions and the overall cell reaction for the following cell: Pt, H 2 |H+(1 m)

F2–, S2–, H+(aq)|Pt

where F2– represents the fumarate ion and S2– the succinate ion. Write the expression for the emf of the cell. Solution 8.5.

Design electrochemical cells in which each of the following reactions occurs: a. Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq) b. Ag+(aq) + Cl–(aq) → AgCl(s) c. HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) In each case, write the representation of the cell and the reactions at the two electrodes. Solution

Thermodynamics of Electrochemical Cells 8.6.

Calculate the equilibrium constant at 25 °C for the reaction 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I 2 (s) using the standard electrode potentials given in Table 8.1. Solution

8.7.

From data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction Sn + Fe2+ → Sn2+ + Fe Solution

8-3

Chapter 8: Electrochemical Cells

8.8.

Thermodynamics of Electrochemical Cells

The standard electrode potential at 25 °C for cytochrome c (Fe3+) + e– → cytochrome c (Fe2+) is 0.25 V. Calculate ∆G° for the process 1 H 2 (g) + cytochrome c (Fe3+) → H+ + cytochrome c (Fe2+) 2

Solution 8.9.

Using the values given in Table 8.1, calculate the standard Gibbs energy change ∆G° for the reaction H2 +

1 O2 → H2O 2 Solution

*8.10. From the data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction 2Cu+ → Cu2+ + Cu What will be produced if Cu 2 O is dissolved in dilute H 2 SO 4 ? Solution 8.11.

For the reaction 3H 2 (g, 1 atm) + Sb 2 O 3 (s, cubic) → 2Sb(s) + 3H 2 O(l), ∆G° = –83.7 kJ [Roberts and Fenwick, J. Amer. Chem. Soc., 50, 2146(1928)]. Calculate the potential developed by the cell Pt|H 2 (g, 1 atm)|H+|Sb 2 O 3 (s, cubic)|Sb(s) Which electrode will be positive? Solution

8-4

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

Nernst Equation and Nernst Potentials 8.12.

Calculate the emf for the following cell at 25 °C: Pt, H 2 (1 bar)|HCl(0.5 m)

HCl(1.0 m)|Pt, H 2 (1 bar) Solution

8.13.

The pyruvate-lactate system has an E°′ value of –0.185 V at 25 °C and pH 7.0. What will be the potential of this system if the oxidation has gone to 90% completion? Solution

8.14.

a. From the data in Table 8.1, calculate the standard electrode potential for the half-reaction Fe3+ + 3e– → Fe b. Calculate the emf at 25 °C of the cell Pt|Sn2+(0.1 m), Sn4+(0.01 m)

Fe3+(0.5 m)|Fe Solution

8.15.

The cell Pt|H 2 (1 bar), H+ KCl(saturated)|Hg 2 Cl 2 |Hg was used to measure the pH of a solution of 0.010 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential expected at 25 °C [K a = 1.81 × 10–5 for acetic acid]. Solution

8.16.

The voltage required to electrolyze certain solutions changes as the electrolysis proceeds because the concentrations in the solution are changing. In an experiment, 500 dm3 of a 0.0500 M solution of copper (II) bromide was electrolyzed until 2.872 g Cu was deposited. Calculate the theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end of the experiment. Solution

8-5

Chapter 8: Electrochemical Cells

8.17.

Nernst Equation and Nernst Potentials

Calculate the concentration of I3– in a standard solution of iodine in 0.5 M KI, making use of the following standard electrode potentials: I 2 + 2e– → 2I–

Eº = 0.5355 V

I3− + 2e– → 3I–

Eº = 0.5365 V

The molality of I– in the standard solution can be assumed to be 0.5 m. Solution 8.18.

Calculate the Nernst potential at 25 °C arising from the equilibrium established in Problem 7.41. Solution

8.19.

It might seem plausible to separate lead and gold by making use of the great difference between their standard electrode potentials (Table 8.1). In order to test this idea, one might electrolyze a solution containing 0.0100 M AuNO 3 and 0.0100 M Pb(NO 3 ) 2 in a well-stirred tank using platinum electrodes at low current density. As the potential difference is slowly increased from zero, which metal will be deposited first? What will be the concentration of this metal ion in solution when the second metal begins to be deposited? Do you think this is an acceptable method of separating the two metals? Solution

8.20.

Calculate the emf of the cell Pt, H 2 (1 bar)|HCl(0.1 m)

HCl(0.2 m)|Pt, H 2 (10 bar) Solution

*8.21. Suppose that the cell in Problem 8.20 is set up but that the two solutions are separated by a membrane that is permeable to H+ ions but impermeable to Cl– ions. What will be the emf of the cell at 25 °C? Solution

8-6

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

*8.22. A typical biological cell has a volume of 10–9 cm3, a surface area of 10–6 cm2, and a membrane thickness of 10–6 cm; the dielectric constant of the membrane may be taken as 3. Suppose that the concentration of K+ ions inside the cell is 0.155 M and that the Nernst potential across the cell wall is 0.085 V. a. Calculate the net charge on either side of the wall, and b. Calculate the fraction of the K+ ions in the cell that are required to produce this charge. Solution *8.23. Calculate the emf at 25 °C of the cell Pt, H 2 (1 bar)|H 2 SO 4 (0.001 m)|CrSO 4 (s)|Cr given the following standard electrode potential: CrSO 4 (s) + 2e– → Cr + SO 2– 4

Eº = –0.40 V

a. First make the calculation neglecting activity coefficient corrections. b. Then make the calculation using activity coefficients estimated on the basis of the Debye-Hückel limiting law. Solution *8.24. Write the individual electrode reactions and the overall reaction for Cu|CuCl 2 (aq)|AgCl(s)|Ag If the emf of the cell is 0.191 V when the concentration of CuCl 2 is 1.0 × 10–4 M and is –0.074 V when the concentration is 0.20 M, make an estimate of the mean activity coefficient in the latter solution. Solution

8-7

Chapter 8: Electrochemical Cells

Nernst Equation and Nernst Potentials

*8.25. a. Write both electrode reactions and the overall reaction for the cell Tl|TlCl(s)|CdCl 2 (0.01 m)|Cd b. Calculate E and E° for this cell at 25 °C from the following information: Tl+ + e– → Tl

Eº = –0.34 V

Cd + 2e → Cd

Eº = –0.40 V

2+

–

The solubility product for TlCl is 1.6 × 10–3 mol2 dm–6 at 25 °C. Solution 8.26.

Problem 7.44 involved calculating the concentrations on each side of a membrane after a Donnan equilibrium had become established. Which side of the membrane is positively charged? Calculate the Nernst potential across the membrane if the temperature is 37 °C. Solution

8.27.

The oxidation of lactate to pyruvate by the oxidized form of cytochrome c—represented as cytochrome c (Fe3+)—is an important biological reaction. The following are the relevant E°′ values, relating to pH 7 and 25 °C: Eº′/V

pyruvate + 2H + 2e → lactate –

+

–

–

–0.185

cytochrome c (Fe ) + e → cytochrome c (Fe ) 3+

–

2+

0.254

Calculate the equilibrium ratio [cytochrome c (Fe 2+ )]2 [pyruvate – ] [cytochrome c (Fe3+ )]2 [lactate – ] at pH 7 and 25 °C. Also calculate the ratio at pH 6. Solution

8-8

Chapter 8: Electrochemical Cells

8.28.

Nernst Equation and Nernst Potentials

Suppose that the cell Ag|AgCl(s)|HCl(0.10 m)

HCl(0.01 m)|AgCl(s)|Ag

is set up and that the membrane separating the two solutions is permeable only to H+ ions. What is the emf of the cell at 25 °C? Solution 8.29.

a. Consider the cell Pt, H 2 (1 bar)|HCl(m 1 )

HCl(m 2 )|Pt, H 2 (1 bar)

in which the solutions are separated by a partition that is permeable to both H+ and Cl–. The ratio of the speeds with which these ions pass through the membrane is the ratio of their transport numbers t + and t – . Derive an expression for the emf of this cell. b. If when m 1 = 0.01 m and m 2 = 0.01 m the emf is 0.0190 V, what are the transport numbers of the H+ and Cl– ions? Solution 8.30.

The metal M forms a soluble nitrate and a very slightly soluble chloride. The cell M|M+(0.1 m), HNO 3 (0.2 m)|H 2 (1 bar), Pt has a measured E = –0.40 V at 298.15 K. When sufficient solid KCl is added to make the solution of the cell 0.20 m in K+, the emf changes to –0.15 V at 298.15 K as MCl precipitates. Calculate the K sp of MCl, taking all activity coefficients to be unity. Solution

8-9

Chapter 8: Electrochemical Cells

8.31.

Temperature Dependence of Cell emfs

The substance nicotinamide adenine dinucleotide (NAD+) plays an important role in biological systems; under the action of certain enzymes it can react with a reducing agent and release a proton to the solution to form its reduced form NADH. With pyruvate the reduced form NADH undergoes the reaction NADH + pyruvate– + H+ NAD+ + lactate– The appropriate E°′ values, relating to 25 °C and pH 7, are pyruvate– + 2H+ +2e– → lactate–

Eº′ = –0.19 V

NAD + H + 2e → NADH

Eº′ = –0.34 V

+

+

–

Use these values to calculate ∆G°′ for the reaction, and also the equilibrium ratio

[lactate – ][NAD + ] [pyruvate – ][NADH] (a) at pH 7, and (b) at pH 8. Solution Temperature Dependence of Cell emfs 8.32.

a. Calculate the standard emf E° for the reaction fumarate2– + lactate– → succinate2– + pyruvate– on the basis of the following information: fumarate2– + 2H+ + 2e– → succinate2–

E°′ = 0.031 V

pyruvate + 2H + 2e → lactate

E°′ = –0.185 V

–

+

–

–

The E°′ values relate to pH 7. The temperature coefficient ∂E/∂T for this cell is 2.18 × 10–5 V K–1. b. Calculate ∆G°, ∆H°, and ∆S° at 25 °C. Solution 8-10

Chapter 8: Electrochemical Cells

8.33.

Temperature Dependence of Cell emfs

The Weston standard cell (see Figure 8.2b) is Cd amalgam|CdSO 4 ·

8 H 2 O(s)|Hg 2 SO 4 (s), Hg 3

(saturated solution) a. Write the cell reaction. b. At 25 °C, the emf is 1.018 32 V and ∂E °/∂T = –5.00 × 10–5 V K–1. Calculate ∆G°, ∆H°, and ∆S°. Solution 8.34.

Salstrom and Hildebrand [J. Amer. Chem. Soc., 52, 4650(1930)] reported the following data for the cell Ag(s)|AgBr(s)|HBr(aq)|Br 2 (g, 1 atm)|Pt

t/ºC

442.3

456.0

490.9

521.4

538.3

556.2

E/V

0.8031

0.7989

0.7887

0.7803

0.7751

0.7702

Find the temperature coefficient for this cell assuming a linear dependence of the cell potential with temperature. What is the entropy change for the cell reaction? Solution 8.35.

The reaction taking place in the cell Mg(s)|Mg2+(aq), Cl–(aq)|Cl 2 (g,1 atm)|Pt is found to have an entropy change of –337.3 J K–1 mol–1 under standard conditions. What is the temperature coefficient for the cell? Solution

8-11

Chapter 8: Electrochemical Cells

Temperature Dependence of Cell emfs

*8.36. a. Estimate the Gibbs energy of formation of the fumarate ion, using data in Problem 8.32 and the following values. ∆ f Gº (succinate, aq) = –690.44 kJ mol–1 ∆ f Gº (acetaldehyde, aq) = 139.08 kJ mol–1 ∆ f Gº (ethanol, aq) = –181.75 kJ mol–1 acetaldehyde + 2H+ + 2e– → ethanol Eº′ = –0.197 V b. If the ∂E °/∂T value for the process fumarate2– + ethanol → succinate2– + acetaldehyde is 1.45 × 10–4 V K–1, estimate the enthalpy of formation of the fumarate ion from the following values. ∆ f Hº (succinate, aq) = –908.68 kJ mol–1 ∆ f Hº (acetaldehyde, aq) = –210.66 kJ mol–1 ∆ f Hº (ethanol, aq) = –287.02 kJ mol–1 Solution *8.37. a. Calculate the emf at 298.15 K for the cell Tl|TlBr|HBr (unit activity)|H 2 (1 bar), Pt b. Calculate ∆H for the cell reaction in the following cell. Tl|Tl+ (unit activity), H+ (unit activity)|H 2 (1 bar), Pt For the half-cell Tl+ + e– → Tl E° = 0.34 V ∂E/∂T = –0.003 V/K and

K sp (TlBr) = 10–4 mol2 dm–6 Solution

8-12

Chapter 8: Electrochemical Cells

Applications of emf Measurements

Applications of emf Measurements 8.38.

Calculate the solubility product and the solubility of AgBr at 25 °C on the basis of the following standard electrode potentials: AgBr(s) + e– → Ag + Br–

E° = 0.0713 V

Ag + e → Ag

E° = 0.7996 V

+

–

Solution 8.39.

The emf of a cell Pt, H 2 (1 bar)|HCl(aq)|AgCl(s)|Ag was found to be 0.517 V at 25 °C. Calculate the pH of the HCl solution. Solution

8.40.

The emf of the cell Ag|AgI(s)I–(aq)

Ag+(ag)|Ag

is –0.9509 V at 25 °C. Calculate the solubility and the solubility product of AgI at that temperature. Solution 8.41.

An electrochemical cell M(s)|MCl(aq, 1.0 m)|AgCl(s)|Ag(s), where MCl is the chloride salt of the metal electrode M, yields a cell potential of 0.2053 V at 25 °C. What is the mean activity coefficient γ ± of the electrolyte MCl? E° for the M(s)|M+ electrode is 0.0254 V. Solution

8.42.

The following thermodynamic data apply to the complete oxidation of butane at 25 °C. C 4 H 10 (g) + (13/2)O 2 (g) → 4CO 2 (g) + 5H 2 O(l) ∆H ° = –2877 kJ mol –1 ∆S ° = –432.7 J K –1 mol –1 Suppose that a completely efficient fuel cell could be set up utilizing this reaction. Calculate (a) the maximum electrical work and (b) the maximum total work that could be obtained at 25 °C. 8-13

Chapter 8: Electrochemical Cells

Essay Questions

Solution *8.43. At 298 K the emf of the cell Cd, Hg|CdCl 2 (aq, 0.01 m), AgCl(s)|Ag is 0.7585 V. The standard emf of the cell is 0.5732 V. a. Calculate the mean activity coefficient for the Cd2+ and Cl– ions. b. Compare the value with that estimated from the Debye-Hückel limiting law, and comment on any difference. Solution

*8.44. The following emf values were obtained by H. S. Harned and Copson [J. Amer. Chem. Soc., 55, 2206(1933)] at 25 °C for the cell Pt,H 2 (1 bar)|LiOH(0.01 m), LiCl(m)|AgCl(s)|Ag at various molalities m of LiCl: m/mol kg–1 E/V

0.01

0.02

0.05

0.10

0.20

1.0498

1.0318

1.0076

1.9888

0.9696

Obtain from these data the ionic product of water. Solution

Essay Questions 8.45.

Explain how emf measurements can be used to obtain ∆G°, ∆H°, and ∆S° for a reaction.

8.46.

Suggest an additional example, giving details, for each of the electrochemical cells listed in Figure 8.8.

8-14

Chapter 8: Electrochemical Cells

Solutions

Solutions 8.1.

Write the electrode reactions, the overall reaction, and the expression for the emf for each of the following reversible cells. a. Pt, H 2 (1 bar)|HCl(aq)|Pt, Cl 2 (1 bar) b. Hg|Hg 2 Cl 2 (s)|HCl(aq)|Pt, H 2 (1 bar) c. Ag|AgCl(s)|KCl(aq)|Hg 2 Cl 2 (s)|Hg d. Pt, H 2 (1 bar)|HI(aq)|AuI(s)|Au e. Ag|AgCl(s)|KCl(c 1 )

KCl(c 2 )|AgCl(s)|Ag

Solution: Given: see above Required: electrode reactions, the overall reaction, and the expression for the emf We calculate the emf, or electromotive force, from Eq. 8.13 u

RT [Y] y [Z]z E = E° − ln where z is the number of electrons. zF [A]a [B]b

Electrode reactions

H 2 → 2H + + 2e − Cl2 + 2e − → 2Cl−

Overall reaction H 2 + Cl2 → 2H + + 2Cl− z=2

2Hg ( l ) + 2Cl− → Hg 2 Cl2 + 2e −

2Hg ( l ) + 2Cl− + 2H + → Hg 2 Cl2 + H 2

2H + + 2e − → H 2

z=2

EMF E= E ° −

(

2 2 RT ln H + Cl− 2F

)

u

RT 1 E= E° − ln 2 F H + 2 Cl− 2 2 2 u RT E= E° + ln H + Cl− 2F

(

u

)

8-15

Chapter 8: Electrochemical Cells

Solutions

Ag ( s ) + Cl− → AgCl ( s ) + e −

2Ag ( s ) + Hg 2 Cl2 ( s ) → 2AgCl ( s ) + 2Hg ( s )

Hg 2 Cl2 ( s ) + 2e − → 2Hg ( s ) + 2Cl−

z=2

1 H 2 ( g ) → H + + e− 2 AuI ( s ) + e − → Au ( s ) + I −

AuI ( s ) +

Ag ( s ) + Cl− ( c1 ) → AgCl ( s ) + e −

Cl− ( c1 ) → Cl− ( c2 )

AgCl ( s ) + e − → Ag ( s ) + Cl− ( c2 )

z =1

Back to Problem 8.1

z =1

1 H 2 ( g ) → Au ( s ) + H + + I − 2

No concentration dependence, therefore E= E ° E= E ° −

(

RT ln H + I − F

RT c1 E= E ° − ln F c2

)

u

u

Back to Top

8-16

Chapter 8: Electrochemical Cells

8.2.

Solutions

At 25 °C and pH 7, a solution containing compound A and its reduced form AH 2 has a standard electrode potential of –0.60 V. A solution containing B and BH 2 has a standard potential of –0.16 V. If a cell were constructed with these systems as half-cells, a. Would AH 2 be oxidized by B or BH 2 oxidized by A under standard conditions? b. What would be the reversible emf of the cell? c. What would be the effect of pH on the equilibrium ratio [B][AH 2 ]/[A][BH 2 ]?

Solution: Given: T = 25 °C, pH = 7, EA° = −0.60 V, EB° = −0.16 V Required: see above a. The reduction potential for each half reaction is: A + 2H + + 2e − → AH 2

E ° = −0.60 V

B + 2H + + 2e − → BH 2

E ° = −0.16 V

We reverse the first reaction since we require a positive potential for the overall reaction to be spontaneous in the forward direction. Therefore AH 2 is oxidized. AH 2 + B → A + BH 2 b. The reversible emf would be = E EB° − EA° E =−0.16 V − ( −0.60 V ) E = 0.44 V c. The equilibrium ratio is given by K =

[ B][ AH 2 ] . There is no dependence of [H O+ ] in the equilibrium expression, and the hydrogen3 [ A ][ BH 2 ]

containing entities cancel in the numerator and denominator. As a result, there is no effect of pH on the equilibrium ratio. Back to Problem 8.2

Back to Top

8-17

Chapter 8: Electrochemical Cells

8.3.

Solutions

Calculate the standard electrode potential for the reaction Cr2+ + 2e– → Cr at 298 K. The necessary E° values are a. Cr3+ + 3e– → Cr E° = –0.74 V b. Cr3+ + e– → Cr2+

E° = –0.41 V

Solution: Given: see above Required: E° for Cr2+ To solve this problem, we follow Example 8.3. We first calculate the ∆G° values for these two reactions using Eq. 8.2 ∆G º = − zFE º 3+ Cr + 3e − → Cr ∆G1o =−3 × 96 485 J mol−1 × (−0.74V) = 2.22 V × 96 485 J mol−1 3+

−

Cr + e → Cr

2+

−1

∆G =−1× 96 485 J mol × (−0.41V) = 0.41 V × 96 485 J mol o 2

−1

(1) (2) The reaction Cr2+ + 2e– → Cr is obtained

by subtracting reaction (2) from reaction (1), and the ∆G° for this reaction is obtained from by subtracting ∆G2º from ∆G1º .

∆G° = ∆G1º − ∆G2º = ∆G° 2.22 V × 96 485 J mol –1 − 0.41 V × 96 485 J mol –1 = ∆G° 1.81 V × 96 485 J mol –1 = ∆G° 1.81 V × F We solve for E° by rearranging Eq. 8.2 and setting z = 2 . ∆G º Eº = − zF 1.81 V F Eº = −2 F E º = −0.905 V E º = −0.91 V Back to Problem 8.3

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8-18

Chapter 8: Electrochemical Cells

8.4.

Solutions

Write the individual electrode reactions and the overall cell reaction for the following cell: Pt, H 2 |H+(1 m) F2–, S2–, H+(aq)|Pt 2– 2– where F represents the fumarate ion and S the succinate ion. Write the expression for the emf of the cell.

Solution: Given: see above Required: E We first write down the half reactions that are occurring in this cell. LHS electrode H 2 → 2H + (1 m ) + 2e − RHS electrode 2e − + 2H + ( aq ) + F2− → S2− The overall reaction is, 2H + ( aq ) + F2− + H 2 → 2H + (1 m ) + S2− The expression for the emf of the cell is determined from Eq. 8.13, RT [Y] y [Z]z E = E° − ln zF [A]a [B]b

u

+ 2 2− RT S H E= E° − ln 2 F F2 − H + 2 aq

u

2 2− RT S (1 m ) E= E° − ln 2 F F2− [ C]2 + Where [H ]=C is the concentration of [H+] on the LHS. We drop the superscript u since we have numerical values for the concentration.

Back to Problem 8.4

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8-19

Chapter 8: Electrochemical Cells

8.5.

Solutions

Design electrochemical cells in which each of the following reactions occurs: a. Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq) b. Ag+(aq) + Cl–(aq) → AgCl(s) c. HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) In each case, write the representation of the cell and the reactions at the two electrodes.

Solution: Given: see above Required: design the electrochemical cells a. In writing the representation of the cell, the oxidation reaction always occurs at the anode, which is placed at the left-hand position of the cell. In this case, Fe2+ is losing electrons, and therefore the oxidation process is: Fe 2+ → Fe3+ + e −

The cathode reaction is written on the right-hand side of the cell and is: Ce 4+ + e − → Ce3+

This is where reduction occurs. The overall reaction is the sum of these two reactions. The cell representation is: Fe3+(aq) | Fe2+(aq)

Ce4+(aq)| Ce3+(aq)

b. Upon examining the standard reduction potentials in Table 8.1, we see that the following half reactions can be combined to give the cited reaction. + Ag (aq) + e − → Ag Eo = 0.7996 V (1)

AgCl(s) + e − → Ag + Cl− (aq)

Eo = 0.22233 V

(2)

Reversal of equation (2), and then addition to equation (1) gives the overall desired equation: Ag + (aq) + Cl – (aq) → AgCl(s)

8-20

Chapter 8: Electrochemical Cells

Solutions

Equation (1) is the reduction reaction and is placed on the right-hand side of the cell. The anode reaction is placed on the left-hand side of the cell representation. Ag|AgCl(s)|Cl-(aq)

Ag+(aq)|Ag(s)

The voltage of this cell is the right-hand reduction potential minus the left-hand reduction potential. E= ° E1° − E2° = E ° 0.7996 V − 0.22233 V E ° =0.577 27 V c. HgO undergoes reduction to Hg and is the cathode. H 2 is oxidized and is the anode. The electrode potentials are obtained from Table 8.1 and the SRP Table. −0.8277 V Eo = 2H 2 O + 2e − → H 2 + 2OH − (3)

HgO + H 2 O + 2e − → Hg + 2OH −

Eo = 0.0977 V

(4)

Reversing the sense of equation (3) and adding to (4) gives, HgO(s) + H 2 (g) → Hg(l) + H 2 O(l) The cell is represented by Pt, H 2 (g)|H 2 O(l),OH-(aq)

HgO (s),H 2 O|OH-(aq) |Hg(l)

The cell potential is E= ° E4° − E3°

E ° 0.0977 V − ( −0.8277 V ) = E ° =0.9254 V

Back to Problem 8.5

Back to Top

8-21

Chapter 8: Electrochemical Cells

8.6.

Solutions

Calculate the equilibrium constant at 25 °C for the reaction 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I 2 (s) using the standard electrode potentials given in Table 8.1.

Solution: Given: Table 8.1 Required: K We can calculate the equilibrium constant from Eq. 8.7 Eo / V =

0.0257 ln K o z

z(E o / V ) K o = exp 0.0257 The half cell reactions are

Fe3+ + e − → Fe 2+

Eo = 0.771 V

I 2 + 2e − → 2I

Eo = 0.5355 V

And the overall emf is = E ° 0.771 V − 0.5355 V E ° =0.2355 V

We solve for K with z = 2 ,

8-22

Chapter 8: Electrochemical Cells

Solutions

2(0.2355) K o = exp 0.0257 K o = exp (18.32684825 ) K o = 91 043 525.2 = K o 9.10 ×107

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8-23

Chapter 8: Electrochemical Cells

8.7.

Solutions

From data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction Sn + Fe2+ → Sn2+ + Fe

Solution: Given: Table 8.1 Required: K We follow the same procedure as we did in problem 8.6 to solve for the equilibrium constant. The half cell reactions are Fe 2+ + 2e − → Fe

Eo = −0.447 V

Sn 2+ + 2e − → Sn

Eo = −0.1375 V

The overall emf is E ° = −0.447 V − ( −0.1375 V ) E ° = −0.3095 V

Solving for K with z = 2 gives, 2(−0.3095 K o = exp 0.0257 = K o exp ( −24.08560311) = K o 3.46 541 679 ×10−11 = K o 3.47 ×10−11

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8-24

Chapter 8: Electrochemical Cells

8.8.

Solutions

The standard electrode potential at 25 °C for cytochrome c (Fe3+) + e– → cytochrome c (Fe2+) is 0.25 V. Calculate ∆G° for the process 1 H 2 (g) + cytochrome c (Fe3+) → H+ + cytochrome c (Fe2+) 2

Solution: Given: E° = 0.25 V Required: ∆G° We calculate ∆G° for this reaction using Eq. 8.2, ∆G º = − zFE º , where z = 1

∆G o =−1× 96 485 mol−1 × 0.25 J ∆G o = −24 121.25 J mol−1 ∆G o = −24 kJ mol−1

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8-25

Chapter 8: Electrochemical Cells

8.9.

Solutions

Using the values given in Table 8.1, calculate the standard Gibbs energy change ∆G° for the reaction H2 +

1 O2 → H2O 2

Solution: Given: Table 8.1 Required: ∆G° From Table 8.1, we write the following half reactions:

O 2 + 4H + + 4e – → 2H 2 O

E1° = 1.229 V

(1)

2H + + 2e – → H 2

E2° = 0

(2)

Subtracting (2) from ½ of (1) gives the desired equation, 1 H 2 + O2 H 2O 2 The overall emf is = E ° 1.229 V − 0 E ° =1.229 V

We calculate ∆G° for this reaction using Eq. 8.2, ∆G o = − zFE , where z=2 ∆G o =−2 × 96 485 mol−1 ×1.229 J ∆G o = −237 160.13 J mol−1 ∆G o = −237.2 kJ mol−1

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Chapter 8: Electrochemical Cells

Solutions

*8.10. From the data in Table 8.1, calculate the equilibrium constant at 25 °C for the reaction 2Cu+ → Cu2+ + Cu What will be produced if Cu 2 O is dissolved in dilute H 2 SO 4 ? Solution: Given: Table 8.1 Required: K We follow the same procedure as we did in problem 8.6 to solve for the equilibrium constant. From Table 8.1, we write the following half reactions:

Cu 2+ + 2e − → Cu

E1o = 0.3419 V (1)

Cu 2+ + e − → Cu +

E1o = 0.153 V (2)

To get the desired overall reaction, we subtract 2×(2) from (1): 2Cu + → Cu 2+ + Cu

E= ° E1° − E2° = E ° 0.3419 V − 0.153 V E ° =0.1889 V We solve for K from Eq. 8.7, with z = 2 , 2(0.1889 K o = exp 0.0257 K o = 2 422 690.131 = K o 2.42 ×106

If Cu 2 O is dissolved in dilute H 2 SO 4 , then half will form Cu2+ and half will form Cu. Back to Problem 8.10

Back to Top 8-27

Chapter 8: Electrochemical Cells

8.11.

Solutions

For the reaction 3H 2 (g, 1 atm) + Sb 2 O 3 (s, cubic) → 2Sb(s) + 3H 2 O(l), ∆G° = –83.7 kJ [Roberts and Fenwick, J. Amer. Chem. Soc., 50, 2146(1928)]. Calculate the potential developed by the cell Pt|H 2 (g, 1 atm)|H+|Sb 2 O 3 (s, cubic)|Sb(s) Which electrode will be positive?

Solution: Given: reaction above, ∆G° = –83.7 kJ Required: E°, positive electrode Note that the ∆G° given is for the reaction of 3 moles of H 2 to form 2 moles of Sb. The half cell reactions may be written as,

3H 2 → 6H + + 6e − Sb 2 O3 + 6H + + 6e − → 2Sb + 3H 2 O

We can solve for E° using Eq. 8.2,

∆G o = − zFE ∆G o ∆E o = − zF −83.7 ×103 J ∆E o = −6 × 96 485 o ∆E = 0.1 445 820 594 V ∆E o = 0.145 V

For this reaction to be spontaneous, the electron flow is from the hydrogen electrode (oxidation) to the antimony electrode (reduction).

Back to Problem 8.11

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Chapter 8: Electrochemical Cells

8.12.

Solutions

Calculate the emf for the following cell at 25 °C: Pt, H 2 (1 bar)|HCl(0.5 m)

HCl(1.0 m)|Pt, H 2 (1 bar)

Solution: Given: T = 25 °C , cell above Required: E We calculate the emf for the cell from Eq. 8.27 where z = 1 ,

E=

RT m2 ln F m1

1.0 m 0.5 m E = 0.017 813 882 5 V

E = 0.0257 ln

E = 0.018 V

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8-29

Chapter 8: Electrochemical Cells

Solutions

The pyruvate-lactate system has an E°′ value of –0.185 V at 25 °C and pH 7.0. What will be the potential of this system if the oxidation has gone to 90% completion?

8.13.

Solution: Given: E°′ = −0.185 V, pH = 7.0, 90% completion Required: E The process is, pyruvate − + 2H + + 2e − lactate − And, the Nernst equation, given by Eq. 8.13 RT = E E − zF o

...[ Y ] y [ Z]z n [ A ]a [ B]b ...

u

For this process, this becomes

RT [lactate − ] E= E° − ln 2 F [pyruvate − ] Solving for E we get, 0.0257 10 ln 2 90 E = −0.156 765 664 2 V E= −0.185 V −

E = −0.157 V

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8-30

Chapter 8: Electrochemical Cells

8.14.

Solutions

a. From the data in Table 8.1, calculate the standard electrode potential for the half-reaction Fe3+ + 3e– → Fe b. Calculate the emf at 25 °C of the cell Pt|Sn2+(0.1 m), Sn4+(0.01 m)

Fe3+(0.5 m)|Fe

Solution: Given: Table 8.1 Required: see above a. To calculate the standard potential we follow example 8.3. We first calculate the ∆G° values for these two reactions using Eq. 8.2 ∆G o = − zFE

Fe3+ + e − → Fe 2+

∆G1o =−1× 96 485 J mol−1 × (0.771 V)=-0.771 V × 96 485 J mol−1

(1)

Fe 2+ + 2e − → Fe 2+

∆G o2 =−2 × 96 485 J mol−1 × (−0.447 V)=0.894 V × 96 485 J mol−1

(2)

Fe3+ + 3e − → Fe

The half-reaction given above is the sum of (1) and (2). The ∆G° value for the given reaction is

∆G o = ∆G1o + ∆G2o G o -0.771 V × 96 485 J mol−1 + 0.894 V × 96 485 J mol−1 ∆ = G o 0.123 V × 96 485 J mol−1 ∆= G o 0.123 V × F ∆=

We solve for E° by rearranging Eq. 8.2 and setting z = 3 .

8-31

Chapter 8: Electrochemical Cells

Solutions

∆G o − zF 0.123 V F Eo = −3 F Eo =

E o = −0.041 V b. The half cell reactions are: Fe3+ + 3e − → Fe

−0.041 V Eo =

−0.151 V Eo = Sn 2+ → Sn 4+ + 2e − The overall reaction is obtained from the sum of 2×(3) and 3×(4)

(3) (4)

2Fe3+ + 3Sn 2+ → 2Fe + 3Sn 4+

The electrical potential would be,

E= ° E3° + E4° E ° = −0.041 V −0.151 V E ° = −0.192 V From the Nernst equation, Eq. 8.13 we can calculate the emf using z = 6. u

RT [Y] y [Z]z ln E= E ° − zF [A]a [B]b 3 0.0257 ( 0.01) E= −0.192 − ln ( 0.1)3 ( 0.5 )2 6 E = −0.168 349 742 4 V E = −0.17 V

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8-32

Chapter 8: Electrochemical Cells

8.15.

Solutions

The cell Pt|H 2 (1 bar), H+ KCl(saturated)|Hg 2 Cl 2 |Hg was used to measure the pH of a solution of 0.010 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential expected at 25 °C [K a = 1.81 × 10–5 for acetic acid].

Solution: Given: cacetic acid = 0.010 M , csodium acetate = 0.0358 M , T = 25 °C, K a = 1.81× 10 –5 Required: E The equilibrium constant, K a is given by, H + CH 3COO − Ka = [CH3COOH ]

Rearranging for [H+], we obtain

K [ CH 3COOH ] H + = a CH 3COO −

(1.81×10 ) ( 0.010 –5

H + =

( 0.0358 M )

M

)

H + 5.055 865 922 ×10 –6 = From Table 8.1, the cell reactions are:

H 2 → 2H + + 2e –

E ° =0

Hg 2 Cl2 + 2e – → 2Hg + 2Cl –

E ° =0.2412 V

However, since the cathode reaction (the reduction reaction), is contained in the standard calomel (Hg 2 Cl 2(s) ) electrode, it is separated from the oxidation of H 2 , and its concentrations are constant throughout the measurement. This is a pH meter, so the only concentration that is relevant is [H+].

8-33

Chapter 8: Electrochemical Cells

Solutions

2 RT ln H + 2F E ° = 0 + 0.2412 V 2 0.0257 0.2412 V − ln ( 5.055 865 922 ×10 –6 ) E= 2 E = 0.554 610 508 5 V

E= E ° −

E = 0.55 V

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8-34

Chapter 8: Electrochemical Cells

8.16.

Solutions

The voltage required to electrolyze certain solutions changes as the electrolysis proceeds because the concentrations in the solution are changing. In an experiment, 500 dm3 of a 0.0500 M solution of copper (II) bromide was electrolyzed until 2.872 g Cu was deposited. Calculate the theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end of the experiment.

Solution: Given: V = 500 dm3, [Cu2+] = 0.0500 M, [Bi] = 0.100 M, m cu = 2.872 g Required: E i , E f The reactions taking place during the electrolysis are:

Cu 2+ + 2e − → Cu

Eo = 0.34190 V

2Br − → Br2 + 2e −

Eo = -1.80730 V

The overall reaction is: Cu 2+ + 2Br – → Cu + Br2

The electrical potential would be the sum of the two potentials, = Eo

0.34190 V − 1.08730 V

E o = −0.74540 V If the reverse reaction were taking place in a galvanic cell, the initial cell voltage would be: 2 RT ln Cu 2+ Br − 2F o E = 0.74540 V 0.0257 2 = Ei 0.74540 V − ln ( 0.0500 )( 0.100 ) 2 Ei = 0.8 430 715 966 V

= E

Eo −

Ei = 0.84 307 V

8-35

Chapter 8: Electrochemical Cells

Solutions

Therefore a minimum voltage of 0.84 307 V would have to be applied at the beginning in order for the electrolysis reaction to occur. At the end of the electrolysis, the concentrations are: m 1 + Cu 2= 0.0500 M − Cu × M Cu V 2+ = Cu 0.0500 M −

2.872 g

( 63.456 g mol ) ( 500 dm ) −1

3

Cu 2+ = 0.0500 M − 9.051 941 503 ×10−5 M 2+ Cu = 0.049 909 480 6 M = Br − 0.100 M − 2 × 9.051 941 503 ×10−5 M Br − = 0.099 818 961 2 M Therefore the final voltage required would be 0.0257 2 ln ( 0.0 499 094 806 M )( 0.0 998 189 612 ) 2 E f = 0.8 431 414 503 V

= Ef

0.74540 V −

E f = 0.84 314 V The E i and E f are close because a small amount of Cu2+ is plated out. Back to Problem 8.16

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8-36

Chapter 8: Electrochemical Cells

8.17.

Solutions

Calculate the concentration of I3– in a standard solution of iodine in 0.5 M KI, making use of the following standard electrode potentials: I 2 + 2e– → 2I–

Eº = 0.5355 V

I3− + 2e– → 3I–

Eº = 0.5365 V

The molality of I– in the standard solution can be assumed to be 0.5 m. Solution: Given: cKI = 0.5 M Required: cI– 3

The overall reaction of interest is obtained by reversing the second equation and adding it to the first. I 2 + I – → I3−

In this case, z = 2, and the standard electrode potential is = E º 0.5355 V − 0.5365 V E º = −0.0010 V

We can solve for the concentration of I3– using Eq. 8.7, where the equilibrium constant for this process is given by,

8-37

Chapter 8: Electrochemical Cells

Solutions

I3− Kc = − I 0.0257 Eo / V = ln K o z z ( Eo / V ) o K = exp 0.0257

z ( Eo / V ) I3− = exp 0.0257 I − o z(E /V ) cI− = cI− exp 3 0.0257

cI− =

2 −0.0010 ( 0.5 mol dm ) exp (0.0257 ) −3

cI− = 0.4 625 649 996 mol dm −3 3

3

cI− = 0.4 626 mol dm −3 3

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8-38

Chapter 8: Electrochemical Cells

8.18.

Solutions

Calculate the Nernst potential at 25 °C arising from the equilibrium established in Problem 7.41.

Solution: Given: Problem 7.41, T = 25 °C Required: ∆Φ ( ∆Φ = ∆E o for concentration cells) The Nernst potential is given by Eq. 8.19 as, ∆Φ =

RT c1 ln zF c2

From problem 7.41, we have the equilibrium concentrations of sodium and chloride as, Palmitate side

Other side

Na + 0.18 Na + 0.12 M = = M Cl− 0.08 Cl− 0.12 M = = M To calculate the Nernst potential, we only consider the Na+ equilibrium, therefore, 0.18 M 0.12 M ∆Φ = 0.010 420 453 3 V ∆Φ = 0.0257 ln

∆Φ = 10 mV

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8-39

Chapter 8: Electrochemical Cells

8.19.

Solutions

It might seem plausible to separate lead and gold by making use of the great difference between their standard electrode potentials (Table 8.1). In order to test this idea, one might electrolyze a solution containing 0.0100 M AuNO 3 and 0.0100 M Pb(NO 3 ) 2 in a well-stirred tank using platinum electrodes at low current density. As the potential difference is slowly increased from zero, which metal will be deposited first? What will be the concentration of this metal ion in solution when the second metal begins to be deposited? Do you think this is an acceptable method of separating the two metals?

Solution: Given:= Table 8.1, cAuNO3 0.0100 = M , cPb( NO3 ) 0.0100 M 2

Required: see above From Table 8.1 the two standard reduction potentials are given by,

Au + + e – → Au

E ° =1.692 V

Pb 2+ + 2e – → Pb

E ° = −0.1262 V

We can see that Au+ has a much higher reduction potential than Pb2+, therefore gold will be deposited first. As the Au+ concentration falls, the lead begins to be deposited. Therefore we have, 2Au(s) + Pb 2+ → 2Au + + Pb(s) With z = 2. The potential for this reaction is given by, E ° = −1.692 V − 0.1262 V E ° = −1.8182 V Following the procedure used in problem 8.17, we can calculate the concentration of Au+.

8-40

Chapter 8: Electrochemical Cells

Solutions

2

Au + Kc = Pb 2+

0.0257 ln K o z z ( Eo / V ) o K = exp 0.0257 Eo / V =

z ( Eo / V ) Au + = exp 0.0257 Pb 2+ 2

cAu + = cAu + = c= Au +

2 ( −1.8182 ) 0.0100 M exp 0.0257 1.88 335 766 ×10−32 M 1.88 ×10−32 M

The conclusion is that only an infinitesimal amount of gold will be left in the solution by the time the lead starts to deposit at the electrode. This is shown by the negligible concentration of gold. Therefore, this is an acceptable way to separate the two metals.

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8-41

Chapter 8: Electrochemical Cells

8.20.

Solutions

Calculate the emf of the cell Pt, H 2 (1 bar)|HCl(0.1 m)

HCl(0.2 m)|Pt, H 2 (10 bar)

Solution: Given: see above Required: E At the right-hand electrode we have the reaction: H + ( 0.2 m ) + e – →

1 H 2 (10 bar ) 2

And at the left-hand electrode we have, 1 H 2 (10 bar ) → H + ( 0.1 m ) + e – 2

The overall reaction, with z = 1, is: 1 1 H 2 (1 bar ) → H 2 (10 bar ) + H + ( 0.1 m ) 2 2

H + ( 0.2 m ) +

To calculate the cell emf, we use Eq. 8.7, = E°

0.0257 ln K °= , where z 1 z 1

E = 0.0257 ln

0.2 × (1 bar ) 2 1

0.1× (10 bar ) 2

E = −0.011 774 335 9 V E = −11.8 mV

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8-42

Chapter 8: Electrochemical Cells

Solutions

*8.21. Suppose that the cell in Problem 8.20 is set up but that the two solutions are separated by a membrane that is permeable to H+ ions but impermeable to Cl– ions. What will be the emf of the cell at 25 °C? Solution: Given: Problem 8.20, a membrane that is only permeable to H+ Required: E From problem 8.20, we have the half reactions at each electrode as: LHS : RHS :

1 H 2 (10 bar ) → H + ( 0.1 m ) + e – 2 1 H + ( 0.2 m ) + e – → H 2 (10 bar ) 2

Every H+ ion produced in the LHS solution will have to pass through the membrane to preserve the electrical neutrality (ion gradient).

H + ( 0.1 m ) → H + ( 0.2 m ) The net reaction is therefore, 1 1 H 2 (1 bar ) → H 2 (10 bar ) 2 2

The cell emf is calculated from Eq. 8.27, = E

RT m2 = ln , where z 1 F m1

We take the ratio of pressures as a measure of the ratio of molalities.

8-43

Chapter 8: Electrochemical Cells

Solutions

1

E = 0.0257 ln

(1 bar ) 2 1

(10 bar ) 2

E = −0.029 588 218 4 V E = −29.6 mV

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8-44

Chapter 8: Electrochemical Cells

Solutions

*8.22. A typical biological cell has a volume of 10–9 cm3, a surface area of 10–6 cm2, and a membrane thickness of 10–6 cm; the dielectric constant of the membrane may be taken as 3. Suppose that the concentration of K+ ions inside the cell is 0.155 M and that the Nernst potential across the cell wall is 0.085 V. a. Calculate the net charge on either side of the wall, and b. Calculate the fraction of the K+ ions in the cell that are required to produce this charge. Solution: 3 2 Given: = V 10 –9 cm= , A 10 –6 cm = , l 10 –6 cm, = ∈ 3,= cK + 0.155 M = , ∆Φ 0.085 V

Required: see above a. The net charge on either side of the wall can be calculated using Q = CV . The capacitance is given by Eq. 8.20 as, C=

∈0∈ A l

We solve for the net charge in the following expression, Q=

∈0∈ A l

V

(8.854 ×10 Q=

−12

C2 N −1 m −2

(10

–8

m)

) (3) (10

–10

m2

) ( 0.085 V )

= Q 2.257 77 ×10−14 C2 N −1 m −1 V = Q 2.257 77 ×10−14 F V where 1 F = 1 C V −1 = Q 2.3 ×10−14 C b. The fraction of K+ ions required to produce this charge is given by

8-45

Chapter 8: Electrochemical Cells

Solutions

Q 2.25 777 ×10−14 C = e 1.602 ×10−19 C Q = 140 900 ions e The number of ions inside the cell is, 0.155 ×10−12 × 6.022 ×1023= 9.3341×1010 Therefore the fraction of ions at the surface is, 140 934.4569 = 1.509 888 012 ×10−6 9.3341×1010 1.51×10−6

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8-46

Chapter 8: Electrochemical Cells

Solutions

*8.23. Calculate the emf at 25 °C of the cell Pt, H 2 (1 bar)|H 2 SO 4 (0.001 m)|CrSO 4 (s)|Cr given the following standard electrode potential: CrSO 4 (s) + 2e– → Cr + SO 2– 4

Eº = –0.40 V

a. First make the calculation neglecting activity coefficient corrections. b. Then make the calculation using activity coefficients estimated on the basis of the Debye-Hückel limiting law. Solution: Given: see above Required: see above a. At the left-hand electrode we have, 1 H2 → H+ + e– 2 and the right-hand electrode, we have, CrSO 4 (s) + 2e – → Cr(s) + SO 42– The overall reaction is given by, CrSO 4 (s) + H 2 → 2H + + Cr(s) + SO 42– With E º = −0.40 V and z = 2 The emf, neglecting the activity coefficients, is given by Eq. 8.13

8-47

Chapter 8: Electrochemical Cells

Solutions

u

RT [Y] y [Z]z ln E= E ° − zF [A]a [B]b u 2 RT ln H + SO 24− E= E ° − 2F 0.0257 2 ln ( 0.002 ) ( 0.001) E= −0.40 V − 2 E = −0.151 519 916 5 V

)

(

(

)

E = −0.152 V

b. To calculate the activity coefficients, we first calculate the ionic strength of the solution from Eq. 7.103, 1 I = ∑ ci zi2 2 i 1 2 1 × 0.002 + 22 × 0.001) I= ( 2 I = 0.003 M Now we rearrange the Debye-Hückel limiting law given by Eq. 7.111, log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 0.003 γ ± = 0.879 290 334 3 Substituting back into the expression for the emf obtained above we can solve for the true emf of the solution.

8-48

Chapter 8: Electrochemical Cells

Solutions

)

( (

u 2 RT ln H + γ ± 2 SO 24− γ ± 2F u 2 RT E= E ° − ln H + SO 24− γ ± 3 2F 0.0257 2 3 −0.40 V − E= ln ( 0.002 ) ( 0.001)( 0.879 290 334 3) 2 E = −0.146 560 839 3 V

E=

E° −

)

(

)

E = −0.147 V

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8-49

Chapter 8: Electrochemical Cells

Solutions

*8.24. Write the individual electrode reactions and the overall reaction for Cu|CuCl 2 (aq)|AgCl(s)|Ag If the emf of the cell is 0.191 V when the concentration of CuCl 2 is 1.0 × 10–4 M and is –0.074 V when the concentration is 0.20 M, make an estimate of the mean activity coefficient in the latter solution. Solution: Given: E1 = 0.191 V, c1 = 1.0 × 10 –4 M , E2 = 0.074 V, c2 = 0.20 M Required: γ ± At the left-hand electrode we have, Cu → Cu 2+ + 2e –

and the right-hand electrode, we have, AgCl(s) + e – → Ag + Cl – The overall reaction is given by, 2AgCl(s) + Cu(s) → 2Ag(s) + 2Cl – + Cu 2+ , where z = 2 To a good approximation, it can be assumed that the activity coefficients at 10-4 M are unity. (The DHLL gives γ ± = 0.988 ) Thus the standard electrical potential is obtained by rearranging Eq. 8.13 u

RT [Y] y [Z]z E= E ° − ln zF [A]a [B]b 2 u RT E °= E + ln Cu 2+ Cl− 2F 2 0.0257 = E ° 0.191 V + ln (10−4 )( 2 × 10−4 ) 2 E ° = −0.146 244 738 8 V

(

)

(

) 8-50

Chapter 8: Electrochemical Cells

Solutions

Suppose that at 0.20 M the activity coefficients are γ + and γ − , then we can solve for the mean activity coefficient from Eq. 8.40. E= E= E=

(

)

u RT ln aCu 2+ aCl− 2 2F 2 RT E° − ln γ + Cu 2+ γ − 2 Cl− 2F

E° −

(

E° −

(

2 RT ln Cu 2+ Cl− γ ±3 2F

−0.074 V = −0.146 244 738 8 V −

)

u

) (

)

0.0257 0.0257 2 ln ( 0.20 )( 0.40 ) − ln γ ±3 2 2

0.0257 ln γ ±3 2 3 ln γ ± = −2.180 139 285 0.028 014 789 9 = −

γ ± = 0.483 495 585 2 γ ± = 0.48

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8-51

Chapter 8: Electrochemical Cells

Solutions

*8.25. a. Write both electrode reactions and the overall reaction for the cell Tl|TlCl(s)|CdCl 2 (0.01 m)|Cd b. Calculate E and E° for this cell at 25 °C from the following information: Tl+ + e– → Tl

Eº = –0.34 V

Cd + 2e → Cd

Eº = –0.40 V

2+

–

The solubility product for TlCl is 1.6 × 10–3 mol2 dm–6 at 25 °C. Solution: Given: cell above, K sp = 1.6 ×10 –3 mol2 dm –6 , T = 25 °C Required: see above a. The left-hand or anode reaction is: Tl(s) + Cl− ( 0.02 m ) → TlCl(s) + e – The right-hand, or cathode reaction is: Cd 2+ ( 0.01 m ) + 2e – → Cd(s) The overall reaction is:

Cd 2+ ( 0.01 m ) + 2Tl(s) + 2Cl− ( 0.02 m ) → Cd(s) + 2TlCl(s) where z = 2 b. To use the electrical potentials given above, we rewrite the overall equation as (Cl- is a spectator ion): Cd 2+ ( 0.01 m ) + 2Tl(s) → Cd(s) + 2Tl+ ( in 0.01 m CdCl2 ) where z = 2 The standard electrical potential is given by, E ° = −0.40 V − ( −0.34 V ) E ° = −0.06 V 8-52

Chapter 8: Electrochemical Cells

Solutions

To solve for the emf, we use Eq. 8.13, + 2 RT Ti E= E ° − ln 2 F Cd 2+ where K sp = Ti + Cl−

u

K sp Ti + = Cl− E=

K sp 2 0.0257 E° − ln Cl− 2 Cd 2+ 2

u

–3 2 0.0257 (1.6 ×10 ) −0.06 V − E= ln ( 0.02 )2 ( 0.01) 2 E = −0.054 265 210 7 V

u

E = −0.054 V

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8-53

Chapter 8: Electrochemical Cells

8.26.

Solutions

Problem 7.44 involved calculating the concentrations on each side of a membrane after a Donnan equilibrium had become established. Which side of the membrane is positively charged? Calculate the Nernst potential across the membrane if the temperature is 37 °C.

Solution: Given: Problem 7.44, T = 37 °C Required: ∆Φ (used for concentration cells) We follow the same procedure as problem 8.18 to solve for the Nernst potential. From problem 7.44, we have the equilibrium concentrations of potassium and chloride as, right-hand side

left-hand side

+ + = = M K 0.04 K 0.16 M − − = = M Cl 0.04 Cl 0.01 M

The diffusible K+ ions are at a higher potential on the right-hand side of the membrane; there is thus a tendency for few of them to cross to the left-hand side and create a positive potential there. The same conclusion can be made by considering the diffusible Cl- ions; they are at a higher potential on the left-hand side, and few tend to cross to the right-hand side and create a negative potential. The Nernst potential is given by Eq. 8.19 as, ∆Φ =

∆Φ =

RT c1 ln zF c2

8.3145 J K −1 mol−1 × 310.15 K 1× 96 485 C mol−1

ln

0.16 M 0.04 M

∆Φ = 0.037 051 310 9 J C−1 where 1 J = 1 C V ∆Φ = 37 mV

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8-54

Chapter 8: Electrochemical Cells

8.27.

Solutions

The oxidation of lactate to pyruvate by the oxidized form of cytochrome c—represented as cytochrome c (Fe3+)—is an important biological reaction. The following are the relevant E°′ values, relating to pH 7 and 25 °C: E o '/V

pyruvate − + 2H + + 2e − → lactate − 3+

-0.185

−

2+

cytochrome c (Fe ) + e → cytochrome c (Fe )

0.254

Calculate the equilibrium ratio 2

cytochrome c (Fe 2+ ) pyruvate − H + 2 − 3+ cytochrome c (Fe ) lactate

2

at pH 7 and 25 °C. Also calculate the ratio at pH 6. Solution: Given: pH = 7, T = 25 °C Required: equilibrium ratio at pH 7 and pH 6 We begin by first writing the overall reaction as:

(

lactate – + 2cytochrome c Fe3+

)

(

)

→ 2cytochrome c Fe 2+ + pyruvate – + 2H +

Where z = 2 and E º ′ 0.254 V + 0.185 V = E º ′ =0.439 V

If K ′ is the equilibrium constant given at pH 7, then

K′ =

[cytochrome c (Fe 2+ )]2 [pyruvate – ][H + ]2 [cytochrome c (Fe3+ )]2 [lactate – ]

And the equilibrium ratio at pH 7 is just K ′ (unitless).

8-55

Chapter 8: Electrochemical Cells

Solutions

From Eq. 8.6 we can obtain K ′ ,

RT ln K o zF 0.0257 E o' = ln K ' 2 Eo =

2 K ' = exp E o' 0.0257 2 K ' = exp (0.439) 0.0257 K ' 6.870 472 098 × 1014 = K ' 6.87 × 1014 = At pH 6, the equilibrium ratio is K ′′ ′[H + ]2 = K true K=

(

[cytochrome c (Fe 2+ )]2 [pyruvate – ] [cytochrome c (Fe3+ )]2 [lactate – ]

)

(

2

′ 10−7 M = K true K= K ′′ 10−6 M = K ′′ 6.870 472 098 ×10

14

(10 × (10

−7 −6

)

2

) M) M

2

2

= K ′′ 6.870 472 098 ×1012 = K ′′ 6.87 ×1012

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8-56

Chapter 8: Electrochemical Cells

8.28.

Solutions

Suppose that the cell Ag|AgCl(s)|HCl(0.10 m)

HCl(0.01 m)|AgCl(s)|Ag

is set up and that the membrane separating the two solutions is permeable only to H+ ions. What is the emf of the cell at 25 °C? Solution: Given: see above, a membrane that is only permeable to H+ Required: E We can solve this problem in a similar manner as problem 8.21. The half reactions for each electrode are: LHS

Ag + Cl – → AgCl(s) + e –

RHS

Ag(s) + Cl – ( 0.01 m ) → AgCl(s) + e −

The electrical neutrality is maintained by the passage of H+ ions from right to left: H + ( 0.01 m ) → H + ( 0.10 m ) The net reaction is therefore,

H + ( 0.01 m ) + Cl− ( 0.01 m ) → H + ( 0.10 m ) + Cl− ( 0.10 m ) The cell emf is calculated from Eq. 8.13,

8-57

Chapter 8: Electrochemical Cells

Solutions

0.0257 ln K ° , where z = 1 and E °= 0 z + − H prod Clprod

E= E° − K° =

− H +react Clreact

E = −0.0257 ln

( 0.10 m ) ( 0.10 m ) ( 0.01 m ) ( 0.01 m )

E = −0.118 352 873 8 V E = −0.12 V

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8-58

Chapter 8: Electrochemical Cells

8.29.

Solutions

a. Consider the cell Pt, H 2 (1 bar)|HCl(m 1 )

HCl(m 2 )|Pt, H 2 (1 bar)

in which the solutions are separated by a partition that is permeable to both H+ and Cl–. The ratio of the speeds with which these ions pass through the membrane is the ratio of their transport numbers t + and t – . Derive an expression for the emf of this cell. b. If when m 1 = 0.01 m and m 2 = 0.01 m the emf is 0.0190 V, what are the transport numbers of the H+ and Cl– ions? Solution: Given: see above Required: see above a. The half reactions for each electrode are: 1 LHS : H 2 → H + ( m1 ) + e – 2 1 RHS : H + ( m2 ) + e – → H 2 2 To maintain electrical neutrality of the solutions, for every mole of H+ produced in the LHS solution, t + mol of H+ will cross the membrane from the left to the right, and t - mol of Cl- ions will pass fromright to left. In the LHS solution, there is therefore a net gain of,

t− mol of H + and of t− mol of Cl− . (1 − t+ ) mol = In the RHS solution, the net loss is t− mol of H + and of t− mol of Cl− (1 − t+ ) mol = The overall process is thus,

t− H + ( m2 ) + t− Cl− ( m2 ) → t− H + ( m1 ) + t− Cl− ( m1 ) The emf is given by Eq. 8.13,

8-59

Chapter 8: Electrochemical Cells

Solutions

0.0257 ln K ° , where z = 1 and E °= 0 z m1t− m1t− E = −0.0257 ln t− t− m2 m2

E= E° −

m2t− E = 0.0257 ln t− m1 = E 0.0257 × 2t− ln

2

m2 m1

b. To calculate the transport numbers we use the expression derived above= with m1 0.01 = m, m2 0.10 m m E 0.0257 × 2t− ln 2 = m1

0.0190 V 0.10 m ÷ ln 0.0257 × 2 0.01 m t− = 0.160 536 870 7

t− =

t− = 0.161 t+ = 1 − 0.160 536 870 7 t+ = 0.839 463129 3 t+ = 0.839

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8-60

Chapter 8: Electrochemical Cells

8.30.

Solutions

The metal M forms a soluble nitrate and a very slightly soluble chloride. The cell M|M+(0.1 m), HNO 3 (0.2 m)|H 2 (1 bar), Pt has a measured E = –0.40 V at 298.15 K. When sufficient solid KCl is added to make the solution of the cell 0.20 m in K+, the emf changes to –0.15 V at 298.15 K as MCl precipitates. Calculate the K sp of MCl, taking all activity coefficients to be unity.

Solution: Given: E1 = −0.40 V, T = −0.15 V, γ ± = 298.15 K, cK + = 0.20 m, E2 = 1 Required: K sp The half reactions for each electrode are: LHS :

M → M + ( 0.1 m ) + e –

RHS :

H + ( 0.2 m ) + e – →

1 H 2 (1 bar ) 2

The overall reaction is given by, M + H + ( 0.2 m ) → M + ( 0.1 m ) +

1 H 2 (1 bar ) 2

with z = 1 and E o EHo + |H − EMo + |M = 2

. E= o

0 − EMo + |M

E o = − EMo + |M

Using Eq. 8.13,

8-61

Chapter 8: Electrochemical Cells

Solutions

u

RT [Y] y [Z]z E= E° − ln , where z = 1 zF [A]a [B]b + RT M γ M + E °= E + ln zF H + γ H+ −0.40 V + 0.0257 ln EM° + |M =

0.1 m 0.2 m

EM° + |M = −0.417 813 882 5 V Upon addition of KCl, almost all of the M+ precipitates, and 0.10 m Cl- is in excess. The value of M+ in solution is found from the K sp , K sp = M + Cl− K sp M + = Cl− Using the Nernst equation, we can solve for the solubility product, E Eo − =

K RT ln + sp − zF H Cl

K E − Eo = ln + sp − −0.0257 H Cl E − Eo K sp = H + Cl− exp −0.0257 −0.15 V-(-0.4 178 138 825 V) K sp = (0.20 m)(0.10 m) exp −0.0257 2 −7 = K sp 5.961 362 163 ×10 m 6.0 ×10−7 m 2 K= sp Back to Problem 8.30

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8-62

Chapter 8: Electrochemical Cells

8.31.

Solutions

The substance nicotinamide adenine dinucleotide (NAD+) plays an important role in biological systems; under the action of certain enzymes it can react with a reducing agent and release a proton to the solution to form its reduced form NADH. With pyruvate the reduced form NADH undergoes the reaction NADH + pyruvate– + H+ NAD+ + lactate– The appropriate E°′ values, relating to 25 °C and pH 7, are pyruvate– + 2H+ +2e– → lactate–

Eº′ = –0.19 V

NAD + H + 2e → NADH

Eº′ = –0.34 V

+

+

–

Use these values to calculate ∆G°′ for the reaction, and also the equilibrium ratio lactate − NAD − pyruvate − [ NADH ] H + (a) at pH 7, and (b) at pH 8. Solution: Given: see above Required: ∆G°′, equilibrium ratio a. The overall reaction that is obtained by reversing the second half reaction and summing the two together: pyruvate – + H + + NADH → lactate – + NAD + where z = 2 and,

E º ′ = −0.19 V − ( −0.34 V ) E º ′ =0.15 V To find ∆G°′ we use Eq. 8.2, ∆G º = − zFE º

8-63

Chapter 8: Electrochemical Cells

Solutions

∆G º′ =−2 × 96 485 C mol−1 × 0.15 V ∆G º′ = −28 945.5 C V mol−1 where 1 J = 1 C V ∆G º′ = −28 945.5 J mol−1 ∆G º′ = −29 kJ mol−1 To solve for the equilibrium ratio, we follow the thought process used in problem 8. 27. If K ′ is the equilibrium constant given at pH 7, then

K′ =

[lactate – ][NAD + ] [pyruvate – ][NADH][H + ]

And the equilibrium ratio at pH 7 is just K ′ (unitless) From Eq. 8.5, we can solve for K ′ . ∆G o ' = − RT ln K' ∆G o ' ) − RT −28945.5 J mol−1 K' = exp −1 −1 (−8.3145 J K mol )(298.15 K) K' = exp(

K' = 117 763.1471 dm3 mol−1 =' 1.2 ×105 K b. At pH 8, the equilibrium ratio is K ′′

8-64

Chapter 8: Electrochemical Cells

Solutions

lactate − NAD − ' H K true K= = pyruvate − [ NADH ] +

K true = K'(10−7 M )= K''(10−8 M )

(10 K''=117 763.1471× (10

−7 −8

M)

M)

K'' = 1177 631.471 '' 1.2 ×106 K=

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8-65

Chapter 8: Electrochemical Cells

8.32.

Solutions

a. Calculate the standard emf E° for the reaction fumarate2– + lactate– → succinate2– + pyruvate– on the basis of the following information: fumarate2– + 2H+ + 2e– → succinate2–

E°′ = 0.031 V

pyruvate + 2H + 2e → lactate

E°′ = –0.185 V

–

+

–

–

The E°′ values relate to pH 7. The temperature coefficient ∂E/∂T for this cell is 2.18 × 10–5 V K–1. b. Calculate ∆G°, ∆H°, and ∆S° at 25 °C. Solution: Given: pH 7, ∂E /∂T= 2.18 × 10 –5 V K –1 Required: see above a. To find the standard emf, we first determine that the overall reaction is: fumarate 2– + lactate – → succinate 2– + pyruvate – where z = 2 Therefore, = E º ′ 0.031 V − ( −0.185 V ) E º=′ E=º 0.216 V (Note that this is also E° , the hydrogen ions having cancelled out.) b. To calculate ∆G°, we use E° from above and substitute into Eq. 8.2

8-66

Chapter 8: Electrochemical Cells

Solutions

∆G º = − zFE º ∆G º =−2 × 96 485 C mol−1 × 0.216 V ∆G º = −41 681.52 C V mol−1 where 1 J = 1 C V ∆G º = −41.7 kJ mol−1 The entropy change is obtained from Eq 8.23, ∂E ∆S = zF ∂T P ∆S = 2 × 96 485 C mol−1 × 2.18 × 10 –5 V K –1 ∆S =4.206 746 C V mol−1 K –1 where 1 J = 1 C V ∆S =4.21 J mol−1 K –1 To calculate enthalpy, we use the relationship between Gibbs energy and entropy we learned earlier as, ∆H =∆G + T ∆ S

(

∆H =− 41 681.52 J mol−1 + 298.15 K

) ( 4.206 746 J mol

−1

K –1

)

∆H =− 40 427.278 68 J mol−1 ∆H =− 40.4 kJ mol−1

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8-67

Chapter 8: Electrochemical Cells

8.33.

Solutions

The Weston standard cell (see Figure 8.2b) is Cd amalgam|CdSO 4 ·

8 H 2 O(s)|Hg 2 SO 4 (s), Hg 3

(saturated solution) a. Write the cell reaction. b. At 25 °C, the emf is 1.018 32 V and ∂E °/∂T = –5.00 × 10–5 V K–1. Calculate ∆G°, ∆H°, and ∆S°. Solution: Given: T= 25 °C, E= ° 1.018 32 V, ∂E /∂T= 5.00 × 10 –5 V K –1 Required: see above a. The half reactions for each electrode are: LHS : Cd(Hg) → Cd 2+ + 2e –

RHS :

Hg 22+ + 2e – → 2Hg

The overall reaction is given by, Cd(Hg) + Hg 22+ → Cd 2+ + 2Hg with z = 2 8 Since the solution is saturated with Hg 2SO 4 H 2 O , the overall reaction can be written as, 3

8 8 Cd(Hg) + Hg 2SO 4 (s) + H 2 O(l) → CdSO 4 H 2 O(s) + 2Hg(l) 3 3

b. We can calculate ∆G°, ∆H°, and ∆S° in a similar manner shown in problem 8.32. From Eq. 8.2,

8-68

Chapter 8: Electrochemical Cells

Solutions

∆G º = − zFE º ∆G º =−2 × 96 845 C mol−1 × 1.018 32 V ∆G º = −196 505.210 4 C V mol−1 where 1 J = 1 C V ∆G º = −196.5 kJ mol−1 The entropy change is obtained from Eq 8.23, ∂E ∆S = zF ∂T P ∆S = 2 × 96 845 C mol−1 × 5.00 × 10 –5 V K –1 ∆S = 9.6485 C V mol−1 K –1 where 1 J = 1 C V ∆S = 9.65 J mol−1 K –1 ∆H =∆G + T ∆ S

(

∆H =− 196 505.2104 J mol−1 + 298.15 K

) ( 9.6485 J mol

−1

K –1

)

∆H =− 199 381.9107 J mol−1 ∆H =− 199 kJ mol−1

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8-69

Chapter 8: Electrochemical Cells

8.34.

Solutions

Salstrom and Hildebrand [J. Amer. Chem. Soc., 52, 4650(1930)] reported the following data for the cell Ag(s)|AgBr(s)|HBr(aq)|Br 2 (g, 1 atm)|Pt

t/ºC

442.3

456.0

490.9

521.4

538.3

556.2

E/V

0.8031

0.7989

0.7887

0.7803

0.7751

0.7702

Find the temperature coefficient for this cell assuming a linear dependence of the cell potential with temperature. What is the entropy change for the cell reaction? Solution: Given: data above Required: ∆S

∂E To solve for ∆S , we need to find the temperature coefficient, from the data above. ∂T P Since we are only interested in the slope of the line best fit, we do not need to convert the temperature data to Kelvin. We can perform a linear regression analysis, using t as the independent variable and E as the dependent variable. The result is: = E 0.930 463 55 − 2.883 37 ×10−4t Differentiation with respect to t gives, ∂E −2.883 37 ×10−4 V °C−1 = ∂T P ∂E −2.883 37 ×10−4 V K −1 = ∂T P Substituting this value into Eq. 8.23 gives the entropy change,

8-70

Chapter 8: Electrochemical Cells

Solutions

∂E ∆S = zF ∂T P = ∆S 96 485 C mol−1 × −2.883 37 × 10−4 V K −1 ∆S =−27.820 195 45 C V K −1 mol−1 where 1 J = 1 C V ∆S =−27.82 J K −1 mol−1

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8-71

Chapter 8: Electrochemical Cells

8.35.

Solutions

The reaction taking place in the cell Mg(s)|Mg2+(aq), Cl–(aq)|Cl 2 (g,1 atm)|Pt is found to have an entropy change of –337.3 J K–1 mol–1 under standard conditions. What is the temperature coefficient for the cell?

Solution: Given: ∆S ° = −337.3J K –1 mol –1

∂E Required: ∂T P The overall reaction we are concerned with is, Mg(s) + Cl2 (g) → Mg 2+ (aq) + 2Cl− (aq), with z=2

Rearranging Eq. 8.23, we can solve for the temperature coefficient,

∂E ∆S = zF ∂T P ∆S ∂E = ∂T P zF −337.3J K –1 mol−1 ∂E = ∂T P 2 × 96 485 C mol−1 ∂E −1.747 940 1×10−3 J C−1 K −1 = ∂T P where 1 J = 1 C V ∂E −1.748 ×10−3 V K −1 = ∂ T P

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8-72

Chapter 8: Electrochemical Cells

Solutions

*8.36. a. Estimate the Gibbs energy of formation of the fumarate ion, using data in Problem 8.32 and the following values. ∆ f Gº (succinate, aq) = –690.44 kJ mol–1 ∆ f Gº (acetaldehyde, aq) = 139.08 kJ mol–1 ∆ f Gº (ethanol, aq) = –181.75 kJ mol–1 acetaldehyde + 2H+ + 2e– → ethanol Eº′ = –0.197 V b. If the ∂E °/∂T value for the process fumarate2– + ethanol → succinate2– + acetaldehyde is 2.18 × 10–5 V K–1, estimate the enthalpy of formation of the fumarate ion from the following values. ∆ f Hº (succinate, aq) = –908.68 kJ mol–1 ∆ f Hº (acetaldehyde, aq) = –210.66 kJ mol–1 ∆ f Hº (ethanol, aq) = –287.02 kJ mol–1 Solution: ∂E Given: problem 8.32, ∆ f Gº, = 2.18 ×10−5 V K −1 , ∆ f Hº ∂T P Required: see above a. The two reactions of interest are, fumarate 2– + 2H + + 2e – → succinate 2– +

acetaldehyde + 2H + 2e

–

→ ethanol

E °′ = 0.031 V

(1)

E º ′ = −0.197 V

(2)

Subtracting (2) from (1) gives the desired reaction: fumarate 2– + ethanol → succinate 2– + acetaldehyde with z = 2 And a standard electrical potential of, 8-73

Chapter 8: Electrochemical Cells

Solutions

= E °′ 0.031 V + 0.197 V E °′ = 0.228 V

The Gibbs energy for the reaction is calculated from Eq. 8.2 ∆G º = − zFE º ∆G º =−2 × 96 845 C mol−1 × 0.228 V ∆G º = −43 997.16 C V mol−1 where 1 J = 1 C V ∆G º = −43 997.16 J mol−1 To find the Gibbs energy of formation of fumarate, we use Eq. 3.91 ∆G o = ∑ ∆ f G o (products) − ∑ ∆ f G o (reactants) o o o o ∆G o = ∆ f Gsuccinate + ∆ f Gacetaldehyde − (∆ f Gethanol + ∆ f Gfumarate ) o o o o ∆ f Gfumarate = ∆ f Gsuccinate + ∆ f Gacetaldehyde − ∆ f Gethanol − ∆G o o ∆ f Gfumarate = −690.44 kJ mol−1 + 139.08 kJ mol−1 − (−181.74 kJ mol−1 ) − (−43.99716 kJ mol−1 ) o ∆ f Gfumarate = −325.61284 kJ mol−1 o ∆ f Gfumarate = −326 kJ mol−1

b. To find the enthalpy of formation of fumarate, we first need to find the enthalpy of the reaction. From Eq. 8.23, we calculate the entropy, and then use it to find the enthalpy.

8-74

Chapter 8: Electrochemical Cells

Solutions

∂E ∆S = zF ∂T P

∆S = 2 × 96 485 C mol−1 × ( 2.18 ×10−5 V K −1 ) ∆S =−4.222 442 C V mol−1 K −1 where 1 J=1 C V ∆S =−4.222 442 J mol−1 K −1 ∆H =∆G + T ∆S ∆H = −43 997.16 + (298.15)(−4.222 442) ∆H = −45 256.08 J mol−1 ∆H = −45.25 608 kJ mol−1 To find the enthalpy of formation of fumarate, we use Eq. 2.53

∆H ° = ∑ ∆ f H ° ( products ) − ∑ ∆ f H ° ( reactants )

(

° ° ° ° ∆H ° = ∆ f H succinate + ∆ f H acetaldehyde − ∆ f H ethanol + ∆ f H fumarate

)

o o o o ∆ f H fumarate = ∆ f H succinate + ∆ f H acetaldehyde − ∆ f H ethanol − ∆H o o ∆ f H fumarate =−908.68 kJ mol−1 − 210.66 kJ mol−1 − (−287.02 kJ mol−1 ) − (−45.25 608 kJ mol−1 ) o ∆ f H fumarate = −787.06 392 kJ mol−1 o ∆ f H fumarate = −787 kJ mol−1

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8-75

Chapter 8: Electrochemical Cells

Solutions

*8.37. a. Calculate the emf at 298.15 K for the cell Tl|TlBr|HBr (unit activity)|H 2 (1 bar), Pt b. Calculate ∆H for the cell reaction in the following cell. Tl|Tl+ (unit activity), H+ (unit activity)|H 2 (1 bar), Pt For the half-cell Tl+ + e– → Tl E° = 0.34 V ∂E/∂T = –0.003 V/K and

K sp (TlBr) = 10–4 mol2 dm–6 Solution: Given: T = 298.15 K, E ° = 0.34 V, ∂E /∂T = −0.003 V K −1 , K sp ( TlBr ) = 10 –4 mol2 dm –6 Required: see above, a. The half reactions for each electrode are: LHS : Tl → Tl+ ( in HBr; a = 1) + e – H + ( a = 1) + e – →

RHS :

1 H 2 (1 bar ) 2

The overall reaction is given by, Tl + H + ( a = 1) → M + ( in HBr; a =+ 1)

1 H 2 (1 bar ) with z = 1 and 2

= E o EHo + |H − ETlo + |Tl 2

E = 0 − ETlo + |Tl o

− ETlo + |Tl = −0.34 V Eo = The emf is given by Eq. 8.13, 8-76

Chapter 8: Electrochemical Cells

Tl+ E = E ° − 0.0257 ln + H

Solutions

u

K sp = Tl+ Br − K sp Tl+ = Br − K sp = E E o − 0.0257 ln Br − H +

u

10−4 mol−2 dm −6 −0.34 − 0.0257 ln E= 1× 1 E = −0.103 294 252 V

u

E = −0.10 V = E 0.0257 × 2t− ln

m2 m1

b. We can calculate the enthalpy using Eq. 8.25 and the data given for the half cell reaction. ∂E ∆H = − zF E − T ∂T

(

(

∆H =−1× 96 485 C mol−1 −0.34 V − ( 298.25 K ) −0.003 V K −1

))

∆H = 53 496.10825 C V mol−1 where 1J = 1 C V ∆H = 53 496.10825 J mol−1 ∆H = 535 kJ mol−1

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Chapter 8: Electrochemical Cells

8.38.

Solutions

Calculate the solubility product and the solubility of AgBr at 25 °C on the basis of the following standard electrode potentials: AgBr(s) + e– → Ag + Br–

E° = 0.0713 V

Ag + e → Ag

E° = 0.7996 V

+

–

Solution: Given: T = 25 °C Required: K sp , s The desired reaction is obtained by subtracting the second reaction from the first. AgBr(s) → Ag + + Br – with z = 1 = E ° 0.0713 V − 0.7996 V E ° = −0.7283 V

We can calculate the solubility product from Eq. 8.13,

(

E o = 0.0257 ln Ag + Br −

)

u

K sp = Ag + Br − E o = 0.0257 ln K sp Eo K sp = exp 0.0257 −0.7283 K sp = exp 0.0257 = K sp 4.92 874 705 ×10−13 mol2 kg −2 = K sp 4.929 ×10−13 mol2 kg −2

The solubility is therefore,

8-78

Chapter 8: Electrochemical Cells

Solutions

= K sp = Ag + Br − s 2 s = K sp = s

4.928 747 05 ×10−13 mol2 kg −2

= s 7.020 503 58 ×10−7 mol kg −1 = s 7.021×10−7 mol kg −1

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8-79

Chapter 8: Electrochemical Cells

8.39.

Solutions

The emf of a cell Pt, H 2 (1 bar)|HCl(aq)|AgCl(s)|Ag was found to be 0.517 V at 25 °C. Calculate the pH of the HCl solution.

Solution: Given: E = 0.517 V, T = 25 °C Required: pH From Table 8.1, the standard emf of the AgCl|Ag electrode is 0.222 33 V and the cell reaction is: 1 AgCl(s) + H 2 → H + + Cl – + Ag(s) where z = 1, 2 To calculate the pH, we first need to find the concentration of H+ ions. Using Eq. 8.13,

(

)

u RT ln H + Cl− F u E − Eo ln H + Cl− = −0.0257 u E − Eo H + Cl− = exp −0.0257

= E Eo −

(

)

(

)

( H

+

Cl−

)

( H

+

Cl−

)

u

u

0.22 233 V − 0.517 V = exp −0.0257 = 95 392.83 548 mol2 dm −6

Since the concentrations of H+ and Cl- are the same,

8-80

Chapter 8: Electrochemical Cells

Solutions

H + = Cl− H + = 95 392.835 48 mol2 dm −6 H + = 308.857 306 mol dm −3

The pH is given by the logarithm of the hydrogen ion concentration pH = log H +

pH = log ( 308.857 306 mol dm −3 ) pH = 2.489 757 879 pH = 2.48

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8-81

Chapter 8: Electrochemical Cells

8.40.

Solutions

The emf of the cell Ag|AgI(s)I–(aq)

Ag+(aq)|Ag

is –0.9509 V at 25 °C. Calculate the solubility and the solubility product of AgI at that temperature. Solution: Given: E = 25 °C −0.9509 V, T = Required: s, K sp To solve this problem, we follow the example set in problem 8.38 The half reactions for each electrode are: Ag(s) → Ag + + e – AgI(s) + e – → Ag(s) + I − The reaction of interest is given by, AgI(s) → Ag + + I – with z = 1 We can calculate the solubility product from Eq. 8.13,

8-82

Chapter 8: Electrochemical Cells

(

E o = 0.0257 ln Ag + I −

)

Solutions

u

K sp = Ag + I − E o = 0.0257 ln K sp Eo K sp = exp 0.0257 −0.9509 K sp = exp 0.0257 = K sp 8.53 304 763 ×10−17 mol2 kg −2 = K sp 8.533 ×10−17 mol2 kg −2

The solubility is therefore, K sp = Ag + I − s 2 = s = K sp

= s

8.533 047 63 ×10−17 mol2 kg −2

= s 9.237 449 662 ×10−9 mol kg −1 = s 9.237 ×10−9 mol kg −1

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8-83

Chapter 8: Electrochemical Cells

8.41.

Solutions

An electrochemical cell M(s)|MCl(aq, 1.0 m)|AgCl(s)|Ag(s), where MCl is the chloride salt of the metal electrode M, yields a cell potential of 0.2053 V at 25 °C. What is the mean activity coefficient γ ± of the electrolyte MCl? E° for the M(s)|M+ electrode is 0.0254 V.

Solution: Given: E= 0.2053 V, T= 25 °C, E= ° 0.0254 V Required: γ ± The half reactions for each electrode are:

M → M+ + e– AgCl(s) + e – → Ag(s) + Cl− And the overall reaction is given by, AgCl(s) + M → Ag(s) + Cl− + M + = E ° 0.222 33 V − 0.0254 V E ° =0.196 93 V To solve for the activity coefficient, we use Eq. 8.43 E+

2 RT 2 RT ln mu = E ° − ln γ ± F F

Since the molality of MCl is exactly 1, this expression simplifies to, 2 RT 2 RT ln1 = E ° − ln γ ± F F 2 RT E = E° − ln γ ± F E+

Rearranging and solving for the mean activity coefficient, we obtain,

8-84

Chapter 8: Electrochemical Cells

Solutions

E = E ° − 2 × 0.0257 ln γ ±

γ± = e

E − E° − 2×0.0257 0.2053 − 0.196 93 − 2×0.0257

γ± = e γ ± = 0.849 726 737 1 γ ± = 0.850 Back to Problem 8.41

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8-85

Chapter 8: Electrochemical Cells

8.42.

Solutions

The following thermodynamic data apply to the complete oxidation of butane at 25 °C. C 4 H 10 (g) + (13/2)O 2 (g) → 4CO 2 (g) + 5H 2 O(l) ∆H ° = –2877 kJ mol –1 ∆S ° = –432.7 J K –1 mol –1 Suppose that a completely efficient fuel cell could be set up utilizing this reaction. Calculate (a) the maximum electrical work and (b) the maximum total work that could be obtained at 25 °C.

Solution: Given: ∆H ° = −2877 kJ mol –1 , ∆S ° = −432.7 J K –1 mol –1 , T = 25 °C Required: see above a. The maximum electrical work for the fuel cell is −∆G° . ∆G° =∆H ° − T ∆ S °

(

∆G° = − 2877 ×103 J mol –1 − 298.15 K

) ( −432.7 J K

–1

mol –1

)

∆G° = −2 747 990.495 J mol –1 ∆G° = −2748 kJ mol –1 electrical work = −∆G° electrical work = 2748 kJ mol –1 b. The maximum total work that can be obtained is −∆A

8-86

Chapter 8: Electrochemical Cells

Solutions

∆G o = ∆Ao − ∑ vRT

∆Ao = ∆G o + ∑ vRT

∑v = ∑v =

4 −1−

13 2

−3.5

∆Ao =−2747 990.495 J mol−1 + ( −3.5 ) ( 8.3145 J K −1 mol−1 ) ( 298.15 K ) ∆Ao = −2756 666.884 J mol−1 ∆Ao = −2758 kJ mol−1 total work = ∆Ao total work = 2758 kJ mol−1

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8-87

Chapter 8: Electrochemical Cells

Solutions

*8.43. At 298 K the emf of the cell Cd, Hg|CdCl 2 (aq, 0.01 m), AgCl(s)|Ag is 0.7585 V. The standard emf of the cell is 0.5732 V. a. Calculate the mean activity coefficient for the Cd2+ and Cl– ions. b. Compare the value with that estimated from the Debye-Hückel limiting law, and comment on any difference. Solution: Given: E 0.7585 = = V, E ° 0.5732 V Required: see above a. To calculate the mean activity coefficients, we follow a similar process as used in problem 8.24. The half reactions at each electrode are,

Cd(s) → Cd 2+ + 2e – AgCl(s) + e – → Ag + Cl – The overall reaction is given by, 2AgCl(s) + Cd(s) → 2Ag(s) + 2Cl− + Cd 2+ , where z = 2 Suppose that at 0. 01 m the activity coefficients are γ + and γ − , then we can solve for the mean activity coefficient from Eq. 8.40.

8-88

Chapter 8: Electrochemical Cells

(

Solutions

)

u RT ln aCd2+ aCl− 2 2F 2 u RT E Eo − = ln γ + Cd 2+ γ − 2 Cl− 2F u 2 RT E Eo − = ln Cd 2+ Cl− γ ±3 2F 0.0257 2 = 0.7585 V 0.5732 V − ln ( 0.01)( 0.02 ) γ ±3 2 −0.01 285ln ( 4 × 10−6 γ ±3 ) 0.1853 =

E Eo − =

)

( (

)

(

)

−14.42 023 346 ln ( 4 × 10−6 γ ±3 ) =

( 4 ×10 ) γ −6

3 ±

5.462 257 621×10−7 =

γ ± = 0.5 149 567 193 γ ± = 0.51 b. To calculate the activity coefficient according to the DHLL, we first calculate the ionic strength of the solution from Eq. 7.103, 1 I = ∑ ci zi2 2 i 1 2 1 × 0.02 + 22 × 0.01) I= ( 2 I = 0.03 m Now we rearrange the Debye-Hückel limiting law given by Eq. 7.111, log10 γ ± = − 0.51z+ z−

γ ± =10−0.51z

+

z−

I /mol dm −3

I /moldm −3

γ ± =10−0.51( 2×1) 0.03 γ ± = 0.665 780 111 6 γ ± = 0.67

8-89

Chapter 8: Electrochemical Cells

Solutions

There is a considerable difference between the two methods.

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8-90

Chapter 8: Electrochemical Cells

Solutions

*8.44. The following emf values were obtained by H. S. Harned and Copson [J. Amer. Chem. Soc., 55, 2206(1933)] at 25 °C for the cell Pt,H 2 (1 bar)|LiOH(0.01 m), LiCl(m)|AgCl(s)|Ag at various molalities m of LiCl: m/mol kg–1 E/V

0.01

0.02

0.05

0.10

0.20

1.0498

1.0318

1.0076

1.9888

0.9696

Obtain from these data the ionic product of water. Solution: Given: emf values above Required: K w In this cell, we see that the LiOH is required for the hydrogen electrode, and the LiCl salt is used to complete the AgCl electrode. Both the Cl- ion and the H+ ion will behave according to their activities in solution. We begin solving the problem by first determining the emf of the cell: Ecell EAgCl − EH2 = ° Ecell = EAgCl −

RT RT ln aCl− − ln aH+ F F

Since K w = aH+ aOH− , we can substitute this into the above expression and develop a relationship involving K w . ° Ecell = EAgCl −

K RT RT ln aCl− − ln w F F aOH−

RT RT RT ln aCl− − ln K w + ln aOH− F F F RT aCl− RT ° EAgCl =− ln − ln K w F aOH− F

° Ecell = EAgCl −

Ecell

Rewriting this expression in terms of activity coefficients and molalities gives, 8-91

Chapter 8: Electrochemical Cells

Solutions

m −γ − RT RT ° Ecell = EAgCl − ln Cl Cl − ln K w F mOH− γ OH− F RT mCl− RT γ Cl− RT ° Ecell − EAgCl = − ln − ln − ln K w F mOH− F γ OH− F ° Ecell − EAgCl

0.0257

m − γ − = − ln Cl − ln Cl − ln K w mOH− γ OH−

° = 0.222 33 V . The molality of OH- is given as 0.01 m, and from Table 8.1, EAgCl

γ − m − Ecell − 0.222 33 + ln Cl = − ln Cl − ln K w γ OH− 0.0257 0.01 We plot the left-hand side of the equation against the ionic strength, which varies with concentration, and extrapolate to zero ionic strength. At zero ionic strength, the activity coefficients approach unity. Then the value of the curve is -lnK w . m − Ecell − 0.222 33 + ln Cl = − ln K w 0.0257 0.01 In the following data, I is based on m + 0.01 m OH − , where 0.01m is constant. m / mol kg −1

0.01

0.02

0.05

0.10

0.20

I / mol kg Ecell − 0.222 33 0.0257 m ln 0.01

0.02

0.03

0.06

0.11

0.21

32.3086

31.5079

30.566

29.834

29.087

0.000

0.693

1.609

2.303

2.996

Ecell − 0.222 33 m + ln 0.0257 0.01

32.209

32.301

32.175

32.137

32.083

−1

8-92

Chapter 8: Electrochemical Cells

Solutions

From the indicated plot shown, the value of -lnK w is 1.010×10-14.

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8-93

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