Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3

July 22, 2017 | Author: Kartik | Category: Electrochemistry, Anode, Cathode, Redox, Chemical Elements
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SOLUTIONS TARGET 5...

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Chapter

3

Electrochemistry Solutions SECTION - A Objective Type Questions (One option is correct) 1.

On passing 3 faradays of electricity through three electrolytic cells connected in series containing Ag+, Ca+2 and Al+3 ion respectively, the molar ratio in which three metal ions are liberated at the electrode is (1) 1 : 2 : 3

(2) 3 : 2 : 1

(3) 6 : 3 : 2

(4) 3 : 4 : 2

Sol. Answer (3) i × t is same for all the electrolytic solutions

⎛W⎞ it it ⎜M⎟ = = (Ag+ + e–  Ag) ⎝ ⎠ Ag nF F ⎛W⎞ it ⎜M⎟ = (Ca2+ + 2e–  Ca) ⎝ ⎠Ca 2F ⎛W⎞ it (Al3+ + 3e  Al) and ⎜ M ⎟ = 3F ⎝ ⎠ Al

 Molar ratio is 1 : 2.

1 1 : 3 2

or

6:3:2

The molar conductances at infinite dilution of BaCl 2, NaCl and NaOH are respectively 280×10 –4 , 126.5 × 10–4, 248 × 10–4 S m2 mol–1. The molar conductance at infinite dilution for Ba(OH)2 is (1) 523 × 10–4 S m2 mol–1

(2) 52.3 × 10–4 S m2 mol–1

(3) 5.23 × 10–4 S m2 mol–1

(4) 65 × 10–4 S m2 mol–1

Sol. Answer (1)    BaCl = Ba2  2Cl– 2

...(i)

   = Na  Cl– NaCl

...(ii)

   = Na  HO– NaOH

...(iii)

for Ba(OH)2 (i) + 2(iii) –2(ii)  (Ba(OH)2 ) = (280 × 10–4) + 2(248 × 10–4) – 2(126.5 × 10–4) = 523×10–4 Sm2 mol–1. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

72 3.

Electrochemistry

Solution of Assignment (Set-2)

During electrolysis of aqueous solution of a salt, pH in the space near one of the electrode is increased. Which of the following salt solution was electrolysed? (1) KCl

(2) CuCl2

(3) Cu(NO3)2

(4) CuSO4

Sol. Answer (1) In KCl solution the reaction at the electrodes are 2H+ + 2e–  H2 2Cl–  Cl2 + 2e–  [H+] decreases in the solution because of which [OH] increases hence increasing the pH. 4.

By how much will the potential of half cell Cu+2/Cu change, if the solution is diluted to 100 times at 298 K? (1) Increases by 59 mV

(2) Decreases by 59 mV

(3) Increases by 29.5 mV

(4) Decreases by 29.5 mV

Sol. Answer (2) For Cu2+ + 2e–  Cu(s) E

Cu2  /Cu

When Cu2+

E'

2

Cu

/Cu

1 0.0591 log 2 [Cu ] 2 solution is diluted to 100 times [Cu2+] decreases to 1/100

= Eº –

= Eº –

E’ = Eº –

100 0.0591 log [Cu2 ] 2

0.0591 [[log 100 – log [Cu2+]] 2

 E’ = Eº –

1 0.0591 0.0591 ×2– log [Cu2 ] 2 2

 E’ = Eº –

1 0.0591 log – 0.0591 [Cu2 ] 2

 E’ = E – 0.0591, Hence, Potential decreases by 59 mV. 5.

The Ecell of the reaction

MnO 4  Fe 2  H  Mn2  Fe 3  H2O is 0.59 V at 25°C. The equilibrium constant for the reaction is (1) 50

(2) 10

(3) 1050

(4) 105

Sol. Answer (3)

Eocell = 0.59 V MnO4– + Fe2+ + H+  Mn2+ + Fe3+ + H2O 0.0591 log Qc 5 At equilibrium, E = 0; Qc = Kc

E = Eocell –

 

Eocell =

0.0591 log Kc 5

5  0.59 = log Kc 0.59 50 = log Kc

 Kc = 1050. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

6.

Electrochemistry

73

A current of 2.0 A when passed for 5 hrs through a molten salt, deposits 22.2 g of metal (of atomic weight 177). The oxidation state of metal in metal salt is (1) +1

(2) +2

(3) +3

(4) +4

Sol. Answer (3) i = 2A, t = 5 hrs = 5 × 60 × 60 s wM = 22.2 g; A = 177 Applying the equation w=

Eit Ait w= F nF

 n=

177  2  5  60  60 A i t = = 2.97 22.2  96500 wF

 n=3 M3+ + 3e–  M  Oxidation state is +3. 7.

Some Indian scientists tried to use a metal x for electroplating iron pillar in Mehrauli but they ended up with Ecell of the reaction to be negative. They concluded that (1) Reaction is spontaneous

(2) Reaction is non-spontaneous

(3) Reaction is reversible

(4) Reaction is non-reversible

Sol. Answer (2) For electroplating Iron a metal ‘x’ is used. Ecell is negative, it means that no reaction takes place and the reaction is non-spontaneous. 8.

In the electrolysis of aqueous solution of NaOH, 2.8 litre of oxygen at NTP was liberated at the anode. How much hydrogen was liberated at cathode? (1) 5.6 litre

(2) 56 ml

(3) 560 ml

(4) 0.056 litre

Sol. Answer (1) NaOH is electrolysed. NaOH  Na+ + HO– H2O  H+ + HO– At cathode: 2H+ + 2e–H2 At anode: 4HO– 2H2O + O2 + 4e–

⎛w⎞ it ⎛ w ⎞ it ⎜M⎟ = ; ⎜ ⎟ = ⎝ ⎠H2 2F ⎝ M ⎠O2 4F 

nH2 : nO2 = VH2 2.8

=

2 1

it it : 2:1 2F 4F 

 Volume ratio VH2 : VO2 = 2 : 1

VH2 = 2.8 × 2 = 5.6 L

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74 9.

Electrochemistry

Solution of Assignment (Set-2)

The equilibrium constant for the reaction Sr (s) + Mg+2 (aq) The E° for a cell made up of Sr/Sr+2 and Mg+2/Mg half cells is (1) 0.3667 V

(2) 0.7346 V

Sr+2 (aq) + Mg (s) is 2.69 × 1012 at 25°C.

(3) 0.1836 V

(4) 3.667 V

Sol. Answer (1) The reaction is Sr(s) + Mg2+(aq)  Mg(s) + Sr2+(aq) Kc = 2.69 × 1012 At equilibrium, E = 0; Q = Kc 0.0591 log Kc 2 0.0591 0.0591  Eº = log Kc = log (2.69 × 1012) = 0.3667 V 2 2

 0 = Eº –

10. Passage of one ampere current through 0.1 M Ni(NO3)2 solution using Ni electrodes bring in the concentration of solution to _________ in 60 seconds. (1) 0.1 M

(2) 0.05 M

(3) 0.2 M

(4) 0.025 M

Sol. Answer (1) The reaction taking place at the electrodes are Anode : Ni  Ni2+ + 2e Cathode : Ni2+ + 2e–  Ni Hence [Ni2+] does not change.  Concentration of Ni2+ is 0.1M. 11. Which species in each pair is a better oxidising agent under standard conditions? (1) Br2 & Au+3

(2) H2 & Ag+

(3) Cd+2 & Cr+3

(4) O2 in acidic medium & O2 in basic medium

Sol. Answer (1) Halogens act as an oxidising agent and in Au3+ the oxidation state of Au is maximum i.e., +3. So, Au3+ also acts as oxidizing agent. 12. At 25°C, the equivalent conductances at infinite dilution of HCl, CH3COONa and NaCl are 426.1, 91.0 and 126.45 cm2 –1eq–1 respectively.  for CH3COOH (in cm2 –1eq–1) is (1) 391.6

(2) 390.6

(3) 380.6

(4) 309.6

Sol. Answer (2) According to given condition: o o  = H + Cl– HCl

...(i)

 o   CH =  CH3COO– Na 3 COONa

...(ii)

o o  = Na + Cl– NaCl

...(iii)

Doing the operation Equation (i) + (ii) – (iii) o o    +  CH3COONa – NaCl =  CH3COO– + H HCl   (426.1 + 91.0 – 126.45) =  CH3COOH



  CH = 390.6 cm2 –1/eq 3 COOH

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Solution of Assignment (Set-2)

Electrochemistry

75

13. When electricity is passed through a solution of AlCl3 13.5 g of Al is deposited. The number of faradays must be (1) 1.0

(2) 1.5

(3) 0.5

(4) 2

Sol. Answer (2) The reaction at cathode is Al3+ + 3e–  Al (n = 3) Applying the equation w=

Eit w ⇒ F F E

13.5 13.5 ⎛  i × t = ⎜ 27 ⎞⎟ × F = F = 1.5 F 9 ⎝ 3 ⎠

14. 0.5 faraday of electricity was passed to deposit all the copper present in 500 ml of CuSO4 solution. What was the molarity of this solution? (1) 1 M

(2) 0.5 M

(3) 0.25 M

(4) 2.5 M

Sol. Answer (2) i × t = 0.5 F Applying the equation w=

⎛w⎞ E  it it 0.5F  ⎜ ⎟ = = = 0.25 moles F 2F 2F ⎝M⎠

V × molarity = No. of moles  500 × x × 10–3 = 0.25  x=

0.25  103 = 0.5 500

Molarity = 0.5 M 15. Cu+ is not stable and undergoes disproportionation E for Cu+ disproportionation ECu 2 / Cu  0.153 V, ECu / Cu  0.53 V

(1) +0.683 V

(2) –0.367 V

(3) 0.754 V

(4) +0.3415 V

Sol. Answer (3) Given EoCu2 /Cu = +0.153 V and EoCu /Cu = 0.53 V The reaction 2Cu+  Cu + Cu2+ 

o o E oCell = ECu /Cu2 + ECu /Cu

o we required ECu /Cu2

Cu2+ + 2e  Cu; 0.153 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

76

Electrochemistry

Solution of Assignment (Set-2)

Cu  Cu+ + e–; – 0.53  Cu2+ + e–  Cu+  –1 × F × Eº = [–2 × F × (0.153)] + [F × 0.53]  –FEº = –2F (0.153) + 0.53 F  Eº = –0.53 + (2 × 0.153) 

EoCu2 /Cu = – 0.224

or

EoCu /Cu2 = 0.224

 Eº = 0.224 + 0.53 = 0.754 V 16. 25 g of a metal is deposited on cathode during the electrolysis of metal nitrate solution by a current of 5 A passing for 4 hours. If atomic weight of the metal is 100. The valency of metal in metal nitrate is (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (3) wM = 25 g, i = 5 A, t = 5 hrs = 4 × 60 × 60 second A = 100 (metal nitrate was electrolysed) Applying the equation, w=

E  it F

 25 =

 n=

100 5  4  60  60  n 96500

100  5  4  60  60 =3 (96500  25)

Hence, valency = 3 17. A well stirred solution of 0.1 M CuSO4 is electrolysed at 25°C using platinum electrodes with a current of 25 mA for 6 hours. If current efficiency is 50%. At the end of the duration what would be the concentration of copper ions in the solution? (1) 0.0856 M

(2) 0.092 M

(3) 0.0986 M

(4) 0.1 M

Sol. Answer (3) 18. 50 ml of a buffer of 1 M NH 3 and 1 M NH4+ are placed in two volatic cells separately. A current of 3.0 amp is passed through both cells for 10 min. If electrolysis of water takes place as 2H2O + O2 + 4e–  4OH– (R.H.S.) 2H2O  4H+ + O2 + 4e– (L.H.S.) then pH of the (1) L.H.S. will increase

(2) R.H.S. will increase

(3) R.H.S. will decrease

(4) Both side will increase

Sol. Answer (2) Because of the reactions of electrolysis, [H+] concentration will decrease as a result of which pH will increase. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electrochemistry

77

19. 1 M aqueous solution of NaCl undergo electrolysis if 50 mA current is passed for 12 hours. Assume current efficiency is 25%. The total volume of gas produced at standard state is (1) 137 ml

(2) 68.5 ml

(3) 125.44 ml

(4) 62.72 ml

Sol. Answer (3) NaCl aq. solution undergoes electrolysis NaCl  Na+ + Cl– H2O  H+ + HO– Reactions : At cathode : 2H+ + 2e  H2(g) At anode : 2Cl–  Cl2 + 2e

⎛w⎞ it it it ⎜M⎟ = + = ⎝ ⎠T 2F 2F F 3  50  10

25 12  60  60  = nT 100 96500

 nT = 0.005595  V = 0.005595 × 22400   125.44 ml. 20. Vanadium electrode is oxidised electrically. If the mass of electrode decreases by 100 mg during the passage of 570 coulomb, the oxidation state of vanadium in the product is (At. wt. of V = 51) (1) 6

(2) 5

(3) 4

(4) 3

Sol. Answer (4) i × t = 570 C w = 100 mg = 100 × 10–3 g w=

E  it F



570 100  10 3 = 96500  n 51

 n=

570  51 100  10 –3  96500

=3

21. The specific conductance of a saturated solution of AgCl is K–1 cm–1. The limiting ionic conductances of Ag+ and Cl– are x and y, respectively. The solubility product of AgCl is 1000 K (1) xy

(2)

⎛ 1000 K ⎞ ⎜⎜ ⎟⎟ ⎝ xy ⎠

2

1000  143.5  K (3) xy

⎛ 10 3  143.5  K ⎞ ⎟ (4) ⎜⎜ ⎟ xy ⎠ ⎝

2

Sol. Answer (2)   AgCl   Ag+ + Cl– Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

78

Electrochemistry

Solution of Assignment (Set-2)

Specific conductance = K–1cm–1 

K  1000 C

o o  AgCl =  Ag  Cl = (x + y)

 (x + y) =

K  1000 C

1000K  C = (x  y)

 Solubility product =

C2

⎡ 1000K ⎤ = ⎢ ⎥ ⎣ (x  y) ⎦

2

22. The correct order of equivalent conductance at infinite dilution of LiCl, NaCl, KCl is (1) KCl > NaCl > LiCl

(2) LiCl > NaCl > KCl

(3) LiCl > KCl > NaCl

(4) LiCl ~  NaCl < KCl

Sol. Answer (1) The ions formed are Li+, Na+ and K+, the hydration is maximum in case of Li+ because of which its mobility is least and has least conductance. Hence, the following order. KCl > NaCl > LiCl 23. The limiting equivalent conductance of NaCl, KCl and KBr are 126.5, 150.0 and 152.0 S cm2 eq–1 respectively. The limiting equivalent ionic conductance of Br– is 76 S cm2 eq–1. The limiting equivalent ionic conductance of Na+ is (1) 25.5

(2) 52.5

(3) 75.5

(4) 57.5

Sol. Answer (2) o o  = Na + Cl– = 126.5 ...(i) NaCl o o  = K  + Cl– = 150 KCl

…(ii)

o o  = K  + Br – = 152 KBr

…(iii)

Adding (i) & (iii) subtract (ii)

o

Na

o o o o o o o + Cl– + K  + Br – – K  – Cl– = Na + Br –

o = (126.5 + 152 – 150) = (76) +  Na

o

Na

= 52.5

 Equivalent ionic conductance for Na+ is 52.5. 24. The equivalent conductances of CH3COONa, HCl and NaCl at infinite dilution are 91, 426 and 126 S cm2 eq–1 respectively at 25°C. The equivalent conductance of 1 M CH3COOH solution is 19.55 S cm2 eq–1. The pH of solution is (pKa = 4.74) (1) 5.3

(2) 4.3

(3) 2.3

(4) 1.3

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Solution of Assignment (Set-2)

Electrochemistry

79

Sol. Answer (4) o o   CH =  CH3 COO–  Na 3 COONa

o o o = H   Cl– HCl

  o NaCl = Na   Cl–     = CH3COONa  HCl– – NaCl  CH 3 COOH   91 + 426 + (–126) = 391 =  CH3COOH

 C  19.55

= [H+]

C 



=

19.55 = 0.05 391

= C = 1 × 0.05 M

pH = – log [H+] = – log (5 × 10–2)  2–log 5 = 2 – 0.7 =1.3 25.

E(Na  /Na)  2.71 V, E(Mg 2 /Mg)  2.37 V E(Fe  2 /Fe)  0.44 V, E(Cr  3 /Cr)  0.41 V Based on this data, which is the poorest reducing agent? (1) Na+

(2) Mg+2

(3) Fe+2

(4) Cr+3

Sol. Answer (4) Cr3+ is the poorest reducing agent because of least value of oxidation potential.

Vol. of KCl

(4)

Vol. of KCl

Conductance

Vol. of KCl

(3)

Conductance

(2)

Conductance

(1)

Conductance

26. Which of following type of plot would you expect from the titration of AgNO3 against KCl solution?

Vol. of KCl

Sol. Answer (3) Fact. 27. The standard reduction potential of Cu+2/Cu and Cu+2/Cu+ are 0.337 V and 0.153 V respectively. The standard reduction potential of Cu+/Cu half cell is (1) 0.521 V

(2) 0.490 V

(3) 0.321 V

(4) 0.290 V

Sol. Answer (1) Given (i) ... Cu2+ + 2e–  Cu(s); E1o = 0.337 V (ii)... Cu2+ + e–  Cu+ Eo2 = 0.153 V Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

80

Electrochemistry

Solution of Assignment (Set-2)

Reversing equation (ii); we get Cu+  Cu2+ + e–

...(iii)

Adding equation (i) and (iii) We get, Cu+ + e–  Cu



o Go  G1o  Go2 = (–1 × F × Eº) = –2  F  E1

 +  –1 F  E  o 2

 – FEº = [– 2× F × (0.337)] + [–F × –0.153] – FEº = – 2 × 0.337 × F + (0.153) F  Eº = (2×0.337) – (0.153) = 0.521 V 28. What is G° for the following reaction? Cu+2(aq) + 2Ag(s)  Cu(s) + 2Ag+ ECu 2 /Cu  0.34 V, E Ag /Ag  0.8 V (1) –44.5 kJ

(2) 44.5 kJ

(3) –89 kJ

(4) 89 kJ

Sol. Answer (4) The reaction given is Cu2+ (aq) + 2Ag(s)  Cu(s) + 2Ag+ o o E oCell = E Ag/ Ag  ECu2 /Cu



E oCell = (– 0.8) + (0.34) = – 0.46

Gº = – n F × Eº = – 2 × F × (–0.46) = – 2 × –0.46 × 96500 = 88780 J or Gº = + 89 kJ. o

29. For the half cell EQuinhydrone  1.30 V

O–H

O +



+ 2H + 2e O–H

O

At pH = 3, electrode potential is (1) 1.48 V

(2) 1.42 V

(3) 1.36 V

(4) 1.3 V

Sol. Answer (1) For the reaction, on applying Nernst equation Ecell = EoCell –

0.0591 log [H+]2 2

Ecell = 1.30 –

0.0591 log (10–3)2 2

=–

0.0591 × (–6) log 10 + 1.30 = 0.0591 × 3 + 1.30 = 1.477  1.48 2

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Solution of Assignment (Set-2)

Electrochemistry

81

30. Emf of the cell Zn | Zn+2(aq) || Cu+2(aq) | Cu is independent of (1) Quantity of Cu+2 and Zn+2 in solution

(2) Concentration of Cu+2

(3) Concentration of Zn+2

(4) Temperature

Sol. Answer (1) For the given cell ⎡ Zn2 ⎤ ⎣ ⎦

E = Eº –

0.0591 log 2 ⎡Cu2 ⎤ ⎣ ⎦

When Zn2+ & Cu2+ quantity is changed the emf does not change because EMF depends upon concentration and not the quantity. 31. Which is correct increasing order of deposition? (1) Na+ < Mg+2 < Zn+2 < Ag+

(2) Ag+ < Zn+2 < Mg+2 < Na+

(3) Mg+2 < Na+ < Zn+2 < Ag+

(4) Mg+2 < Zn+2 < Na+ < Ag+

Sol. Answer (1) Increasing order of deposition is related to the order of reduction and oxidation potential (in accordance with preferential discharge theory)  Na+ < Mg2+ < Zn2+ < Ag+ 32. Which is the correct order of deposition of anion? (1) SO4–2 > OH– > Cl– > Br– > I–

(2) SO4–2 < OH– < CI– < Br– < I–

(3) SO4–2 > Cl– > Br– > I– > OH–

(4) Br– > Cl– > I– > SO4–2 > OH–

Sol. Answer (2) It is in the order of discharge potential  In anion order of deposition is – – – – SO2– 4 < HO < Cl < Br < I

33. Which metal oxide is thermally unstable? (1) Al2O3

(2) Na2O

(3) BaO

(4) Ag2O

Sol. Answer (4) Ag2O decomposes as 1  Ag2O   2Ag + 2 O2.

34. Rate of corrosion is maximum when (1) An electrolyte is present in water

(2) Metal has low S.R.P.

(3) Metal has high standard oxidation potential

(4) All of these

Sol. Answer (4) When metal has high standard oxidation potential, it has more tendency to undergo oxidation. In presence of electrolyte, rate of Corrosion is maximum. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

82

Electrochemistry

Solution of Assignment (Set-2)

35. H2(1 atm) | 2.26 M HCOOH || 0.222 M CH3COOH| (1 atm) H2 Ka(HCOOH) = 1.77 × 10–4, Ka(CH3COOH) = 1.8 × 10–5 Emf of the cell is (Neglect the liquid-liquid junction potential) (1) 0.0591 V

(2) –0.0591 V

(3) 0.02955 V

(4) –0.02955 V

(3) –418.424

(4) 229.284

Sol. Answer (2)   HCOOH   HCOO– + H+

C1(1 – ) C1 C1 [H+]L = C1 =

(Ka)1  C1

  CH3 COOH   CH3COO– + H+ C2(1 – )C2  [H+]R =

C2 (Ka)2  C2 [H ]L 0.0591 log [H ]R 1

E = Eº –

E=0–

or E = –

E=–

K1  C1

0.0591 log 1

K 2  C2

1.77  10 4  2.26 0.0591 log 2 1.8  10 5  0.222

0.0591 log 100 = – 0.0591 2

36. Given that NiO2 + 4H+ + 2e–  Ni2+ + 2H2O, E° = 1.678 V NiO2 + 2H2O + 2e–  Ni(OH)2 + 2OH–, E° = –0.49 V For the following reaction Ni(OH)2 + 2H+  Ni2+ + 2H2O Gibb’s free energy change (in kJ mol–1) is (1) 418.424

(2) –229.284

Sol. Answer (3)

NiO2

 4H   2e –

  Ni 2  2H2O

, E1o  1.678 V

Ni(OH)2  2OH –  2e –

  NiO2  2H2O  2e – , Eo2   0.49 V

Ni(OH)2  2H 

  Ni 2  2H2O

, Eo  x V

Go = G1o  Go2  – nF(E1o  Eo2 ) = –2 × 96500 × (1.678 + 0.49) J mol–1 = –418.424 kJ mol–1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electrochemistry

83

37. Zn amalgam is prepared by electrolysis of aqueous ZnCl2 using 9 gram Hg cathode. How much current is to be passed through ZnCl2 solution for 1000 seconds to prepare a Zn amalgam with 25% by weight? (Atomic mass, Zn = 65.4 g) (1) 5.6 A

(2) 7.2 A

(3) 6.64 A

(4) 11.2 A

Sol. Answer (3) Let, x gram of Zn deposited on 9 gram of Hg. % of Zn in amalgam 

x  100  25 9x

 x = 3 gram Equivalent of Zn  Current 

32 65.4

6 96500   8.85 A 65.4 1000

38. Which of the following cannot be extracted by electrolysis from aqueous solution of their salts? (1) Zn

(2) Ag

(3) Cu

(4) Pt

Sol. Answer (1) As EoZn2 / Zn is less than Eo  H /H

2

39. Emf of cell given, Ag(s), AgCl(s)||KCl(aq)|Hg2Cl2(s)|Hg(s) is 0.05 V at 300 K and temperature coefficient of the cell is 3.34 × 10–4 VK–1. Calculate the change in enthalpy of the cell. (1) 965

(2) 9650

(3) 96500

(4) 96.5

Sol. Answer (2) 2Ag  2Ag+ + 2e–

:

(anode)

Hg22+ + 2e–  2Hg

:

(cathode)

∵ n=2 ⎛ Ecell ⎞ H = –nFEcell  nFT ⎜ ⎟ ⎝ T ⎠P

= 2 × 96500(300 × 3.34 × 10–4 – 0.05) = 9650 J mol–1 40. Given : Ag+ + e–  Ag ; E° = 0.799 V Dissociation constant for [Ag(NH3)2]+ into Ag+ and NH3 is 6 × 10–14. Then for the following half-cell reaction: [Ag(NH3)2]+ + e–  Ag + 2NH3, calculate E°. (1) 0.019 V

(2) 0.03 V

(3) 0.014 V

(4) 0.19 V

Sol. Answer (1)

  Ag  e – ;

o EOP  – 0.799 V

Ag(NH3 )2  e –

  Ag  NH3 ;

o ERP ?

Ag(NH3 )2



Ag

Ag  2NH3

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Electrochemistry

o  Ecell  Ecell

Solution of Assignment (Set-2)

[Ag(NH3 )2 ] 0.0591  0 at equilibrium log10 1 [Ag ] [NH3 ]2

 Eocell  0.0591  log10 K C  0.0591  log10 (6  10–14 ) = –0.780 V o  EoOP( Ag/ Ag )  ERP( Ag(NH

 3 )2 / Ag)

 EoAg(NH

 – 0.780  0.799

 3 )2 / Ag

= +0.019 V 41. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time [IIT-JEE 2008] required to liberate 0.01 mol of H2 gas at the cathode is (1 faraday = 96500 C mol–1) (1) 9.65 × 104 s

(2) 19.3 × 104 s

(3) 28.95 × 104 s

(4) 38.6 × 104 s

Sol. Answer (2)

W it  E F

W 10  103  t  0.01 2  E 96500 t = 19.3 × 104 s 42. AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. [IIT-JEE 2011] The plot of conductance () versus the volume of AgNO3 is





volume (P) (1) (P)



volume (Q) (2) (Q)



volume (R) (3) (R)

volume (S) (4) (S)

Sol. Answer (4) Ag+ and K+ have nearly same ionic mobility AgNO3 + KCl  AgCl(s) + KNO3



conc. of KCl Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electrochemistry

85

43. Consider the following cell reaction:  2  2H O(l) 2Fe(s)  O2(g)  4H(aq)  2Fe(aq) ; 2

E° = 1.67 V

At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is (1) 1.47 V

(2) 1.77 V

(3) 1.87 V

[IIT-JEE 2011] (4) 1.57 V

Sol. Answer (4) Ecell  1.67 

0.0591 [Fe2 ]2 log 4 pO2  [H ]4

 1.67 

0.0581 (103 )2 log 4 0.1 (103 )4

 1.67 

0.0591 10 6 log 4 1013

 1.67 

0.0591 log107 4

 1.67 

0.0591  7  1.57 4

SECTION - B Objective Type Questions (More than one options are correct) 1.

1.0 L of 0.1 M aqueous solution of KCl is electrolysed. A current of 96.50 mA is passed through the solution for 10 hours. Which is/are correct? (Assume volume of solution remains constant during electrolysis) (1) After electrolysis molarity of K+ is 0.064 and molarity of Cl– is 0.064 (2) After electrolysis molarity of K+ is 0.1 and molarity of Cl– is 0.064 (3) At S.T.P. 202 ml of Cl2 produced when current efficiency is 50% (4) At S.T.P. 606 ml of total gases produced when current efficiency is 50%

Sol. Answer (2, 3) i = 96.50 A, t = 10 × 60 × 60 s Solution is 1.0 L and 0.1 M Moles present = 1 × 0.1 = 0.1 moles Reactions : 2H+ + 2e  H2 : 2Cl–  Cl2 + 2e–

w it w 96.50  10  60  60  10 –3 = = = = 0.018 2  96500 M nF M For Cl– = 0.036; Molarity = 0.1 – 0.036 = 0.064

VCl2 =

0.018  22.4 = 0.202 L or 202 ml. 2

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Electrochemistry

Solution of Assignment (Set-2)

1000 ml 2 M CuSO4 is electrolysed by a current of 9.65 amp for 2 hours. Which is/are correct? (1) After electrolysis remaining concentration of Cu+2 is 1.64 M using Cu electrode (2) After electrolysis remaining concentration of Cu+2 is 1.64 M using Pt-electrode (3) When remaining concentration of Cu+2 is 1.822 then volume of solution is reduced by 10% using Ptelectrode (4) 17.15 g copper deposit when current efficiency is 75% using copper electrode

Sol. Answer (2, 3, 4) No. of moles of CuSO4 = 1000 × 2 = 2000 millimoles = 2 moles i = 9.65 A; t = 2 hrs = 2 × 60 × 60 s Cu deposited is w =

 w=

n=

E  it F

63.5 9.65  2  60  60 × = 22.86  96500

W = 0.36 M

 2 – 0.36 = 1.64 Hence, molarity = 1.64 M using Pt electrode w=

3.

75 (9.65)  2  60  60 63.5 × = 17.15 g 100 96500 2

For the electrolysis of CuSO4 solution which is/are correct? (1) Cathode reaction : 2H+ + 2e  H2 using Pt electrode (2) Cathode reaction : Cu+2 + 2e–  Cu using Cu electrode (3) Anode reaction : Cu  Cu+2 + 2e– using Cu electrode (4) Anode reaction : Cu  Cu+2 + 2e– using Pt electrode

Sol. Answer (2, 3) CuSO4(aq) forms the ions Cu2+, H+, HO– and SO2– 4 Using Pt electrode At cathode; Cu2+ + 2e  Cu Using Cu electrodes At anode : Cu  Cu2+ + 2e– Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

4.

Electrochemistry

87

Daniell cell : Zn|Zn 2(aq) || Cu 2(aq)|Cu operates as electrolysis cell for 60 min and a current of ( 50 ml 1M)

( 50 ml 1M)

0.965 amp is passed. Which is/are correct?

(E  Cu 2 /Cu  0.34 V, E  Zn 2 /Zn   0.76 V) (1) After electrolysis Zn+2 concentration is 1.36 M

(2) After electrolysis Cu+2 concentration is 0.64 M

(3) After electrolysis Zn+2 concentration is 0.82 M

(4) After electrolysis Cu+2 concentration is 1.18 M

Sol. Answer (1, 2) WCu (deposited) =

31.75 × 0.965 × 60 × 60 = 1.143 g 96500

Total weight of copper = 50 × 10–3 × 1 × 63.5 = 3.175 g Left weight of copper

5.

= 3.175 – 1.143 = 2.032 g 2.032 1000  = 0.64 M 63.5 50

Molarity of Cu2+ solution

=

Molarity of Zn2+ solution

= 1 + 0.36 = 1.36 M

Which compounds have maximum conductivity? (1) 0.2 M [Cr(NH3)3Cl3]

(2) 0.15 M [Cr(NH3)4Cl2]Cl

(3) 0.1 M [Cr(NH3)5Cl]Cl2

(4) 0.07 M [Cr(NH3)6]Cl3

Sol. Answer (2, 3) [Cr(NH3)4Cl2]Cl  [Cr(NH3)4Cl2]2+ +Cl–  0.15 × 2 = 0.30 and for [Cr(NH3)5Cl]Cl2  [Cr(NH3)5Cl]2+ + 2Cl–  0.1 × 3 = 0.30 6.

Molar conductance of 2 M H2A acid is 10 S cm2 mol–1. Molar conductance of H2A at infinite dilution is 400 S cm2 mol–1. Which statement is/are correct? (1) Degree of dissociation is 2.5% and pH of solution is 1.3 (2) Degree of dissociation is 4 and pH of solution is 1.4 (3) Dissociation constant of H2A is 6.24 × 10–5 (4) Dissociation constant of H2A is 2.56 × 10–4

Sol. Answer (1) 7.

Which of following is/are correct? (1) The metallic conduction is due to the movement of electrons in the metal (2) The electrolytic conduction is due to the movement of ions in the solution (3) The metallic conduction increases with increase in temperature whereas electrolytic conduction decreases with increase in temperature (4) None of these

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Electrochemistry

Solution of Assignment (Set-2)

Sol. Answer (1, 2) The metallic conduction is due to the presence of electrons in the metal and electrolytic conduction is due to the movement of ions in the solution. 8.

For electrolyte AxBy which is/are not correct relation between molar conductivity (M) and equivalent conductivity (eq) (1) M = xy eq

(2) eq = xy M

(3) xM = y eq

(4) yM = x eq

Sol. Answer (2, 3, 4) For the electrolyte Ax By n-factor = xy M = (xy) eq

Only 1st option is correct & others are incorrect option. 9.

The cell constant of a conductivity cell is defined as ( = cell constant, l = length between the electrode, A = area, R = resistance, G = conductance, K = conductivity) (1)  

l A

(2)



 R

(3)  = (Gr)–1

(4)  

G K

Sol. Answer (1, 3) R= 

1 1 l l    RA A

K = C



l A

RA l ;  and   (G)1 l A

Vol. of NaOH

Vol. of NaOH

(4)

Conductance

(3)

Conductance

(2)

Conductance

(1)

Conductance

10. Which of following plots will not be obtained for a conductometric titration of HCl and NaOH?

Vol. of NaOH

Vol. of NaOH

Sol. Answer (2, 3, 4) In the conductometric titration of HCl and NaOH conductance first decreases, reaches a minimum value and then increases. 11. Zn | Zn+2 (1M) || Ni+2 (1 M) | Ni, antilog (0.7411) = 5.5

E Zn 2 /Zn  0.75 V, ENi 2 /Ni  0.24 V Which statement is/are correct for above cell? (1) Emf of cell is 0.51 V and cell reaction is spontaneous (2) Emf of cell is –0.51 V and cell reaction is non-spontaneous (3) Emf of cell is zero when concentration of Ni+2 is 5.5 × 10–18 M (4) Cell reaction is non-spontaneous when concentration of Ni+2 is less than 5.5 × 10–18 M Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electrochemistry

89

Sol. Answer (1, 3, 4) The given cell is Zn|Zn2+(1M)||Ni2+(1M)|Ni Eº = (0.75) + (–0.24) = 0.51V and cell reaction is spontaneous. E = Eº –

[Zn2 ] 0.0591 log [Ni2 ] 2

 E = 0; [Ni2+] = 5.5 × 10–18 M The cell reaction is Non-spontaneous when concentration of M2+ is less than 5.5 × 10–18 M. 12. Which statement is correct about electrolysis of CuSO4? (1) At cathode Cu will deposit and at anode O2 will be produced using Pt-electrode (2) At cathode Cu will not deposit but Cu dissolve at anode using Cu-electrode (3) At cathode Cu will deposit and at anode O2 will be produced using Cu-electrode (4) At cathode Cu will deposit and at anode Cu will dissolve using Cu-electrode Sol. Answer (1, 4) Using Pt electrodes CuSO4  Cu2+ + SO2– 4 H2O  H+ + HO– At cathode : Cu2+ + 2e–  Cu Anode : 4HO–  2H2O + O2 + 4e– Products are Cu and O2 Using Cu electrodes Anode : Cu  Cu2+ + 2 e– Cathode : Cu2+ + 2 e–  Cu 13. Aqueous solution of which electrolyte produces H2 gas at cathode? (1) NaCl

(2) MgCl2

(3) CuCl2

(4) AgCl

Sol. Answer (1, 2) H+ has lower discharge potential as compared to Na+ and Mg2+ Hence, in case of NaCl and MgCl2 reaction is 2H+ + 2e–  H2. 14. Which is/are correct? (1) If temperature coefficient is greater than zero, cell reaction is endothermic (2) If temperature coefficient is less than zero, cell reaction is endothermic (3) If temperature coefficient is less than zero, cell reaction is exothermic (4) If Ecell is negative then G is negative and cell reaction is spontaneous Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Electrochemistry

Solution of Assignment (Set-2)

Sol. Answer (1, 3) It is known fact that H and temperature coefficient are related as,

⎛ E ⎞ H > 0 for ⎜ ⎟ >0 ⎝ T ⎠ ⎛ E ⎞ and H < 0 for ⎜ ⎟ 0 is spontaneous for Fe3+ + Ni  Ni2+ + Fe o o Eº = ENi/Ni2  EFe3  /Fe = (0.24) + (–0.04) > 0 i.e. spontaneous

and for reaction Cd2+ + Zn  Zn2+ + Cd o o Eo = EZn/Zn2  ECd2 /Cd = (0.76) + (– 0.40) > 0 is spontaneous.

21. Which of the following cells give the cell potential to their standard values? (1) Zn|Zn2+(0.01 M)||H3O+(0.1 M)|H2(1 atm), Pt

(2) Cu|Cu2+(0.25 M)||Ag+(0.5 M)|Ag

(3) Cd|Cd2+(0.01 M)||pH = 1|H2(1 atm), Pt

(4) Zn|Zn2+(0.1 M)||pH = 1|H2(1 atm), Pt

Sol. Answer (1, 2, 3) o For Ecell  Ecell , K C  1.

22. Which solution(s) become(s) more acidic after the electrolysis using inert electrodes? (1) NaCl solution

(2) CuSO4 solution

(3) AgNO3 solution

(4) Na2SO4 solution

Sol. Answer (2, 3) In the electrolysis of CuSO4 solution and AgNO3 solution, H2SO4 and HNO3 are formed respectively. 23. In which of the following cells, reaction quotient is equal to one? (1) Pb|PbC2O4, CaC2O4, CaCl2(0.1 M)||CuSO4(0.1 M) | Cu (2) Zn|ZnSO4(0.1 M)||CuSO4(0.1 M)|Cu (3) Zn|ZnSO4(0.1 M)||Hg2Cl2, KCl(0.1 M)|Hg, Pt (4) Cu|CuSO4(0.1 M)||SnCl2(0.1 M)|SnCl4(0.1M), Pt Sol. Answer (1, 2) In (3),  = 10 In (4),  = 0.1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electrochemistry

93

24. Saturated solution of KNO3 is used to make ‘salt-bridge’. Then incorrect option(s) is/are (1) Velocity of K+ is zero (2) Velocity of NO3– is zero (3) Velocity of both K+ and NO3– are nearly the same (4) KNO3 is highly soluble in water Sol. Answer (1, 2) Fact. 25. When a lead-storage battery is discharged, then incorrect option(s) is/are (1) H2SO4 is consumed

(2) Pb is formed

(3) SO2 is evolved

(4) PbSO4 is consumed

Sol. Answer (2, 3, 4) Pb is consumed and PbSO4 is formed. SO2 is not evolved. 26. For the reduction of NO3– ion in an aqueous solution, Eo is +0.96 V. Values of Eo for some metal ions are given below

[IIT-JEE 2009]

V2+ (aq) + 2e–  V

Eo = – 1.19 V

Fe3+ (aq) + 3e–  Fe

Eo = –0.04 V

Au3+ (aq) + 3e–  Au

Eo = +1.40 V

Hg2+ (aq) + 2e–  Hg

Eo = +0.86 V

The pair(s) of metals that is(are) oxidized by NO3– in aqueous solution is(are) (1) V and Hg

(2) Hg and Fe

(3) Fe and Au

(4) Fe and V

Sol. Answer (1, 2, 4) Oxidation of V Eo = 0.96 – (–1.19) = 2.15 V For Fe, Eo = 0.96 – (–0.04) = 1.0 V For Au, Eo = 0.96 – 1.4 = – 0.044 V (not feasible) For Hg Eo = 0.96 – 0.86 = 0.1 V 27. In a galvanic cell, the salt bridge

[JEE(Advanced)2014]

(1) Does not participate chemically in the cell reaction (2) Stops the diffusion of ions from one electrode to another (3) Is necessary for the occurrence of the cell reaction (4) Ensures mixing of the two electrolytic solutions Sol. Answer (1, 2) In a galvanic cell, the salt bridge does not participate in the cell reaction, stops diffusion of ions from one electrode to another and is not necessary for the occurrence of the cell reaction. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

94

Electrochemistry

Solution of Assignment (Set-2)

SECTION - C Linked Comprehension Type Questions Comprehension-I An electrochemical cell is constructed by immersing a piece of copper wire in 50 ml of 0.1 M CuSO4 solution and zinc strip in 50 ml of 0.1 M ZnSO4 solution [E Cu 2 /Cu  0.34 V, E Zn 2 /Zn  0.76 V]

1.

The emf of cell is (1) 1.07 V

(2) 1.1 V

(3) 1.3 V

(4) 1.13 V

Sol. Answer (2) EoCu2 /Cu = 0.34 V and EoZn2 / Zn = – 0.76 V

ECell

[Zn2 ] 0.0591 = Eº – log [Cu2 ] 2

[Zn2+] = [Cu2+] = 1M o o  E = Eº = EZn/Zn2  ECu2 /Cu

 E = (0.76) + (0.34)  E = Eº = 1.1 V 2.

The emf of cell increases when small amount of concentrated NH3 is added to (1) ZnSO4 solution

(2) CuSO4 solution

(3) Both (1) & (2)

(4) Can't say

Sol. Answer (1) When NH3 is added to ZnSO4 solution, NH3 reacts with Zn2+ in the following manner : 2+   Zn2+ + 4 NH3   [Zn(NH3)4]

i.e., [Zn2+] decreases. In the equation

E = Eo –

0.0591 [Zn2 ] log 2 [Cu2 ]

If [Zn2+] decreases then log

3.

[Zn2 ] [Cu2 ]

decreases hence, EMF of cell increases.

In a separate experiment, 50 ml of 1.5 M NH3 is added to CuSO4 solution. Emf of the cell is [Kf ([Cu(NH3)4]+2) = 5.88 × 1013] (1) 0.933 V

(2) 1.327 V

(3) 1.467 V

(4) 0.61 V

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Solution of Assignment (Set-2)

Electrochemistry

95

Sol. Answer (4) Due to the complex formation, [Cu2+] decreases & it can be calculated by the reaction,   Cu2   4HN3   [Cu(NH3)4]2+

50  0.1

50  1.5

[Cu2+] =

50  0.1 100 = = 9.3 × 10–15 M (0.55)4  5.88  1013

[Cu (NH3 )4 ]2 [NH3 ]4  k f

Ecell = 1.1 –

⎛ 0.0591 0.1 log ⎜⎜ –15 2 ⎝ 9.3  10

⎞ ⎟⎟ = 0.715 V ⎠

Thus, the e.m.f. of cell decreases. Comprehension-II The ionic mobility for some ions in water at 298 K is given as following

1.

ions

ionic mobility

K+

7.616 × 10–4

Ca+2

12.33 × 10–4

Br–

8.09 × 10–4

SO4–2

16.58 × 10–4

The equivalent conductance of CaSO4 at infinite dilution is (2) 28.51 × 10–4

(1) 279

(3) 31.82 × 10–4

(4) 306

Sol. Answer (1) Equivalent conductance of CaSO4 is the sum of ionic conductance of Ca2+ & SO2– 4 .     CaSO =  Ca2    SO2– 4 4







= UCa2 



= USO2–

Ca2 

SO2– 4

U

Ca2 



4

F

F

& USO24 are ionic mobilities

  CaSO = F {12.33 + 16.58} × 10–4 4   CaSO = 96500 × 10–4 × 28.91 = 278.98  279 4

 Equivalent conductance of CaSO4 is 279 2.

If degree of dissociation is 10% then equivalent conductance of CaSO4 is (1) 27.9

(2) 2.851 × 10–4

(3) 3.182 × 10–4

(4) 30.6

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96

Electrochemistry

Solution of Assignment (Set-2)

Sol. Answer (1) We know that =

C



 (0.1) =

C (279)

C = 279 × (0.1) = 27.9

 Equivalent conductance = 27.9 3.

If the equivalent conductance of 0.01 M CaSO4 solution is 13.95, then equilibrium constant is (1) 5.26 × 10–4

(2) 5.26 × 10–5

(3) 2.63 × 10–4

(4) 2.63 × 10–5

Sol. Answer (4) C = 0.01M; C = 13.95 =

C 



 K=

=

13.95 = (0.05) 279

 2C (0.05)2  (0.01) = (1   ) (1– 0.05)

 K = 2.63 × 10–5. Comprehension-III Given below are a set of half-cell reactions (in acidic medium) alongwith their E° (in volt) values.

1.

I2  2e   2I

E  0.54

Cl2  2e   2CI

E  1.36

Mn3  e   Mn2

E  1.50

Fe3  e   Fe2

E  0.77

O2  4H  4e   2H2 O

E  1.23

Among the following, identify the correct statement (1) Cl– is oxidised by O2

(2) Fe+2 is oxidised by iodine

(3) I– is oxidised by chlorine

(4) Mn+2 is oxidised by chlorine

Sol. Answer (3) 2.

While Fe+3 is stable, Mn+3 is not stable in acid solution because (1) O2 oxidises Mn+2 to Mn+3 (2) O2 oxidises both Mn+2 to Mn+3 and Fe+2 to Fe+3 (3) Fe+3 oxidises H2O to O2 (4) Mn+3 oxidises H2O to O2

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Solution of Assignment (Set-2)

3.

Electrochemistry

97

The strongest reducing agent in aqueous solution is (1) I–

(2) Cl–

(3) Mn+2

(4) Fe+2

Sol. Answer (1) Comprehension-IV Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200, 1 faraday = 96500 coulomb). [IIT-JEE 2007] 1.

The total number of moles of chlorine gas evolved is (1) 0.5

(2) 1.0

(3) 2.0

(4) 3.0

Sol. Answer (2) nNaCl =  2.

4  500 2 1000

nCl2  1.

If the cathode is a Hg electrode, then the maximum weight (g) of amalgam formed from this solution is (1) 200

(2) 225

(3) 400

(4) 446

Sol. Answer (4) nNa deposited = 2  nNa–Hg formed = 2  Mass = 2 × 223 = 446. 3.

The total charge (coulomb) required for complete electrolysis is (1) 24125

(2) 48250

(3) 96500

(4) 193000

Sol. Answer (4) Total charge required = 2F = 2 × 96500 = 193000 C. Comprehension-V The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+(aq; 0.05 molar) || M+ (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV.

[IIT-JEE 2010]

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98 1.

Electrochemistry

Solution of Assignment (Set-2)

For the above cell (1) Ecell < 0; G > 0

(2) Ecell > 0; G < 0

(3) Ecell < 0; Gº > 0

(4) Ecell > 0; Gº < 0

Sol. Answer (2) Cell reaction M   M (1 M)

(0.05 M)

Apply Nernst equation E  Eº –

2.

0.059 0.05 log 1 1

E–

0.059 log 5  10 –2 1

E–

0.059 ⎡ –2   log5 ⎤⎦ 1 ⎣





If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be (1) 35 mV

(2) 70 mV

(3) 140 mV

(4) 700 mV

Sol. Answer (3) E1 log0.05  E2 log0.0025 E1 log5  10 –2  E2 log25  10 –4

E1  70(given) 70 –1.3 1   E2 –2.6 2

Comprehension-VI The electrochemical cell shown below is a concentration cell. M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. [IIT-JEE 2012] 1.

The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V) (1) 1 × 10–15

(2) 4 × 10–15

(3) 1 × 10–12

(4) 4 × 10–12

Sol. Answer (2) 0.059  

0.059 0.001 log 2  2 (M )

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Solution of Assignment (Set-2)

log

Electrochemistry

99

0.001 2 [M2  ]

0.001  100 K sp = 4  (10 5 )3 = 4  10 15 [M2  ] [M2  ]  105 2.

The value of G (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1) (1) –5.7

(2) 5.7

(3) 11.4

(4) –11.4

Sol. Answer (4) G = –nFE = –2 × 96500 × 0.059 = –11387 joule mol–1  –11.4 kJ mol–1

SECTION - D Assertion-Reason Type Questions 1.

STATEMENT-1 : The molar conductivity of strong electrolyte decreases with increase in concentration. and STATEMENT-2 : At high concentration, migration of ion is slow.

Sol. Answer (1) Molar conductance is given by the following expression  = (K × V) =

K  1000 c

Here ‘c’ is the concentration More is the concentration lesser is the molar conductance Hence, both statements are correct and statement-2 is the correct explanation of statement-1. 2.

STATEMENT-1 : Electrolysis of molten PbBr2 using platinum electrodes produces Br2 at anode. and STATEMENT-2 : Br2 is obtained in gaseous state at room temperature.

Sol. Answer (3) PbBr2  Pb2+ + 2Br– At cathode : Pb2+ + 2e–  Pb At anode : 2Br–  Br2 + 2e– Br2 obtained in liquid state at room temperature.  Statement-1 is correct and statement-2 is false. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

100 3.

Electrochemistry

Solution of Assignment (Set-2)

STATEMENT-1 : For the concentration cell, Zn(s) Zn  2 (aq) Zn  2 (aq) Zn for spontaneous cell reaction C1 < C2. C1

C2

and STATEMENT-2 : For concentration cell, E cell 

C RT loge 2 for spontaneous reaction Ecell = +ve  C2 > C1. nF C1

Sol. Answer (1) The given cell is Zn|Zn2+(C1) ||Zn2+ (C2)|Zn Zn(s)  Zn2+ + 2e– E

E

Zn/Zn2 

2

Zn

/Zn

o = E Zn/ Zn2 –

= EZn2 /Zn –

0.0591 log (C1) 2

⎛ 1 ⎞ 0.0591 log ⎜ C ⎟ 2 ⎝ 2⎠

o o  E = (E Zn/ Zn2  E Zn2 / Zn ) –

⎛ C1 ⎞ 0.0591 log ⎜ C ⎟ 2 ⎝ 2⎠

 EMF of cell E=

⎛C ⎞ – 0.0591 log ⎜ 1 ⎟ 2 ⎝ C2 ⎠

⎛ C1 ⎞ log ⎜ C ⎟ < 0 for spontaneity ⎝ 2⎠ ⎛ C1 ⎞ log ⎜ C ⎟ < log 1 ⎝ 2⎠

 C1 < C2.  Statement-1 and statements-2 is correct and it is also the correct explanation. 4.

STATEMENT-1 : A saturated solution of KCl is used to make salt bridge in concentration cells. and STATEMENT-2 : Mobility of K+ and Cl– are nearly same.

Sol. Answer (1) Mobilities of ions involved in salt bridge is same which is used in concentration cells. 5.

STATEMENT-1 : The molar conductance of weak electrolyte at infinite dilution is equal to sum of molar conductances of cation and anion. and STATEMENT-2 : Kohlrausch’s law is applicable for both strong and weak electrolytes.

Sol. Answer (3)   AB =  A   B–

Kohlraush law is applicable for weak electrolyte and not for strong electrolyte. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

6.

Electrochemistry

101

STATEMENT-1 : When a copper wire is placed in a solution of AgNO3, the solution acquires blue colour. and o of Cu+2/Cu is lesser than E o STATEMENT-2 : ERP

Ag / Ag

.

Sol. Answer (1) Cu + AgNO3  Cu2+ + Ag Eo for reaction is positive because Eo

Cu2  /Cu

7.

 Eo

Ag / Ag

STATEMENT-1 : G° = –nFE°. and STATEMENT-2 : E° should be positive for a spontaneous reaction.

Sol. Answer (2) Go = – nFEo But Go does not decide the spontaneity only G decides the spontaneity of reaction.  Both statements are correct but statement-2 is not the correct explanation. 8.

STATEMENT-1 : One coulomb of electric charge deposits the weight that is equal to electrochemical equivalent of substance. and STATEMENT-2 : One faraday deposits one mole of substance.

Sol. Answer (3) One faraday deposits one equivalent of substance. 9.

STATEMENT-1 : If an aqueous solution of NaCl is electrolysed, the product obtained at the cathode is H2 gas and not Na. and STATEMENT-2 : Gases are liberated faster than metals.

Sol. Answer (3)   NaCl   Na+ + Cl– + –   H2O   H + OH

Among cations, hydrogen has higher standard electrode potential and among anions chlorine has low standard electrode potential. Thus, at cathode preferentially H2 gas is evolved, and at anode Cl2 gas is evolved. 10. STATEMENT-1 : H2 + O2 fuel cell gives a constant voltage throughout its life. and STATEMENT-2 : In this fuel cell, H2 reacts with OH– ions, yet the overall concentration of OH– ions does not change. Sol. Answer (1) In H2 + O2 fuel cell, Anode : 2H2(g) + 4OH–(aq)  4H2O() + 4e– Cathode : O2(g) + 2H2O() + 4e–  4OH–(aq) OH– consumed is reformed, so [OH–] does not change. Hence, fuel-cell gives constant voltage throughout its life. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

102

Electrochemistry

Solution of Assignment (Set-2)

SECTION - E Matrix-Match Type Questions 1.

Match the following Column-I

Column-II Molar Conductivity ( –1)

Complex (A) CoCl3.6NH3

(p)

97

(B) CoCl3.5NH3

(q)

0

(C) CoCl3.4NH3

(r)

404

(D) CoCl3.3NH3

(s)

229

Sol. Answer A(r), B(s), C(p), D(q) [Co(NH 3 ) 6 ] Cl 3 will give maximum number of ions(4) because of which conductivity is maximum i.e. 404. In [Co(NH3)3Cl3] no ions are given Hence molar conductivity is zero. [Co(NH3)5Cl]Cl2 & [Co(NH3)4Cl2] Cl forms 3 & 2 ions. 2.

Match the following Column-I

Column-II (Amount of charge used for diposition/liberation)

(A) 1 mol Al+3

(p)

F

(B) 2.3 gm of Na+

(q)

3F

(C) 3.6 gm of Mg+2

(r)

0.1 F

(D) 11.2 L H2 at S.T.P.

(s)

0.3 F

Sol. Answer A(q), B(r), C(s), D(p) (A) 1 mole Al3+ w it wF =  it = E F E

⎛ wF ⎞ wF it = 3 ⎜ = 3F ⎟ = E ⎝ M ⎠ (B) 2.3 g Na+

it =

2.3 w F= F = 0.1 F 23 E

(C) 3.6 g of Mg2+

it =

3.6 F = 0.3 F 12

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Solution of Assignment (Set-2)

(D)

11.2 1 = 22.4 2

Electrochemistry

103

nH  2

w it w =  it = F E F E

⎛w⎞ 1 = 2⎜ ⎟ F = 2  F = F M 2 ⎝ ⎠ 3.

Match the following Column-I

Column-II c m

(A) , specific conductance

(p)

(B)  m , molar conductance

(q)

 m

(C) , degree of dissociation

(r)

Decreases with dilution

(D) Kohlrausch law

(s)

Decreases with increase in concentration of strong electrolytes

 m

Sol. Answer A(r), B(s), C(p), D(q) (A) Specific conductance decreases with dilution (B) Molar conductance decreases with increase in concentration of electrolyte (C)  

m and decreases with dilution m

(D) Resistance  4.

l and decreases with dilution A

Match the following Column-I

Column-II

(A) Calomel electrode

(p)

Electrolyte concentration cell

(B) Zn-Cd(C1) |CdCl2| Zn-Cd(C2)

(q)

Metal-insoluble anion half cell

(C) Quinhydrone electrode

(r)

Electrode concentration cell

(D) Pt|H2(1 atm)|H+(C1)||H+(C2)|H2(1 atm)|Pt

(s)

Redox half cell

Sol. Answer A(q), B(r), C(s), D(p) 5.

Match the following Column-I

Column-II

(Electrolysis)

(Observation)

(A) Aqueous solution of NaCl using inert

(p)

Metal loss at anode

(q)

Chlorine gas evolved at anode

(C) CuSO4 using copper electrodes

(r)

Oxygen gas evolved at anode

(D) 50% H2SO4 solution

(s)

A compound with peroxide bond is formed

electrodes (B) Very dilute aqueous solution of NaCl using mercury cathode

Sol. Answer A(q), B(r), C(p), D(s) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

104

Electrochemistry

Solution of Assignment (Set-2)

SECTION - F Integer Answer Type Questions 1.

The half cell potentials of a half cell | A  x n  , A x  | Pt were found to be as follows: % of reduced form

24.4

48.8

Half cell potential (V)

0.101

0.115

Determine the value of 'n'. Sol. Answer (2) A

2.

 x n 

 ne –  A x 

o 0.101  ERP 

0.059 ⎛ 75.6 ⎞ log ⎜ ⎟ n ⎝ 24.4 ⎠

o 0.115  ERP 

0.059 ⎛ 51.2 ⎞ log ⎜ ⎟ n ⎝ 48.8 ⎠

⇒n2

o o The standard reduction potential of EBi3  /Bi and ECu2 /Cu are 0.226 V and 0.344 V respectively. A mixture of

salts of Bi and Cu at unit concentration each is electrolysed at 25°C. At what value of  log Cu2  does Bismuth starts to deposit during electrolysis. Sol. Answer (4) The passage of current would initially deposit Cu2+ till ECu2 /Cu becomes 0.266 V because then only Bi3+ will be deposited. o Thus, ECu2 /Cu  Ecu  2 /Cu

0.266  0.344  3.

0.059 log Cu2   2

0.059 log  Cu2   ⇒  log Cu2    4 2

A cell is containing two H electrodes. The negative electrode is in contact with a solution of pH = 6. EMF of the cell is 0.118 V at 25°C. Calculate pH at positive electrode.

Sol. Answer (4) Ecell  0.059log

H  cathode H  anode

= 0.059 [pH anode – pH cathode] 0.118 = 0.059 [6 – pH] pH = 4 4.

How many faradays of electricity is required to deposit 2 mol copper from CuSO4 solution?

Sol. Answer (4) Equivalent weight of copper = 63.5/2 Hence, 2 mol require 4 F electricity. 5.

A current of 3 ampere has to be passed through a solution of AgNO3 solution to coat a metal surface of 80 cm2 with 0.005 mm thick layer for a duration of approximately (y)3 seconds. What is the value of y? (Density of Ag is 10.5 g/cm3)

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Solution of Assignment (Set-2)

Electrochemistry

105

Sol. Answer (5) Volume of surface

= 80 × 0.0005 = 0.04 cm2

WAg = 0.04 × 10.5 = 0.42 gram 

 0.42 

Eit 96500

108  3  t ⇒ t  125.09 s  y 3 96500

 y5 6.

The cost at 5 paise per kWh of operating an electric motor for 8 hours, which takes 15 ampere at 110 V, is 11y paise. Calculate y.

Sol. Answer (6) Total energy consumed for 8 hours = iVt = 15 × 110 × 8 × 10–3 kWh = 13.2 kWh Total cost = 5 × 13.2 = 66 paise 11y = 66  y=6

SECTION - G Multiple True-False Type Questions 1.

STATEMENT-1 : Corrosion of iron is essentially an electrochemical phenomenon. STATEMENT-2 : Corrosion reaction at anode : 2Fe  s  2Fe3   6e STATEMENT-3 : Corrosion reaction at cathode : O 2  g  4H  aq  4e   2H2 O l (1) T T T

(2)

TFT

(3)

TFF

(4)

FFT

Sol. Answer (2) Facts about corrosion Statement 2 : 2Fe  s   2Fe2  4e 2.

STATEMENT-1 : Using Kohlrausch's law of independent migration of ions, it is possible to calculate 0 for any electrolyte from the ° of individual ions. STATEMENT-2 : Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. STATEMENT-3 : When concentration approaches zero, molar conductivity reaches the lowest limit. (1) T T F

(2)

TTT

(3)

FTT

(4)

FFT

Sol. Answer (1) Fact. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

106 3.

Electrochemistry

Solution of Assignment (Set-2)

STATEMENT-1 : Electrolysis of acidulated water using inert electrodes results in evolution of gases at cathode and anode both. STATEMENT-2 : Al3+ discharges more readily than Zn2+ at cathode. STATEMENT-3 : In an electrolytic cell, cations move towards anode. (1) F T T

(2)

TTT

(3)

FTF

(4)

TFF

Sol. Answer (4) In electrolysis of acidulated water, H2 and O2 is evolved in 2 : 1 volume ratio, at cathode and anode respectively. EoAl 3 /Al  EoZn2 / Zn  Zn2+ discharges more readily at cathode.

In an electrolytic cell, cations move towards cathode, i.e., negative electrode.

SECTION - H Aakash Challengers Questions 1.

The standard potential of the following cell is 0.23 V at 15°C and 0.21 V at 35°C. Pt|H2(g)|HCl (aq)||AgCl(s)|Ag(s) (i) Write the cell reaction. (ii) Calculate H° and S° for the cell reaction by assuming that these quantities remain unchanged in the range 15°C to 35°C. [Given SRP of Ag+(aq)|Ag(s) is 0.80 V at 25°C]

Sol. (i)

1 H2 (g)  AgCl(s)  H (aq)  Ag(s)  Cl– (aq) 2

⎛ E ⎞ (ii) S  nF ⎜ T ⎟ , n  1, F  96500 coulombs ⎝ ⎠P E = 0.21 – 0.23 = – 0.02 V, T = 35 – 15 = 20°C

⎛ –0.02 ⎞ –1 –1 S  1  96500  ⎜ ⎟  – 96.5 JK mol 20 ⎝ ⎠ o G15  –1  0.23  96500  – 22195 J mol–1 o H15  G – TS

= –22195 – 288 × (–96.5) = –49987 J mol–1

⇒ S  – 96.5 J k –1 mol–1 , H  – 49987 J mol–1 2.

Calculate Gro of the following reaction Ag+(aq) + Cl–(aq)  AgCl(s) Given: Gro (AgCl)  –109 kJ mol–1; Gro (Cl– )  –129 kJ mol–1; Gro (Ag )  77 kJ mol–1 (i) Represent the above reaction in form of a cell. (ii) Calculate E° of the cell. (iii) Find log10Ksp of AgCl.

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Solution of Assignment (Set-2)

Electrochemistry

107

Sol. (i) Ag(s) | Ag+ | | AgCl | | Cl– | Cl2, Pt o o o o (ii) Gr  GAgCl – GAg – GCl– (Ag+(aq) + Cl–(aq)  AgCl(s))

= –109 – (–129) – 77 = –57 kJ mol–1

G  – nFE ⇒ Eocell 

–57000 –1  96500

⇒ Eocell  0.59 V (iii) G° = –2.303 RT logK ⇒ logK 

–57000  10 –2.303  6.314  298

K = 1010 K sp 

3.

1  10 –10 ⇒ log10 K sp  –10 K

The following electrochemical cell has been set-up: Pt1|Fe3+|Fe2+(1M)||Ce4+ | Ce3+ (1M)|Pt2 o o EFe  0.77 V and ECe  1.61 V. 3 4 /Fe2 /Ce3

If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time? Sol. SRP of Fe > SRP of Ce  Fe acts as anode and Ce acts as cathode o Eocell  Ecathode – Eoanode  1.61 – 0.77  0.84 V

Hence, flow of current is from cathode to anode (right to left) Current will decrease with time. 4.

An excess of liquid mercury is added to an acidified solution of 1.0 × 10–3 M Fe3+. It is found that 5% of Fe3+ o remains at equilibrium at 25°C. Calculate EHg assuming that the only reaction that occurs is 2 /Hg

2Hg + 2Fe3+  Hg22+ + 2Fe2+ [Given Eo 3 2  0.77 V ] . Fe /Fe

2Fe3   2Hg

Sol.

At t  0

1  10 –3 M

at t(eq)

0.05  10 –3 M

 2Fe2



0 0.95  10 –3 M

Hg22 0 0.95  10 –3 M 2

2

Ecell  Eocell

⎡Fe2 ⎤ [Hg22 ] 0.0591 ⎦ – log10 ⎣ n [Fe3  ]2

o At eqb., 0  0.77 – EHg – 2 / Hg

0.0591 (0.95  10 –3 )2  (0.475  10 –3 ) log10 2 (0.05  10 –3 )2

o ⇒ EHg  0.792 V 2 /Hg

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108 5.

Electrochemistry

Solution of Assignment (Set-2)

Chromium metal can be plated out from an acidic solution containing CrO3 according to the following equation. CrO3(aq) + 6H+(aq) + 6e–  Cr(s) + 3H2O. (i) How many grams of chromium will be plated out by 24000 coulombs? (ii) How long will it take to plate out 1.5 g of chromium by using 12.5 ampere current?

Sol. 52 gram of Cr is deposited by passing 6 × 96500 coulomb (i) Amount of Cr deposited =

52  24000 = 2.155 g 6  96500

(ii) M = Zit

⇒ 1.5 

52  12.5  t 6  96500

 t = 1336.15 s 6.

The specific conductivity of a saturated solution of AgCl is 2.30 × 10–6 ohm–1 cm–1 at 25°C. Calculate the solubility of AgCl at 25°C if    61.9 ohm–1 cm2 mol–1 and  –  76.3 ohm –1 cm2 mol–1. Ag Cl

1000 Sol. Dilution = Solubility

oAgCl   Ag   Cl –  61.9  76.3  138.2 ohm–1 cm2 mol–1 o Sp. Conductivity × dilution =  AgCl

2.30  10 –6 

1000  138.2 s

Solubility (gram per litre) = 2.382 × 10–3 g lt–1 7.

Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at 25°C. H2(1 atm) | 0.5 M HCOOH | | 1 M CH3COOH | (1 atm)H2 (Ka for HCOOH and CH3COOH are 1.77 × 10–4 and 1.8 × 10–5 respectively)

Sol. [H ]HCOOH  K a  C  1.77  10–4  0.5  0.9407  10–2 M

[H ]CH3COOH  1.8  10 –5  1  4.2426  10 –3 M Ecell  0.0591 log10

[H ]RHS [H ]LHS

 0.0591 log10

4.2426  10 –3 0.9407  10 –2

 Ecell = –0.0204 V 8.

For the cell reaction, Mg | Mg2+(aq) | | Ag+(aq) | Ag. Calculate the equilibrium constant at 25°C and maximum work that can be obtained by operating the cell. o [ Given EMg  – 2.37 V, EoAg / Ag   0.80 V ] 2 /Mg

Sol. Eocell  0.80  2.37  3.17 V log K c 

nEocell 2  3.17   107.2758 0.0591 0.0591

Kc = 1.89 × 10107 Maximum work = – G  nFEocell = 2 × 96500 × 3.17 = 611.81 kJ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

9.

Electrochemistry

109

Under standard conditions (i) Will Cu reduces Ag+ to Ag? (Eo   0.799 V, Eo 2  – 0.337 V) Ag /Ag Cu /Cu o o (ii) Will Fe3+ be reduced to Fe2+ by Sn2+? (EFe3 /Fe2  0.771 V, ESn2 / Sn4  – 0.13 V)

(iii) Would you use a silver spoon to stir a solution of Cu(NO3)2? Sol. (i) Yes, Eocell of Cu | Cu2+ | | Ag+ | Ag is positive (ii) Yes, Eocell of Pt, Sn4+ | Sn2+ | | Fe3+ | Fe2+, Pt is positive (iii) Yes, the reaction between silver and Cu2+ does not occur 10. A weak monobasic acid is 5% dissociated in 0.01 mol/lt solution. The limiting molar conductivity at infinite dilution is 4 × 10–2 ohm–1 cm2 mol–1. Calculate conductivity of 0.05 molar solution of acid. Sol. Ka = C2 = 0.01 × (0.05)2 = 2.5 × 10–5 For C = 0.05 M,  

Ka  C

2.5  105 0.05

  = 0.0223 



M o M

⇒ M  0.0223  4  10 2

11. 500 ml CuSO4 solution was electrolysed using a current of 2 amp (efficiency = 75%) for 60 min. Calculate the pH of solution at the end of electrolysis. (Assume initial pH = 7) Sol. When CuSO4 solution is electrolysed then H2SO4 is formed. Equivalents of H2SO4 =

NH2SO4 =

1 75 2 × 60 × 60 = 0.0559 96500 100

0.0559 = 1.12 × 10–1 500 1000

Now, [H+] = 1.12 × 10–1 pH = – log [H+] = – log 1.12 × 10–1 pH = 1 – 0.05 = 0.95 12. A constant current flowed for 30 min through a solution of KI oxidising the iodide ion to iodine. At the end of experiment, the iodine was titrated with 10 ml 0.075 M Na2S2O3 solution. Calculate the strength of current. Sol. t = 30 min = 30 × 60 = 1800 s KI  K++I– H2O  H++HO– (Cathode) 2H+ + 2e H2 (Anode) 2I–  I2 + 2e– eq. (I2) = eq. (Na2S2O3) = 10–3 × 10 × 0.075 = 0.75 × 10–3 

w it = E F



i=

wF (0.75  10 –3 ) (96500) = Et 30  60

i = 0.04 A Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

110

Electrochemistry

Solution of Assignment (Set-2)

13. A direct current of 3.0 amp (efficiency 75%) was passed through 400 ml 0.2 M Fe2(SO4)3 solution for a period of 60 min. The resulting solution in cathode chamber was analysed by titrating against acidic KMnO4 solution 20 ml of KMnO4 required to reach the end point. Determine the molarity of KMnO4 solution. Sol. Equivalents of Fe2+ formed =

1 75 3 × 60 × 60 = 0.0839 96500 100

Equivalents of Fe2+ = Equivalents of KMnO4 0.0839 = NKMnO4 × 20 × 10–3

0.0839 × 103 20

NKMnO4 =

NKMnO4 = 4.195 In acidic medium, n-factor for KMnO4 is 5.

MKMnO4 =

4.195 = 0.84 M 5

14. H2O2 can be produced by successive reactions 2NH4HSO4  H2 + (NH4)2S2O8 60% (NH4 )2 S 2O 8  2H2O   2NH4HSO 4  H2O 2

The first reaction is an electrolytic reaction and second is steam distillation. What amount of current would have to be used in first reaction to produce enough (NH4)2S2O8 to yield 100 gm H2O2 per hour? (Assume current efficiency is 75%) Sol. Given (NH4)2S2O8 + H2O 2NH4HSO4 + H2O2 228 g

34g

34 gm of H2O2 is produced by 228 g (NH4)2 S2O8 100 g of H2O2 will be produced by

228 × 100 = 670 g 34

Since efficiency is 60% hence mass of (NH4)2 S2O8 required will be =

670  100 = 1116.66 gm 60

Equivalent mass of (NH4)2 S2O8 may be calculated 2NH4 SO4–  (NH4)2S2O8+2e– ENH  S O = M = 228 = 114 4 2 2 8 2 2

From I- law w=

Ei t F

1116.66 =

i

75  3600  114 100 96500

i = 350.09 A Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

15

Electrochemistry

111

During discharge of a lead storage battery the density of sulphuric acid fell from 1.3 to 1.15 gm/ml. Sulphuric acid of density 1.3 gm/ml is 40% H2SO4 by wt and that of density 1.15 gm/ml is 20% by wt. The battery holds 4.0 L of the acid and volume remained practically constant during the discharge. Calculate the number of amphours for which the battery must have been used. Pb + SO4–2  PbSO4 + 2e– (discharging) PbO2 + 4H+ + SO4–2 + 2e–  PbSO4 + 2H2O (discharging)

Sol. Adding the charging and discharging reactions. We get Pb + PbO2 + 4H+ + 2SO2– 4  2PbSO4 + 2H2O

NH2SO4 = MH2SO4 (since 2SO2– 4 requires 2-electrons) i.e., Normality = Molarity Before discharge 40  1.3  1000 = 5.3 M 98  100

MH2SO4 =

Moles of H2SO4 = 5.3 × 4 = 21.22 After discharge MH2SO4(ii) =

20  1.15  1000 = 2.34 M 98  100

Moles of H2SO4 = (2.34) (4) = 9.38 Moles or equivalents of H2SO4 Used = 21.22 – 9.38 = 11.84 i × t = (11.84) × 96500 = 317 ampere s. 16. A dilute solution of NaCl was placed between two Pt-electrodes 8 cm apart, across which a potential of 4 V was applied. How far would the Na+ move in 2.5 hours? Ionic conductance of Na+ at infinite dilution at 25°C is 50.11 mho cm2. Sol. º Na+ = 50.11 cm2/eq Ionic mobility =

50.11  ºK  = 96500 F

5.19×10–4 cm2 sec–1 volt–1 Potential gradient applied =

4 = 0.5 volt/cm 8

Speed of Na+ = ionic mobility × potential gradient = (5.19 × 10–4) × 0.5 = 0.259 × 10–3 cm/sec Distance traveled by Na+ in 2.5 hr = 0.259 × 10–3 × 2.5 × 60 × 60 = 2.33 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

112

Electrochemistry

Solution of Assignment (Set-2)

17. The resistance of a conductivity cell filled with 0.01 N KCl at 25°C was found to be 500 . The specific conductance of 0.01 N KCl at 25°C is 1.41 × 10–3 –1 cm –1. The resistance of same cell filled with 0.3 N ZnSO4 at 25°C was found to be 69 . Calculate the cell constant, equivalent and molar conductivities of ZnSO4 solution. Sol. K =

1 l × R A

 (1.41 × 10–3) =

1 l × 500 A

l = 1.41 × 10–3 × 500 = 0.705 A

m = =

K  1000 1 l 1000 = × × M R A M 1 1000 × 1.41 × 10–3 × 500 × 69 0.3

= 34.1 –1 cm–1 eq–1.  Molar conductivity = 68.2 –1cm–1mol–1. 18. The equivalent conductance of 0.1 N of H3PO4 at 18°C is 96.5 –1 cm2 eq–1. If °HCl = 378.3, °NaCl = 109, Λ NaH2PO 4  70 Ω –1 cm 2 eq –1 respectively, calculate the degree of dissociation and dissociation constant for

the reaction : H+ + H2PO4–

H3PO4

o o o o Sol. H3PO4 = HCl  NaH2PO4 – NaCl = 378.3 + 70 – 109 = 339.3

H3PO4 96.5  = o = = 0.2844 339.3 H3PO4 K=

C.2 0.1  (.2844)2 = = 1.13 × 10–2 (1 –  ) (1 – .28844)

19. Ag(s) | AgCl (saturated salt), KCl (C = 0.025) || KNO3, AgNO3 (C = 0.2) | Ag The emf of above cell is 0.43 V. (a) Write down the cell reaction. (b) Calculate the solubility product of AgCl. (Antilog (4.2758) = 1.887 × 104) Sol. The given cell Ag(s) |AgCl (Saturated Salt), KCl (C = 0.025 M)||AgNO3(C=0.2)|Ag E = 0.43V (a) Reaction in L.H.C. and R.H.C. Ag Ag+ + e– (Anode) Ag+ + e– Ag (Cathode) o (b) E = E Ag/ Ag – Ag/ Ag

E

Ag / Ag

o = E Ag / Ag –

0.0591 log [Ag+]L 1 0.0591 1 log 1 [Ag ]R

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

EMF of cell = E (Eo

Ag/ Ag

 E=

 E= –

Ag/ Ag

Electrochemistry

E

 Eo

Ag / Ag

Ag / Ag

)



0

[Ag ]L 0.0591 – log [Ag ]R 1

[Ag ]L 0.0591 log (0.2) 1

[Ag ] 0.43 = log 0.0591 (0.2)



113

0.2 [Ag ]

0.2 0.43 = log [Ag ] 0.0591

or

= 1.887 × 106 ;

[Ag+] = 1.05 × 10–10 M.

Therefore, [Ag+] = 1.05 × 10–10 M K   Cl– KCl  0.025 0.025

 Ksp (AgCl) = (0.025) (1.05 × 10–10)  Ksp = 2.64 × 10–10 20. Pt | H2(1 bar), H+ || KCl (1.0 M saturated) | Hg2Cl2 | Hg was used to measure the pH of 0.05 M acetic acid in 0.04 M CH3COONa. Calculate the cell potential.

K a(CH3 COOH)  1.8  10 5 , EHg Cl 2

2

/Hg, Cl–

 0.28 V

[Acid] 0.05 = 1.8 × 10–5 × = 2.25 × 10–5 [Salt] 0.04

Sol. [H+] = Ka × o ECell = (EHg

2Cl2 /Hg, Cl



– Eo 

H /H2

)–

0.059 [H ] log 1 1

0.059 log 2.25 × 10–5 1 = 0.28 + 0.275 = 0.555 V

ECell = 0.28 – ECell 21.

V

Ag

0.5 M AgNO3

Zn

Salt bridge

Zn(NO3)2 (A)

(B)

(a) If the cell emf is – 1.58 V, what is the concentration of Zn+2? (b) If NH3 added to half cell A, how emf of cell will change? E Ag /Ag  0.8 V, E Zn 2 /Zn  0.76 V

Antilog (0.6768) = 1.4768

Sol. (a) [Zn+2] = 0.054 M; (b) emf increases Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

114 22.

Electrochemistry

Solution of Assignment (Set-2)

ECu2 /Cu  0.34 V, EZn/Zn2  0.76 V A cell formed by the combination of Cu and Zn. (a) When CuSO4 is added to Cu+2 compartment what is the effect on emf of cell? (b) When NH3 is added to Cu+2 compartment what is the effect on emf of cell? (c) When ZnSO4 is added to Zn+2 compartment what is the effect on emf of cell? (d) When Zn+2 is diluted what is the effect on emf of cell?

o o Sol. ECu2 /Cu = 0.34V; E Zn/ Zn2 = 0.76 V

 EMF of the cell is o o Eº = E Zn/ Zn2 + ECu2 /Cu

 (0.34) + (0.76) = 1.1V (a) According to Nernst equation

[Zn2 ] 0.0591 log E = Eº – [Cu2 ] 2 [Cu2+] increases Hence log terms will decreases and therefore EMF increases. (b) When NH3 is added it combines with Cu2+ and, hence, [Cu2+] decreases 4NH3 + Cu2+  [Cu (NH3)4]2+. With decrease in [Cu2+], log tem increase hence EMF decreases. (c) [Zn2+] will increase as result of which EMF decreases. (d) Zn2+ dilution means concentration decreases and EMF increases.







Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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