fuel cell

Electrochemistry

CHAPTER

4

ELECTROCHEMISTRY

LEARNING OBJECTIVES (i) Describe an electrochemical cell and differentiate between galvanic and electrolytic cells. (ii) Apply Nernst equation for calculating the emf of galvanic cell and define standard potential of the cell. (iii) Derive relation between standard potential of the cell, Gibbs energy of cell reaction and its equilibrium constant. (iv) Define resistivity , conductivity and molar conductivity of ionic solutions. (v) Justify the variation of conductivity and molar conductivity of solutions with change in their concentration and define  0m (molar conductivity at zero concentration or infinite dilution). (vii) Enunciate Kohlrausch law and learn its applications. INTRODUCTION Electrochemistry deals with the chemical changes produced by electric current and with the production of electricity by chemical reactions. Many metals are purified or are plated onto jewelry by electrochemical methods. Digital watches, automobile starters, calculators, and pacemakers are just a few devices that depend on electrochemically produced power. Corrosion of metals is an electrochemical process. We learn much about chemical reactions from the study of electrochemistry. The amount of electrical energy consumed or produced can be measured quite accurately. All electrochemical reactions involve the transfer of electrons and are therefore, oxidation-reduction reactions. The sites of oxidation and reduction are separated physically so that oxidation occurs at one location while reduction occurs at the other. Electrochemical processes required some method of introducing a stream of electrons into a reacting chemical system and some means of withdrawing electrons. In most applications the reacting system is contained in a cell, and an electric current enters or exists by electrodes. Electrodes are surfaces upon which oxidation or reduction half-reactions occur. The cathode is defined as the electrode at which reduction occurs as electrons are gained by some species. The anode is the electrode at which oxidation occurs as electrons are lost by some species. ELECTROCHEMICAL CELLS : Cell is a system or arrangement in which two electrodes are fitted in the same electrolyte or in two different electrolytes which are joined by a salt bridge. Cells are of two types. (a) Electrolytic cell (b) Galvanic or voltaic cell (a) Electrolytic cell : It is a device in which electrolysis (chemical reaction involving oxidation and reduction) is carried out by using electricity or in which conversion of electrical energy into chemical energy is done. (b) Galvanic or voltaic cell : It is a device in which a redox G e reaction is used to convert chemical energy into Key electrical energy, i.e., electricity can be obtained with the help of oxidation and reduction reaction. The Å Current chemical reaction responsible for production of Salt bridge Cl¯ K+ Copper rod Zinc rod electricity takes place in two separate compartments. (Cathode) (Anode) Each compartment consists of a suitable electrolyte KCl in solution and a metallic conductor. The metallic Agar-agar conductor acts as an electrode. The compartments containing the electrode and the solution of the Zinc Copper is Cotton electrolyte are called half-cells. When the two Dissolves deposited plugs compartments are connected by a salt bridge and electrodes are joined by a wire through galvanometer 1 M ZnSO4 1 M CuSO4 the electricity begins to flow. This is the simple form of voltaic cell. Anode : Zn  Zn2+ + 2e– Cathode : Cu+2 + 2e–  Cu (Ist half cell reaction) (IInd half cell reaction) Total cell reaction : Zn(s) + Cu+2(aq)  Zn+2(aq.) + Cu(s) Gyaan Sankalp

1

Electrochemistry Salt bridge : It is U-shaped glass tube filled with a gelly like substance, agar-agar, mixed with an electrolyte like KCl, KNO3, NH4NO3, etc. It performs following functions. (i) It prevents voltage drops, i.e. prevents junction potential. (ii) It allows flow of current by completing the circuit, i.e., migration of anion from anode to cathode half cell. (iii) It prevents accumulation of charges and maintains the electrical neutrality of solutions. (iv) Nature of junction potential is opposite to EMF of cell. (v) Salt bridge form a electrolyte which have almost same movability of cation and anion. Representation of Galvanic cell : Galvanic cell is a combination of two half cells, namely, oxidation half cell and reduction half cell. If M represents the symbol of the element and Mn+ represents its cation (i.e., its oxidised state) in solution, then Oxidation half cell is represented as M/Mn+(C) Conventional Current Reduction half cell is represented as Mn+(C)/M e Electron Flow In both the notations C refers to the molar concentration of the ions in solution. Conventionally, a cell is represented by writing the cathode on the right hand side and anode on the left hand side. The two vertical lines are put between the two half cells which indicate salt 2+ 2+ Saturated bridge. Sometimes the formula of the electrolyte Zn / Zn (1M) Cu (1M)/ Cu KCl used in the salt bridge is also written below the vertical lines. For example, zinc-copper sulphate cell is represented as follows : ANODE (—) CATHODE (+) 2 2 Zn (s) | Zn (aq) || Cu (aq) | Cu (s) anode

salt bridge

Cathode

OXIDATION HALF

ELECTRODE POTENTIAL : The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. EMF = Eright(cathode) – Eleft(Anode) or simply as EMF = Ecation – Eanion The potential of individual half-cell cannot be measured.

SALT BRIDGE

REDUCTION HALF

H2 gas under 1 atm pressure

Electrode 1MHCl solution Platinum coated with platinum black

Fig. Standard hydrogen electrode (SHE) We can measure only the difference between the two half-cell potentials that gives the emf of the cell. If we arbitrarily choose the potential of one electrode (half-cell) then that of the other can be determined with respect to this. According to convention, a halfcell called standard hydrogen electrode represented by Pt (s) | H2(g) | H+(aq), is assigned a zero potential at all temperatures corresponding to the reaction 1 H (g) 2 2 The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity. This implies that the pressure of hydrogen gas is one atm. and the concentration of hydrogen ion in the solution is one molar. Thus, we can define standard electrode potential as the potential of an electrode relative to a standard hydrogen electrode at 298K, 1 atm pressure when the concentration of the ion taking part in the electrode reaction is 1 mol L–1. If we consider left half cell as SHE than the measured emf of the cell equals the standard electrode potential of the right half cell. The measured emf of the cell : Pt (s) | H2(g , 1 atm) | H+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction : Cu2+ (aq, 1M) + 2e–  Cu(s)

H+ (aq) + e– 

2

Gyaan Sankalp

Electrochemistry Similarly, the measured emf of the cell : Pt(s) | H2(g , 1 atm) | H+ (aq, 1 M) || Zn2+ (aq, 1M) | Zn is -0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e–  Zn(s) The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid it is oxidised by nitrate ion and not by hydrogen ion. The negative value of the standard electrode potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce hydrogen ions). Following above convention, the half reaction for the zinc-copper cell can be written Zn(s) Zn2+ (aq, 1M) + 2e– (oxidation, anode) left Cu2+ + 2e– Cu(s) (reduction, cathode) right (aq. 1M) –––––––––––––––––––––––––– Cu2+ + Zn (s) Zn2+ (aq) + Cu (s) (overall reaction) emf of the cell = Ecell  ER  E L  0.34  (0.76)  1.10V

Electrode Reduction Reaction Li+ + e–  Li K+ + e–  K Ba2+ + 2e– Ba Ca2+ + 2e– Ca Na+ + e–  Na Mg2+ + 2e–  Mg Al3+ + 3e–  Al Mn2+ + 2e–  Mn Zn2+ + 2e–  Zn Cr3+ + 3e–  Cr Fe2+ + 2e–  Fe Cd2+ + 2e–  Cd Ni2+ + 2e–  Ni Sn2+ + 2e–  Sn Pb2+ + 2e–  Pb 2H+ + 2e–  H2 Cu2+ + 2e–  Cu I2 + 2e–  2I– Hg2+2 + 2e–  2Hg Ag+ + e–  Ag Hg+2 + 2e–  Hg Br2 + 2e–  2 Br– Cl2 + 2e–  2 Cl– Au3+ + 3e–  Au F2 + 2e–  2F–

Standard Reduction Potential E° (in Volt) – 3.05 – 2.93 – 2.90 – 2.87 – 2.71 – 2.37 – 1.66 – 1.18 – 0.76 – 0.74 – 0.44 – 0.40 – 0.25 – 0.14 – 0.13 0.00 0.34 0.53 0.79 0.80 0.85 1.08 1.36 1.50 2.87

Increasing strength as reducing agent

Element Li K Ba Ca Na Mg Al Mn Zn Cr Fe Cd Ni Sn Pb H2 Cu I2 Hg Ag Hg Br2 Cl2 Au F2

Increasing strength as oxidizing agent

ELECTRO CHEMICAL SERIES : The arrangement of various elements in order of increasing values of standard reduction potential is called electro chemical series. Standard reduction potentials at 298 K (Electrochemical Series)

Strength of oxidising and reducing agents : Standard electrode potentials are useful in determining the strengths of oxidizing and reducing agents under standard-state conditions. Because electrode potentials are reduction potentials those reduction half-reactions in the table with the larger (that is more positive) electrode potentials have the greater tendency to go left to right as written. A reduction half-reaction has the general form. Oxidised species + ne–  reduced species. The oxidized species acts as an oxidizing agent. Consequently, the strongest oxidizing agents in a table of standard electrode potentials are the oxidized species corresponding to half-reactions with the largest (most positive) E° values. Gyaan Sankalp

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Electrochemistry Those reduction half-reactions with lower (that is, more negative) electrode potentials have a greater tendency to go right to left. That is, Reduced species  oxidized species + ne– The reduced species acts as a reducing agent. Consequently, the strongest reducing agents in a table of standard electrode potentials are the reduced species corresponding to half-reactions with the smallest (most negative) E° values. It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. Applications of Electrochemical Series : Reactivity of metals : (i) Alkali metals and alkaline earth metals having high –ve values of SRP which are chemically active. These react with cold water and evolve hydrogen and readily dissolve in acids. (ii) Metals like Fe, Pb, Sn, Ni, Co, etc. do not react with cold water but react with steam to evolve hydrogen. (iii) Metals Li, Be, Cu, Ag, and Au which lie below hydrogen are less reactive and do not evolve hydrogen from water. (b) Electropositive character of metals : Electropositive character of metals decreases from top to bottom. (c) Displacement reactions : To predict whether a given metal will displace another, from its salt solution. The metal having low SRP will displace the metal from its salt’s solution which has higher value of SRP. (d) Reducing power of metals : Reducing nature decreases from top to bottom in the electrochemical series. (e) Oxidizing nature of non-metals : Oxidizing nature increases from top to bottom in the electrochemical series. (f) Thermal stability of metallic oxides : The thermal stability of the metal oxide decreases from top to bottom. (g) Products of electrolysis : The ion which is stronger oxidizing agent is discharged first at cathode. (h) Corrosion of metals : Corrosion is defined as the deterioration of a substance because of its reaction with its environment. The corrosion tendency increases from top to bottom. (i) For a reaction that is spontaneous at standard conditions, E° must be positive. Example 1 : Electrode potential of the metals in their respective solution are provided. Arrange the metals in their increasing order of reducing power. K+/K = – 2.93V, Ag+/Ag = + 0.80V, Hg+/Hg = +0.79V, Mg2+/Mg = – 2.37V, Cr3+/Cr = –0.74V Sol. We know that the reducing power of a metal depends upon its tendency to lose electrons. Thus lower the reduction potential, more the tendency to get oxidized and thus more will be the reducing power. Hence increasing order of reducing power is: Ag < Hg < Cr < Mg < K Example 2 : Using the standard electrode potentials predict the reaction, if any, that occurs between the following : (a) Fe3+(aq) and I–(aq) (b) Ag+(aq) and Cu(s) (c) Ag(s) and Fe3+(aq) (d) Br2(aq) and Fe2+(aq) (a)

  Given : EFe3 /Fe2  0.77V, E I

Sol. (a) Here I–(aq) loses electrons and

 2/I Fe3+

 0.54V ; E   0.80V, E 2+  0.34V ; E Ag / Ag Cu /Cu Br

2 / Br

(aq)

gains electrons. Thus

Oxidation half cell reaction 2I–  I2 + 2e–, E° = – 0.54V Reduction half cell reaction [Fe3+ + e–  Fe2+] × 2 ; E° = +0.77V ––––––––––––––––––––––––––––––––––––––––––––––––––––– Overall reaction 2I– + 2Fe3+  I2 + 2Fe2+, E0cell =  0.23V Since E°cell is +ve, the reaction is spontaneous i.e., the reaction does take place. (b) Here Cu(s) loses electrons and Ag+(aq) gains electrons. Thus Oxidation half cell reaction:

Cu  Cu2+ + 2e–,

E° = – 0.34V

[Ag+ + e–  Ag] × 2, E° = +0.80V –––––––––––––––––––––––––––––––––––– Overall reaction Cu + 2Ag+  Cu2+ + 2Ag, E°cell= 0.46V Since E°cell is +ve, the reaction is spontaneous Reduction half cell reaction

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Gyaan Sankalp



 1.08V

Electrochemistry (c) Here Ag loses electrons and

Fe3+

(aq)

gains electrons. Thus

Ag  Ag+ + e–, E° = – 0.80V

Oxidation half cell reaction

Fe3+ +e–  Fe2+, E° = +0.77V –––––––––––––––––––––––––––––––––––– Overall reaction Ag + Fe3+  Ag+ + Fe2+, E°cell = – 0.03V Since E°cell is –ve, the reaction is non-spontaneous, i.e., reaction does not take place. Reduction half cell reaction

(d) Here Fe2+(aq) loses electrons and Br2(aq) gains electrons. Thus 2Fe2+  2Fe3+ + 2e–,

Oxidation half cell reaction

E° = – 0.77

Br2 + 2e–  2Br–, E° = +1.08V –––––––––––––––––––––––––––––––––––––– Overall reaction 2Fe2+ + Br2  2Fe3+ + 2Br–, E°cell = +0.31V Since E°cell is +ve, the reaction is spontaneous, i.e., the reaction does not take place. Reduction half cell reaction

TRY IT YOURSELF Q.1 Calculate e.m.f. of the half cell given below : (a) Fe | FeSO4 ; Eº = 0.44 V (b) Zn | ZnSO4 Eº = 0.76 V (c) Cu | Cu(NO3)2 ; Eº = – 0.34 V 0.1M 0.2 M Q.2 A galvanic cell is constructed with 2 metals P and Q . Electrolysis used in the galvanic cell are sulphates of the metals. If EºP+n | P = a and EºQ+n | Q = b and a > b then find out : (a) Anode of the cell (B) Cathode of the cell (C) Reaction at anode (d) Reaction at cathode Q.3 Which one of the following is different from others – (A) Daniel cell (B) Voltaic cell (C) Galvanic cell (D) Electrolytic cell Q.4 The standard reduction potentials of 4 elements are given below. Which of the following will be the most suitable reducing agent– I = – 3.04 V II = – 1.90 V III = 0 V IV = 1.90 V (A) III (B) II (C) I (D) IV Q.5 The reaction Zn2+ + 2e– Zn has a standard potential of – 0.76 V. This means – (A) Zn can't replace hydrogen from acids (B) Zn is reducing agent (C) Zn is an oxidising agent (D) Zn2+ is a reducing agent Q.6 Choose the correct statement from the following which is related to the electrochemical series(A) Electrochemical series is not the arrangement of metals and ions according to their reactivity. (B) The metal ions at the top of the electrochemical series are highly electronegative (C) Strongly electropositive metals can displace weakly electropositive metals from their salt solution (D) All metals above hydrogen in the series do not displace hydrogen from dilute acids. Q.7 Which one of the following reaction occurs at the cathode – (A) 2OH–  H2O + O + 2e– (B) Ag  Ag+ + e– (C) Fe2+  Fe3+ + e– (D) Cu2+ + 2e–  Cu

ANSWERS (1) (a) 0.47 V (b) 0.78 V (c) – 0.33 V (3) (D) (4) (C)

(2) (a) Q (b) P (c) Q  Q+n + ne– (d) P+m + me–  P (5) (B) (6) (C) (7) (D)

NERNST EQUATION : The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as m1A + m2B +..................  n1X + n2Y + ........... .............. (1) occurring in the cell, the Gibbs free energy change is given by the equation a nx1  a ny 2

G = G° + 2.303 RT log10

m

m

a A1  a B 2

.............. (2) Gyaan Sankalp

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Electrochemistry where ‘a’ represents the activities of reactants and products under a given set of conditions and G° refers to free energy change for the reaction when the various reactants and products are present at standard conditions. The free energy change of a cell reaction when the various reactants and products are present at standard conditions. The free energy change of a cell reaction it related to the electrical work that can be obtained from the cell, i.e., G = –nFEcell and G° = –FE°. (F is Faraday constant  96500C) On substituting these values in Eq. (2), we get –nFEcell = –nF E 0cell

or Ecell = E 0cell –

+ 2.303RT log10

a nx1  a ny 2 ... m

m

a A1  a B 2 ...

a nx1  a ny 2 ... 2.303RT log10 m m nF a A1  a B 2 ...

.............. (3)

.............. (4)

2.303RT [Products] log10 .............. (5) nF [Reactants] This equation is known as Nernst equation. Putting the values of R = 8.314 JK–1 mol–1,T = 298 K and F = 96500 C, Eq.(5) reduces to

E = E0 –

E = Eº –

0.0591 [Products] log10 n [Reactants]

.............. (6)

0.0591 [P] log10 n [R] Potential of single electrode (Anode) : Consider the general oxidation reaction, M  Mn+ + ne–

E = Eº –

Applying Nernst equation, Eox = E 0ox –

[M n  ] 0.0591 log10 [M] n

where Eox is the oxidation potential of the electrode (anode), E 0ox is the standard oxidation potential of the electrode. The concentration of pure solids and liquids are taken as unity. 0.0591 log10 [Mn+] n Potential of single electrode (Cathode) : Consider the general reduction reaction. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Mn+ + ne–  M Applied Nernst equation,

Eox = E 0ox –

EReduction= E0Reduction –

0.0591 0.0591 [M] log10 [M n  ] log10 = E0Reduction + n n n [M ]

Emf of the cell : Cell potential depend upon potential of cathode and anode. Ecell = Eanode + Ecathode Ecell = Eox + Ered  concentration of Anode  0.0591 log10   n  concentration of Cathode  Let us consider an example, in which the values of n for the two ions in the two half-cells are not same. For example, in the cell Cu | Cu2+ || Ag+ | Ag The cell reaction is Cu(s) + 2Ag+  Cu2+ + 2Ag The two half-cell reactions are: Cu Cu2+ + 2e– Ag+ + e–  Ag The second equation is multiplied by 2 to balance the number of electrons. 2Ag+ + 2e–  2Ag

= (E 0ox  E 0red ) –

Eox = E 0ox – 6

0.0591 log10[Cu2+] 2

Gyaan Sankalp

Electrochemistry Ered = E 0red

0.0591 log10[Ag+]2 2

Ecell = Eox + Ered = E 0ox + E 0red –

[Cu 2  ] [Cu 2  ] 0.0591 0.0591 0 E log10 = – log cell 10 [Ag  ]2 [Ag  ]2 2 2

Example 3 : The standard electrode potentials of the electrode Cu2+|Cu and Ag+|Ag are 0.34V and 0.7991V respectively. What would be the concentration of Ag+ in a solution containing 0.06M of Cu2+ ion such that both the metals can be deposited together. Assume that activity coefficients are unity and both silver and copper do not dissolve among themselves. Sol. The individual reactions are : Cu2+ + 2e–  Cu(s) ; Ag+ + e–  Ag(s) The electrode potentials given by Nernst equation E(Cu 2  | Cu)  E 0 

(0.0591) 1 0.0591 1 log log = 0.037 – = 0.037 – 0.036 = 0.301 2  2 0.06 2 [Cu ]

E(Ag  | Ag)  0.7991 

0.0591 1 log 1 [Ag  ]

Two metals will be deposited together when the electrode potentials are equal i.e. 0.301  0.7991  0.0591log 1 

 108.428

[Ag ]

or

1 

[Ag ]

i.e. log

1 

[Ag ]



0.7991  0.301  8.428 0.0591

[Ag+] = 10–8.428 = 0.37 × 10–8 mol dm–3

THERMODYNAMICS OF THE CELLS : The e.m.f. of the cell is related to free energy by equation G = – nFE

…...(1)

 G  Now,  T   S P

So

 G   E   E      nF    S or S  nF   T P T P  T  P

...…(2)

The enthalpy of the cell reaction will be  E  H = G + TS = nFE  TnF    T  P

...…(3)

The thermodynamic quantities of the cell reaction can be calculated by equations (1), (2) and (3) provided one knows the emf of cell and its dependence on temperature. The heat effects in the system can be calculated as follows. If the process is irreversible (i.e. by mixing the reactants together), heat flow to the system can be given by the reaction , H = QP. If the process is reversible the heat flow to system is given by QP = TS. CONDITION OF EQUILIBRIUM : When Ecell = 0.0 V, i.e., no potential difference is obtained between the terminals of cell (battery), the cell reaction in such a state is said to be in equilibrium. So in such cases , When, Q = Keq = equilibrium constant. 0 E cell  E cell 

E 0cell 

0.059 0.059 log Q ; 0.0  E 0cell  log K eq n n

0.059 log K eq n Gyaan Sankalp

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Electrochemistry SOLUBILITY PRODUCT : The solubility product of sparingly soluble salt MX can be discussed in terms of the equilibrium of the kind   MX(s)   M(aq)  X (aq)

Since the activity of the pure solid is always unity, the equilibrium constant of the solubility product can be written. K sp  a

a

…...(1)

M x

E  E 

RT a M a X  ln nF a MX

Under equilibrium conditions the emf of the cell will be zero i.e. , E = 0 and also the activity of pure solid is unity. Further Ksp  a

M

a

X

, the above equation at 25°C can be written as :

E  n ....…(2) 0.0591 The solubility product thus can be calculated from the standard emf of one cell, formed in such a way that the final reaction is the type given above. Example 4 : (a) Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 298 K ; assuming CuSO4 to be completely dissociated. The standard electrode potential of Cu2+ | Cu system is + 0.34 volts at 298 K. (b) At what concentration of copper ions will this electrode have a potential of zero volt? Sol. (a) The reduction electrode reaction is Cu2+(aq.) + 2e–  Cu(s) Thus here, n = 2 ; E° = + 0.34 volts, [Products] = 1 , [Cu2+] = 0.1 M Substituting these values in the Nernst equation, log K sp 

[Products] 0.059 0.059 1 log = 0.34 – log = 0.34 – 0.0295 log 10 = 0.3105 volt [Reactants] 2 2 0.1 (b) Here, E = 0 Volt, [Cu2+] = ? Applying again Nernst equation E = E° –

[Cu] 1 0.059 0.059 0.34 log ; 0 = 0.34 – log ; log [Cu2+] = – 2 [Cu ] n 2 [Cu 2  ] 0.0295 On solving, we get [Cu2+] = 2.95 × 10–12 M Example 5 : Calculate the equilibrium constant for the reaction , Zn2+ + 4NH3  [Zn(NH3)4]2+ E° (Zn2+/Zn) = – 0.763 and E° [Zn(NH3)4]2+/Zn + NH3) = 1.03V Sol. The electrode reactions for the given electrodes, can be written as

E = E° –

Zn2+ + 2e–  Zn

E1 = – 0.763V

............ (1)

[Zn(NH3)4]2+ + 2e–  Zn + 4NH3

E1 = – 1.03V

............ (2)

Reaction (1) – (2) Zn2+ + 4NH3  [Zn(NH3)4]2+ The standard emf of this reaction = E1  E2 = – 0.763V – (–1.03) = 0.267V According to Nernst equation E  E 

a RT [Zn(NH3 )4 ]2  ln 2F a 2  a NH3 Zn

If the process is equilibrium, E = 0 at 25°C,

log K  8

0.0591 log K  0.267 2

(0.267)(2)  9.036 ; K  100.36109  1.09  109 (0.059)

Gyaan Sankalp

Electrochemistry Example 6 : Calculate the Ksp of AgI by forming proper cell. Given E

I / Ag(s) / Ag

 0.151V and E

Ag + / Ag

 0.7991V

Sol. The cell can be written as : Ag | Ag+ || I– | AgI | Ag At left electrode Ag(s)  Ag+ + e– At right electrode AgI(s) + e–  Ag(s) + I– AgI(s)  Ag+ + I–

E° = 0.7991V E° = – 0.151V

The standard emf of the cell is E  ER  EL  0.151  0.7991 = – 0.9501V log K sp  

(0.9501)(1)  16.1 ; Ksp  1016.1  7.94  1017 0.059

TRY IT YOURSELF Q.1 The standard electrode potentials of the two half cells are given below : Ni2+ + 2e– = Ni ; Eº = – 0.25 Volt Zn2+ + 2e– = Zn ; Eº = – 0.77 Volt The voltage of cell formed by combining the two half cells would be (A) – 1.02 volt (B) + 0.52 volt (C) + 1.02 volt (D) – 0.52 volt Q.2 Calculate the standard free energy change for the reaction 2Ag + 2H+  H2 + 2Ag+ , Eº for Ag+ + e–  Ag is 0.80 V (A) + 154.4 kJ (B) + 308.8 kJ (C) – 154.4 kJ (D) – 308.8 kJ Q.3 The correct representation of Nernst's equation is – (A) E M n  / M = Eº (C) E M n  / M = Eº

Mn / M

Mn / M





n log (Mn+) 0.0591

Q.4 The nernst equation, E = Eº –

Q.5 Q.6 Q.7 Q.8

0.0591 log (Mn+) n

(B) E M n  / M = Eº

Mn / M



0.0591 log (Mn+) n

(D) None of these

RT ln Q indicates that the equilibrium constant Kc will be equal to Q when nF

RT =1 (C) E = 0 (D) Eº = 1 nF The cell Pt(H2) (1atm) | H+ (pH= X) | | Normal calomel electrode has e.m.f. of 0.67 V at 25ºC. Calculate pH of solution. The oxidation potential of calomel electrode on H scale is – 0.28 V . The e.m.f. of the cell obtained by combining Zn and Cu electrodes of a Daniel cell with N calomel electrodes are 1.083 V and – 0.018V respectively at 25ºC. If the potential of N calomel electrode is – 0.28 V , find e.m.f. of Daniel cell. What is the e.m.f. of a cell containing two H electrodes, the negative one in contact with 10–8 molar OH– and the positive one in contact with 10–3 molar H+ ? Calculate the potential of a cell in which H electrodes are immersed in a solution with a pH of 3.5 and in a solution with a pH of 10.7 at 30 ºC .

(A) E = Eº

(B)

(1) (B) (5) 6.6

(2) (C) (6) 1.1 V

ANSWERS (3) (A) (7) – 0.295 V

(4) (C) (8) 0.425 V

CONDUCTANCE OFELECTROLYTIC SOLUTIONS : Conductance (C) : The amount of electric current can be passed through the solution is called conductance. Conductance is inverse to resistance. 1 1 or C = ; Unit of Conductance is inverse to ohm it represent as mho. Resistance R Factors affecting electrolytic conduction : Nature of the electrolyte : The conductance of solution is depend on nature of electrolyte. Generally strong electrolytes ionize almost completely in the solution and hence conduct electricity to a large extent whereas weak electrolytes ionize to a small extent.

Conductances = (i)

Gyaan Sankalp

9

Electrochemistry (ii) (iii) (iv) (v) (vi) (viii)

Concentration of the solution : The conductance of solution increase with increase the dilution. (because rate of dissocation increase with increase the dilution for weak electrolyte so no. of ions in solution increases and movability of ion also increase). Temperature : Conductances increase with increase the temperature because the all attraction force will be decrease. Degree of ionization : Conductance of solution increase with increase the degree of ionization. Interionic attractions : Movability of ion decrease with increase the interionic attractions so conductance of electrolyte decrease. Viscosity : Movability of ion decreases with increase the viscosity so conductance of electrolyte decrease. Solvation of ions : Movability of ion decrease with increase the solvation so conductance of electrolyte decreases. Specific Conductivity (or simply called conductivity) : Ohm’s law is valid for electrolytic solution so that resistance of electrolyte R is directly proportional to its length () and inversely proportional to its area of cross section (a).   or R =  a a where  = distance between two electrode and a= area of cross-section of electrode where  is a constant of proportionality, called specific resistance or resistivity. Its value depends upon the material of the conductor. Inverse of resistance is called conductance or observed conductivity while inverse of specific resistance is called specific conductivity or conductivity , we can write

i.e.,

R

1 1    Observed conductivity Specific conductivity a

 a or Specific conductivity = conductivity × cell constant

or Specific conductivity = conductivity ×

 for a cell is constant and it known as cell constant it denoted by x a Now if  = 1 cm and a = 1 sq. cm, then Specific conductivity = Observed conductivity If the volume of the solution is V cm3, the specific conductivity of such a solution at this dilution V is written as . Units :

ratio

(i) Resistivity () = R

a (cm 2 ) = ohm = ohm cm or  cm  cm

(ii) Specific conductivity (v) =

1 1 = = ohm–1 cm–1 or –1 cm–1 or S cm–1  ohm cm

Cell constant : We know that R =

 1  1  = a  Ra

1  1   ( = cell constant cm–1)   G, conductance R Ra a The conductivity of the solution is measured in a cell known as conductivity cell. Since in such cells the electrodes may be exactly 1 cm apart or may not have an area of 1 cm². Therefore we calculate a factor called constant. (/a) for these cells. Also, Specific conductivity = cell constant × observed conductivity

v =

conductivity conductance Equivalent conductivity : Equivalent conductivity is the conducting power of all the ions produced by one g-equivalent i.e. one equivalent of an electrolyte in a given solution. The equivalent conductivity may, therefore, be defined as the conductivity which is observed when two sufficiently large electrodes are dipped into solution at such a distance so as to enclose in between them the entire volume of solution containing one equivalent of the electrolyte. It is denoted by the symbol . Let one equivalent of an electrolyte is dissolved in V mL solution. Then all the ions produced by 1 equivalent of the electrolyte will be present in this V mL solution. So, the conductance of this V c.c. solution will be the equivalent conductance of the electrolyte i.e. eq = Conductivity of V c.c. solution containing one equivalent of the dissolved electrolyte. = Conductivity of 1 c.c solution  V =   V

or cell constant =

10

Gyaan Sankalp

Electrochemistry Where V = volume of solution in c.c containing 1 equivalent of the electrolyte If C be the normality of solution i.e. concentration of electrolytic solution in equivalent/L, then 1000 1000   = C C Unit of : Ohm–1 cm–1  cm3 i.e. Ohm–1 cm2 or –1cm2 Molar Conductivity : The recent trend is to describe electrolytic conductivity in terms of molar conductivity which is defined as the conductivity of solution due to all the ions produced by one mole of the dissolved electrolyte in a given solution. It is denoted by the symbol m m and  are inter-related as

V=

1000 C Where V = Volume of solution in c.c. containing one mole of the electrolyte and C = Concentration of solution in mole L–1 i.e. molarity The above inter-relationship may also be expressed as Unit of m : –1cm2mol–1 In SI system it is S m2 mol–1 Relation between eq and  m : m = n factor × eq M Where n = n-factor of the electrolyte = total charge carried by either ion = E

m =   V m =

Determination of Conductance (, eq and m)  = Observed conductivity 

 a

For a given conductivity cell in a given experiment ,

 = constant called cell constant (x). a

1 x Observed resistance The resistance of a solution is determined by Wheatstone bridge method using a meter bridge the conductivity cell remains dipped in the test solution. The current used is AC. Variation of molar conductivity with concentration : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in  on dilution of a solution is more than compensated by increase in its volume. When concentration approaches zero, the molar

Thus,  = conductance  x =

conductivity is known as limiting molar conductivity and is represented by the symbol  0m . The variation in m with concentration is different for strong and weak electrolytes. The m vs C plot of strong electrolyte being linear it can be extrapolated to zero concentration. Thus, m values of the solution of the test electrolyte are determined at various concentrations the concentrations should be as low as good. m values are then plotted against C when a straight line is obtained. This is the extrapolated to zero concentration. The point where

m0 Stron g el ectrol yte

m Weak e lectro lyt e

C

the straight line intersects m axis is  0m of the strong electrolyte. However, the plot in the case of weak electrolyte being non linear, shooting up suddenly at some low concentration and assuming the shape of a straight line parallel to m axis. Hence extrapolation in this case is not possible. Thus ,  of a weak electrolyte cannot be determined experimentally. It can , however , be done with the help of Kohlrausch’s law. KOHLRAUSCH LAW OF INDEPENDENT MIGRATION OF IONS : Kohlrausch determined  values of pairs of some strong electrolytes containing same cation say KF and KCl , NaF and NaCl etc., and found that the difference in  values in each case remains the same . 0m (KCl)  0m (KF)  0m (NaCl)  0m (NaF) Gyaan Sankalp

11

Electrochemistry He also determined  values of pairs of strong electrolytes containing same anion say KF and NaF , KCl and NaCl etc. and found that the difference in  values in each case remains the same . 0m (KF)  0m (NaF)   0m (KCl)  0m (NaCl) This experimental data led him to formulate the following law called Kohlrausch’s law of independent migration of ions. At infinite dilution when dissociation is complete , every ion makes some definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion which with it is associated and that the molar conductance at infinite dilution for any electrolyte is given by the sum of the contributions of the two ions . Thus 0m   0   0

Where  0 is the contribution of the cation and  0 is the contribution of the anion towards the molar conductance at infinite dilution. These contributions are called molar ionic conductances at infinite dilution. Thus,  0 is the molar ionic conductance of cation and  0 is the molar ionic conductance of anion, at infinite dilution. The above equation is, however, correct only for binary electrolyte like NaCl, MgSO4 etc. For an electrolyte of the type of AxBy, we have : 0m  x 0  y 0 Application of Kohlrausch’s Law :

(i)

Determination of 0m of a weak electrolyte : In order to calculate 0m of a weak electrolyte say CH3COOH , we determine experimentally 0m values of the following three strong electrolytes : (a) A strong electrolyte containing same cation as in the test electrolyte , say HCl (b) A strong electrolyte containing same anion as in the test electrolyte , say CH3COONa (c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl. 0m of CH3COOH is then given as : 0m (CH3COOH) = 0m (HCl) + 0m (CH3COONa) – 0m (NaCl) Proof : 0 0m (HCl) =  H   Cl

…I

0 0 0m (CH3COONa) =  CH3COO    Na 

… II

0 0 0m (NaCl) =  Na    Cl 

… III

Adding equation (I) and equation (II) and subtracting (III) from them : 0(HCl)   0(CH3COONa)  0(NaCl)   0

(H  )

(ii)

 0

(CH3COO0 )

 0(CH 3COOH)

Determination of degree of dissociation () : 

 No. of molecules ionised  m total number of molecules dissolved 0m

(iii) Determination of solubility of sparingly soluble salt : The specific conductivity of a saturated solution of the test electrolyte (sparingly soluble) made in conductivity water is determined. From this the specific conductivity of conductivity water is deducted. The molar conductance of the saturated solution is taken to be equal to 0m as the saturated solution of a sparingly soluble salt is extremely dilute. Hence from equation. 0m =

1000 , Where C is the molarity of solution and hence the solubility.. C

(iv) Determination of ionic product of water : From Kohlrausch’s law , we determine 0m of H2O where 0m is the molar conductance of water at infinite dilution when one mole of water is completely ionised to give one mole of H+ and one mole of OH– ions i.e. 12

Gyaan Sankalp

Electrochemistry

0H + 0OH 

0m (H2O) =

Again using the following m 

 m 

  1000 , where C = molar concentration i.e. mol L–1 or mol dm–3 C  , where C = concentration in mol m–3 C

Assuming that  m differs very little from 0m 0m =

  C= 0 C m

Specific conductance () of pure water is determined experimentally. Thereafter, molar concentration of dissociated water is determined using the above equation . Kw is then calculated as : Kw = C2 Example 7 : The specific conductivity of 0.02 M KCl solution at 25 ºC is 2.768  10–3 ohm–1 cm–1. The resistance of this solution at 25 ºC when measured with a particular cell was 250.2 ohms. The resistance of 0.01 M CuSO4 solution at 25 ºC measured with the same cell was 8331 ohms. Calculate the molar conductivity of the copper sulphate solution.

2.768  103 Sp. cond. of KCl = = 2.768  10–3  250.2 1/ 250.2 Conductance of KCl For 0.01 M CuSO4 solution

Sol. Cell constant =

Sp. conductivity = Cell constant  conductance = 2.768  10–3  250.2  Molar conductance = Sp. cond. 

1 8331

1000 1000 2.768  103  250.2 =  = 8.312 ohm–1 cm2 mole–1 C 1/100 8331

Example 8 : The specific conductivity of a saturated solution of silver chloride is 2.30  10–6 mho cm–1 at 25 ºC . Calculate the solubility of silver chloride at 25 ºC if  Ag = 61.9 mho cm2 mol–1 and 

Cl

= 76.3 mho cm2 mol–1.

Sol. Let the solubility of AgCl be s gram mole per litre Dilution =

 0AgCl  

1000 S Ag 



Cl

= 61.9 + 76.3 = 138.2 mho cm2 mol–1

Sp. conductivity  dilution =  0 AgCl = 138.2 2.30  10–6 

S=

1000 = 138.2 S

2.30  103 = 1.66  10–5 mole per litre = 1.66  10–5  143.5 gL–1 = 2.382  10–3 gL–1 138.2

Example 9 : The equivalent conductances of sodium chloride , hydrochloric acid and sodium acetate at infinite dilution are 126.45 , 426.16 and 91.0 ohm–1 cm2 equiv–1, respectively at 25 ºC. Calculate the equivalent conductance of acetic acid at infinite dilution. Sol. According to Kohlrausch’s law,  0CH3COONa =  CH3COO    Na  = 91.0

.......(i)

 0HCl  

......(ii)

H

 0 NaCl  

  Cl¯ = 426.16

Na 

  Cl¯ = 126.45

.......(iii) Gyaan Sankalp

13

Electrochemistry Adding equations (i) and (ii) and substracting (iii),  



CH3COO 

CH3COO 

+

Na 

H

+  H   Cl¯ –  Na    Cl¯ = 91.0 + 426.16 – 126.45 = 

0 CH3COOH

= 390.7 ohm–1 cm2 equiv–1

TRY IT YOURSELF Q.1 The specific conductance of a salt of 0.01M concentration is 1.061 × 10-4. Molar conductance of the same solution will be : (A) 1.061 × 10-4 (B) 1.061 (C) 10.61 (D) 106.1 Q.2 The specific conductances of a 0.1 N KCl solution at 23ºC is 0.0112 ohm-1 cm-1. The resistance of the cell containing solution at the same temperature was found to be 55 ohm. The cell constant will be : (A) 0.142 cm-1 (B) 0.918 cm-1 (C) 1.12 cm-1 (D) 0.616 cm-1 -1 -1 Q.3 The conductivity of 0.25 M solution of univalent weak electrolyte XY is 0.0125  cm . The value of  0m of XY is 500 -1 cm2 mol-1. the value of Ostwald dilution constant of AB is : (A) 2.5 × 10-3 (B) 2.5 × 10-4 (C) 2.8 × 10-3 (D) 2.8 × 10-4 Q.4  CH3COOH = 20 ohm-1 cm2 eq-1 and CH3COOH = 400 ohm-1 cm2 equiv-1 , then pH of 1 M CH3COOH solution is : (A) 1.3 (B) 0 (C) 1.7 (D) 4 Q.5 If molar conductance at infinite dilution of (NH4)2SO4, NaOH and Na2SO4 solutions are x1, x2 and x3 respectively, then molar conductance of NH4OH solution is x1  2x 2  x 3 x1  x 2  x3 (B) x1  2x 2  x 3 (C) (D) x1  x 2  x 3 2 2 Q.6 The resistance of 1 N solution of acetic acid is 250 ohm, when measured in a cell of cell constant 1.15 cm-1. The equivalent conductance (in ohm-1 cm2 equiv-1) of 1 N acetic acid is : (A) 4.6 (B) 9.2 (C) 18.4 (D) 0.023 Q.7 Which of the following solutions of KCl has the lowest value of specific conductance ? (A) 1M (B) 0.1 M (C) 0.01 M (D) 0.001 M

(A)

ANSWERS (1) (C) (5) (A)

(2) (D) (6) (A)

(3) (A) (7) (D)

(4) (A)

FARADAY’S LAW OF ELECTROLYSIS : On passing an electric current through electrolyte chemical change take place is called electrolysis. Faraday's First Law : When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte. If W be the mass of the substance deposited by passing Q coulomb of charge, then according to the law, we have the relation. WQ A coulomb is the quantity of charge when a current of one ampere is passed for one second. Thus, amount of charge in coulombs. Q = current in amperes × time in seconds =  × t W×t W=Z××t where Z is a constant, known as electro-chemical equivalent, and is characteristic of the substance deposited. When a current of one ampere is passing for one second, i.e., one coulomb (Q = 1), then W=Z Thus, electrochemical equivalent can be defined as the mass of the substance deposited by one coulomb of charge or by one ampere of current passed for one second. equivalent wt.of element 96500 (the electrochemical equivalent (Z) of an element is directly proportional to its equivalent weight (E), i.e. E  Z or E = FZ Where F is again a proportionality constant and has been found to be 96500 coulombs. It is called Faraday. Thus E = 96500 × Z Therefore, when 96500 coulombs of electricity is passed through an electrolyte, one gram equivalent of its ions is deposited at the respective electrode. This quantity of electricity which liberates one gram equivalent of each element is called one Faraday and is denoted by F.  1 Faraday = 96500 coulombs)

Electro-chemical equivalent (Z) =

14

Gyaan Sankalp

Electrochemistry Faraday's second Law : It states that when same quantity of electricity is passed through different electrolytes then the quantity of deposit is directly proportional to its equivalent weight. (Equivalent wt. of electrolytes). WE

G

WA WB WC   EA EB EC

A C B Example 10 : Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of each metal will be deposited assuming only cathodic reaction in each cell. Sol. The cathodic reaction in the cell are respectively, Ag+ + e–  Ag Cu2+ + 2 e–  Cu Fe3+ + 3 e–  Fe 1 mole 1 mole 1 mole 1 mole 1 mole 1 mole 108 g 1F 63.5 g 2F 56 g 3F Hence, Ag deposited = 108 × 0.4 = 43.2 g Cu deposited =

63.5 56 × 0.4 = 12.7 g and Fe deposited = × 0.4 = 7.47 g 2 3

Example 11 : Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 minute. It was found that after electrolysis, the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with. Sol. Equivalent of Cu2+ lost during electrolys is =

it 2  10 –3  16  60  = 1.989 × 10–5 96500 96500

1.989  10 –5 2 This value is 50% of the initial concentration of solution or mole of Cu2+ lost during electrolysis =

Thus, initial mole of CuSO4

=

2  1.989  10 –5 = 1.989 × 10–5 2

1.989  10–5  1000 250 [CuSO4] = 7.95 × 10–5 M

Thus, initial concentration of CuSO4=

Example 12 : A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (at. mass of copper = 63.5). Sol. The electrode reactions are : Cu2+ + 2 e–  Cu (cathode) 1 mole 2 × 96500 C Cu  Cu2+ + 2 e– (Anode) Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolved. Charge passed through cell = 2.68 × 60 × 60 coulomb 63.5 = 2.68 × 60 × 60 = 3.174 g 2  96500 Increase in mass of cathode = Decrease in mass of anode = 3.174 g

Copper deposited or dissolved =

Gyaan Sankalp

15

Electrochemistry FUELCELL : It is an electrochemical device for continuous conversion of the portion of free energy change into electrical energy. Such cell converts 74% of chemical energy into electrical energy. The fuel used is in the gaseous state, H2 –O2 fuel cell is a common example. In the cell hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide. Catalysts are incorporated in the electrode. The electrode reactions are: Anode : 2[H2(g) + 2OH– (aq)  2H2O() + 2e–] Cathode : O2(g) + 2H2O() + 4e–  4OH– (aq) Overall reaction : 2H2(g) + O2(g) 2H2O () Nickel-cadmium storage cell. It consists of cadmium anode and a metal grid containing NiO2 acting as cathode. The electrolyte in this cell is is KOH. Following reactions take place during discharging. Cd(s) + 2OH– (aq)  Cd(OH)2(s) + 2e– NiO2(s) + 2H2O() + 2e–  Ni(OH)2(s) + 2OH– (aq) Cd(s) + NiO2(s) + 2H2O() Cd(OH)2 (s) + Ni(OH)2(s) As after discharging products formed are solid. Hence, the reaction can be reversed during charging. Further, as no gases are produced during charging or discharging the cell can be sealed. It produces a potential of 1.4 V. The cell has a longer life than lead storage cell and is used in cordless appliances (Phones, pagers, mobile phones, electric shavers etc.) Advantages of fuel cells over ordinary batteries : The fuel cells convert the energy of the fuel directly into electricity, while the conventional method of generating electricity by burning hydrogen, carbon fuels first convert fuels in to thermal energy and then in to electrical energy although theoretically, fuel cells are expected to have an efficiency of 100% practically only 60–70% efficiency has been attained, efficiency of the conventional method is only about 40%. CORROSION : If any household item of iron is left outside, then it gets rusted and it becomes Water O2(g) rough and porous. Almost all useful metals except gold, platinum, aluminium, Dropled Rust etc., are destroyed slowly. Decay process of the metals is called corrosion of metals. Corrosion takes place rapidly at bends, seratches, nicks and cuts in Fe2 O 3 .xH 2 O the metal. Anode Cathode Electrochemical Theory of rusting : According to this theory follow of electric 2+ – e– Fe Fe +2e current between separate anode and cathode bar as reason because of this corrosion is held. In the anodic reason metal is destroyed by the formation of + – O2+4H +4e 2H2O their ion or combined state (ex. oxide) in the oxidation reaction. So corrosion is + – O +2H O +4e 4OH¯ always held in the anodic part. 2 2 Overall reaction of corrosion cell : IRON 1 2+ O2 Fe + H2O ; E°cell = 1.67V 2 The ferrous ions so formed move through water and come at the surface of iron object where these are further oxidised to ferric state by atmospheric oxygen and constitute rust which is hydrated iron (III) oxide. +

Fe + 2H +

1 + O + 2H2O  Fe2O3 + 4H 2 2 Fe2O3 + xH2O  Fe2O3.xH2O To prevent corrosion the metal surface is coated with paint which keeps it out of contact with air, moisture etc. till the paint layer develops cracks. 2+

2Fe +

TRY IT YOURSELF Q.1 How long will it take for a current of 3 amperes to decompose 36 g of water. (Eq. wt. of hydrogen is 1 and that of oxygen 8) – (A) 36 hours (B) 18 hours (C) 9 hours (D) 4.5 hours Q.2 If m represents mass of a substance (equivalent weight E) consumed or produced when quantity Q of electricity is passed then,– (A) m  Q (B) m  (1/Q) (C) m  (I/E) (D) m  Q.E. Q.3 A current of 0.5 ampere when passed through AgNO3 solution for 193 sec deposited 0.108g of Ag. The equivalent weight of silver is – (A) 108 (B) 54 (C) 5.4 (D) 10.8 Q.4 A current of 9.65 ampere flowing for 10 minutes deposits 3.0 g of the metal which is monovalent the atomic mass of the metal is (A) 10 (BB) 50 (C) 30 (D) 96.5 16

Gyaan Sankalp

Electrochemistry Q.5 On passing electric current into a solution of a salt of metal, M, the reaction at the cathode takes place as : M2+ + 2e–  M. Atomic weight of M is 65. The equivalent weight of metal is – (A) 65 (B) 32.5 (C) 130 (D) 200 Q.6 The electrolytic cell, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be(A) 3 : 1 (B) 2 : 1 (C) 1 : 1 (D) 3 : 2 Q.7 10800 C of electricity through the electrolyte deposited 2.977 g of metal with atomic mass 106.4 g mol–1. The charge on the metal cation is – (A) + 4 (B) + 3 (C) + 2 (D) + 1

ANSWERS (1) (A) (5) (B)

(2) (A) (6) (D)

(3) (A) (7) (A)

(4) (B)

USEFUL TIPS 1.

m = Z.I.t

2.

Degree of dissociation :  =

3.

Kholrausch’s law :  0m  x 0A  y 0B

4.

Nernst Equation E = Eº –

 eq  0eq

=

Equivalent conductance at given concentration equivalent conductance at infinite dilution

[Products]  nEº  0.0591 log10 & E°cell = Eºright + Eºleft & Keq. = antilog   [Re ac tants] n  0.0591 

 G  G = – nFEcell & Gº = –nFEº cell & Wmax= +nFEº & G = H + T  T   P

5.

Calculation of pH of an electrolyte by using a calomel electrode : pH 

E cell  0.2415 0.0591

MISCELLANEOUS SOLVED EXAMPLES Example 1 : Would H2O2 behave as oxidant or reductant with respect to the following couples at standard concentrations ? (a) I2 / I– : Eº = 0.533 V (b) S2O82–/SO42–; Eº = 2.0 V (c) Fe3+ / Fe2+; Eº = 0.771 V + – Given : H2O2  O2 + 2H + 2e Eº = – 0.69 V .......... (P) H2O2 + 2H+ + 2e  2H2O Eº = 1.77 V .......... (Q) Sol. (a) Given 2e– + I2 2I– Eº = 0.533 V on adding with eq. (P)  I2 + H2O2 O2 + 2H+ + 2I– Eº = – 0.157 V  On reversing direction 2I– I2 + 2e– Eº = – 0.535 V on adding with eq. (Q)  + – H2O2 + 2H + 2I  2H2O + I2 Eº = 1.235 V So H2O2 acts as an oxidizing agent in the reaction for which its Eº value is positive. (b) Given S2O82– + 2e– 2SO42– Eº = 2.0 V on adding with eq. (P)  2– H2O2 + S2O8 2SO42– + O2 + 2H+ Eº = 1.31 V  On reversing direction 2SO42–  S2O82– + 2e– Eº = – 2.0 V on adding with eq. (Q) 2SO42– + H2O2 + 2H+  S2O82– + 2H2O Eº = – 0.23 V So H2O2 is the reducting agent in the reaction with a positive Eº value. (c) Given Fe3+ + e– Fe2+ Eº = 0.771 V on adding with (P)  3+ 2Fe + H2O2 2Fe2+ + O2 + 2H+ Eº = 0.081 V  On reversing direction Fe2+  e– + Fe3+ Eº = – 0.771 V on adding with (Q) 2Fe 2  H 2 O2  2H 2   2Fe3  2H 2 O E  0.999V Since both potentials are positive, H2O2 will act as an oxidizing agent and a reducing agent. In fact, iron (II) or ion (III) salts catalyze the self-oxidation-reduction of H2O2. Gyaan Sankalp

17

Electrochemistry Example 2 : A T+ / T couple was prepared by saturating 0.1 M KBr with TBr and allowing the T+ from the relatively insoluble bromide to equilibrate. This couple was observed to have a potential of – 0.443 V with respect to Pb2+ / Pb couple in which Pb2+ was 0.1 M. What is Ksp of TBr. (Report answer in multiplication of 10–8) ( E o

Pb 2  / Pb

o = – 0.126, ET  / T = – 0.336 V)

2.303 RT = 0.059) F

(Take antilog (0.5509) = 3.55,  Pb Sol. Pb2+ + 2e– 

E0

= E0

Pb 2  / Pb

Ecell = E E E

T / T

T / T

Pb 2  / Pb

Pb 2  / Pb

– E

–

1 0.0591 log 0.1 = – 0.1555 volt 2

; 0.443 = – 0.1555 – E

T / T

T+ + e–  T

= – 0.5985V ; = E0

T / T

T / T

– 0.059 log

1 (T  )

– 0.5985 = – 0.336 + 0.059 log (Tl+) ; T+ = 3.55 × 10–5 M Ksp = (T+) (Br–) = 3.55 × 10–5 × 0.1 = 3.55 × 10–6 = 355 × 10–8 Example 3 : Calculate the volume of gas liberated at anode at NTP from the electrolysis of CuSO4 solution by a current of 2 ampere passed for 10 minutes. Sol. At cathode : Cu+2 + 2e  Cu At anode : 2H2O  4H+ + 4e + O2  EO2 =

32 = 8. 4

E.i.t. 32  2  10  60  w O2 = 96500 = 4  96500 = 0.0995 g

 At NTP : Volume of O2 =

0.0995  22.4 = 0.0696 litre. 32

Example 4 : Formulate a cell in which the following reaction 2Cu+(aq)   Cu++ (aq) + Cu can occur. Given that the standard electrode reduction potential of Cu|Cu+ and Cu|Cu++ electrodes are 0.52 and 0.34 volts respectively at 25°C, calculate the standard EMF of the cell constructed and also find out the equilibrium constant of the reaction. Sol. Cell : Cu|Cu++ (C = 1.0m) || Cu++ (C = 1.0 m) | Cu E°cell = E°R – E°L = 0.52 – 0.34 = 0.18 V RT 0.18  2 ln K Using n = 2, loge K = = 6.1 ; K = 106.1 nF 0.0591 Example 5 :

E° =

Determine potential for the cell Pt

Fe2 Fe 3

Cr2 O72 , Cr 3 , H  Pt; in which [Fe+2] and [Fe+3] are 0.5 M and 0.75 M respectively

and [Cr2O7–2], [Cr+3] and [H+] are 2M, 4M and 1M respectively. Given : Fe+3 + e  Fe+2; Eº = 0.770 V and 14H+ + 6e + Cr2O7–2  2Cr+3 + 7H2O; Eº = 1.35 V o Sol.  E Fe2 / Fe3 > E o

Cr 3 /Cr26

Anode : Fe+2  Fe+3 + e o Eºcell = E Fe2 / Fe3 – E o

Cr 3 /Cr26

18

Gyaan Sankalp

 Redox changes will be Cathode : 14H+ + 6e + Cr2O7–2  2Cr+3 + 7H2O = – 0.770 + 1.35 = 0.58 V

Electrochemistry Ecell = Eºcell –

[Fe 3 ]6 [Cr 3 ]2 (0.75)6 (4) 2 0.059 0.059 log log 2 6 2  14 = 0.58 – 6 [Fe ] [Cr2O7 ] [H ] 6 (2) (1)14 (0.5)6

= 0.58 –

0.059 log (91.125) = 0.58 – 0.0192 = 0.56 V.. 6

Example 6 : On the basis of the following Eºvalues, predict whether ferricyanide ion is stronger or weaker oxidising agent than ferric ions. [Fe(CN)6]4–  [Fe(CN)6]3– + e; Eº = – 0.36 V Fe2+  Fe3+ + e ; Eº = – 0.77 V Sol. The given reactions are oxidation reactions. On writing them as reduction reactions we get : Oxidising agent Reducing agent [Fe(CN)6]3– + e [Fe(CN)6]4–; Eºred = + 0.36 V Fe3+ + e Fe2+ ; Eºred = + 0.77 V  Now since the value of SRP for Fe3+/Fe2+ couple is greater than the value of SRP for [Fe(CN)6]3– / [Fe(CN)6]4– couple. Fe3+ is stronger oxidising agent than [Fe(CN)6]3–. Example 7 : Write the electrode reactions and the net cell reactions for the following cells. Which electrode would be the positive terminal in each cell ? (a) Zn | Zn2+ || Br–, Br2 | Pt (b) Cr | Cr3+ || I–, I2 | Pt (c) Pt | H2, H+ || Cu2+ | Cu (d) Cd | Cd2+ || Cl–, AgCl | Ag Sol. (a) Oxidation half cell reaction, Zn Zn2+ + 2e–  Reduction half cell reaction, Br2 + 2e– 2Br–  Net cell reaction Zn + Br2 Zn2+ + 2Br– Positive terminal : Cathode Pt  (b) Oxidation half reaction, Reduction half reaction, Net cell reaction

[Cr [I2 + 2e– 2Cr + 3I2

  

Cr3+ + 3e–] × 2 2I–] × 3 2Cr3+ + 6I–

Positive terminal : Cathode Pt

(c) Oxidation half reaction, Reduction half reaction, Net cell reaction,

H2 Cu2+ + 2e– H2 + Cu2+

  

2H+ + 2e– Cu Cu + 2H+

Positive terminal : Cathode Cu

(d) Oxidation half reaction, Reduction half reaction, Net cell reaction Example 8 :

Cd [AgCl + e– Cd + 2AgCl

  

Cd2+ + 2e– Ag + Cl–] × 2 Cd2+ + 2Ag + 2Cl–

Positive terminal : Cathode Ag

HA Pt (H ) Calculate emf of the following cell 1atm2 pK  5 a

Sol. Left half cell E E

H/ H 

H/ H 

= Eºoxid –

HB Pt (H 2 ) pK a  3 1atm

0.0591 log [H] 1

= E ooxid – 0.0591 pH (HA)

1 Right half cell E (H  /H) = E oRe d – 0.0591 log  [H ]HB

E

(H  /H)

= E oRe d – 0.0591 pH (HB)

For HA (H+) = (pH)HA =

CK a or – log [H+] = –

1 1 log C – log Ka 2 2

1 1 (pKa)HA – log C 2 2

Gyaan Sankalp

19

Electrochemistry Similarly for HB (pH)HB =

1 1 (pKa)HB – log C 2 2

1 1  0.0591 Ecell = Eright – Eleft = 0.0591  pKa (HA)  pKa (HB)  = [5 – 3] = 0.0591. 2 2  2 Example 9 : Write a cell diagram for each of the following reactions and calculate Eº for each cell : (a) 2Fe3+ + Sn2+  2Fe2+ + Sn4+ (b) AgBr(s) + ½H2(g)  Ag(s) + H+ + Br– Given than E o

Fe3 / Fe 2

= + 0.77 volt;

Eo

Sn 4  /Sn 2

Eo 

E oAgBr /Ag(s), Br = 0.10 volt;

= 0.15 volt

= 0.00 volt

H /H 2 Sn2+ is oxidised to Sn4+ is oxidised to Sn4+ and Fe3+ is reduced to Fe2+. Hence the cell diagram is :

Sol. (a) From the given reaction (Pt) Sn2+ | Sn+4 || Fe3+ | Fe2+ (Pt) Eºcell = Eºright – Eºleft= (0.77 – 0.15) volt = 0.62 volt. (Platinum inert electrode was used to make electrical contact). (b) From the reaction we see that H2 is oxidised to H+ and AgBr is reduced to Ag(s) The cell diagram is : Pt|H2(g) | H+ || Br– | AgBr(s), Ag(s) Eºcell = Eºright – Eºleft = (0.10 – 0.0) volt = 0.10 volt. Example 10 : Small spherical ball of silver metal used in jewellery having diameter 0.1 cm, which is obtained by the electrolytic deposition. If total number of balls in jewellery is 10,000, then calculate the applied amount of electricity in coulombs, which is used on the deposition on electrodes having entire surface area 0.12 m2. [Density of Ag = 10.47]. Also determine the thickness of deposited metal. It is assumed that 10% electricity consumed as wastage during electrolysis and 60% of electrode body immersed in electrolyte. 0.1 = 0.05 cm ; Total number of balls = 10,000 2 Surface area of electrodes = 0.12 m2 = 1200 cm2 of 60% = 2000 cm2 Density of Ag metal = 10.47

Sol. Here, radius of ball =

4 3 4 Weight of jewellery =  r  × 10,000 × density = × 3.14 × (0.05)3 × 10,000 × 10.47 = 5.23 × 10.47 g 3 3   Therefore, Surface area × thickness × density = Total weight of Ag 2000 × thickness × 10.47 = (5.23 × 10.47) g 5.23 = 2.615 × 10–3 cm 2000 Amount of electricity in Faraday = Number of equivalent  Thickness =

=

Weight of Ag 5.23 10.47 = g Equivalent weight of Ag 108

= 0.50 F = 48250 coulomb; then applied electricity amount =

48250  100 = 53611.11 coulomb 90

Example 11 : The standard oxidation potential of Ni/Ni+2 electrode is 0.236 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution will the measured emf be zero at 25ºC. Assume [Ni+2] = 1M. Sol. Ni  Ni+2 + 2e; Eº = 0.236 V 2H+ + 2e  H2; Eº = 0.0 V  Eºcell = 0.236  Ecell = Eºcell +

[H  ]2 0.059 log [Ni 2 ] 2

or – log H+ = 1 × 10–4 

20

Gyaan Sankalp

pH = 4.

or

0 = 0.236 +

0.059 log [H+]2 2

Electrochemistry Example 12 : Write the Nernst equation and e.m.f. of the following cells at 298 K. (i) Mg(S) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s) (ii) Fe(s) | Fe2+ (0.001 M) || H+(1M) | H2(g) (1 bar) | Pt(s) (iii)Sn(s) | Sn2+ (0.050 M) || H+ (0.020 M) | H2(g) (1 bar), Pt(s) (iv) Pt(s) | Br2(), Br– (0.010 M)|| H+(0.030 M) | H2(g) (1 bar), Pt(s) Sol. (i) The cell reaction is At anode : Mg(s) Mg2+(0.001 M) + 2e–  + – At cathode : Cu (0.0001 M) + 2e Cu(s)  Net cell reaction : Mg(s) + Cu2+ (0.0001 M)  Mg2+(0.001 M) + Cu(s) The Nernst equation is, Ecell = Eºcell –

[Mg 2 ] 2.303 RT log [Cu 2  ] nF

2.303  8.314  298   o 0.001 o Ecell =  (E Cu 2  /Cu  E Mg 2  / Mg )  log = {0.34 – (– 2.37) – 0.0295 × log 10} V   2  96500 0.0001 So,Ecell = (2.71 – 0.0295) V = 2.68 V (ii) The cell reaction is At anode : Fe(s)  Fe2+(0.001 M) + 2e– At cathode : 2H+(1 M) + 2e–  H2 (1 bar)

Net cell reaction : Fe(s) + 2H+ (1M)  Fe2+ (0.001 M) + H2 (1 bar) The Nernst equation for the cell emf is, [Fe2 ] pH 2 2.303 RT Ecell = Eºcell – log 2F [H  ]2

0.001 1 2.303  8.314  298 log = {0.0 – (– 0.44 ) – 0.0295 × log 0.001}V 2  96500 12 Ecell = 0.44 V + 0.09 V = 0.53 V (iii)The cell reaction is At anode : Sn(s)  Sn2+ (0.05 M) + 2e– At cathode : 2H+ (0.02 M) + 2e–  H2 (1 bar) o o = E H /H – E Fe2  /Fe – 2

Net cell reaction : Sn(s) + 2H+(0.02 M)  Sn2+ (0.05 M) + H2(1 bar) The Nernst equation for this cell at 298 K is [Sn 2 ] pH 2 0.05  1 0.0591 Ecell = Eºcell – log = (Eºcathode – Eºanode) – 0.0295 log  2 (0.02) 2 2 [H ] 0.05

Ecell = {0.0 – (– 0.14 ) – 0.0295 log

(0.02) 2

}V

or Ecell = 0.14 V – 0.062 V = 0.078 V (iv) For the given cell, the cell reaction can be written as follows : At anode : 2Br– (aq) (0.01 M)  Br2(l) + 2e– At cathode : 2H+ (0.03 M) + 2e–  H2 (1 bar) Net cell reaction : 2Br– (0.01 M) + 2H+ (0.03 M)  Br2(l) + H2 (1 bar)

Gyaan Sankalp

21

Electrochemistry The corresponding Nernst equation at 298 K is Ecell = Eºcell –

pH2 1 0.0591  Eo   Eo log   – 0.0295 log 2   2 =  H / H Br / Br (0.01) (0.03) 2  2 2  [Br ] [H ] 2

This gives, Ecell = {(0 – 1.09 ) – 0.0295 × log

1

}V 9 108 Ecell = – 1.09 V – 0.21 V = – 1.30 V Example 13 : Chromium metal can be plated out from an acidic solution containing CrO3 according to the following equation CrO3(aq) + 6H+ + 6e+  Cr(s) + 3H2O Calculate, (i) How many grams of chromium will be plated out by 24000 coulombs ? (ii) How long will it take to plate out 1.5 gm of Cr using 12.5 ampere current ? Sol. (i) Eq. wt. of Cr =

Atomic weight 52 = Change in oxidation number per mole 6

52  96500 coulomb deposit = 6 gm (Cr)  24000 coulomb will deposit =

52 24000 × gm = 2.1554 gm of Cr 6 96500

(ii) WCr = 1.5 gm, i = 12.5 ampere ECr =

52 Eit 52  12.5  t ,t=? ; W=  1.5 =  t = 1336.15 sec. 6 96500 6  96500

Example 14 : Calculate the potential of an indicator electrode, versus the standard hydrogen electrode, which originally contained 0.1 M MnO4– and 0.8 M H+ and which has been treated with 90% of the Fe2+ necessary to reduce all the MnO4– to Mn2+. Sol. MnO4– + 8H+ + 5e–  Mn2+ + 4H2O Eº = 1.51 V 5Fe2+  5Fe3+ + 5e– MnO4– + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O [MnO4–] = 0.10 [Fe3+] = 0.450 [H+] = 0.080 [Mn2+] = 0.0900 Since the Mn7+/Mn2+ and Fe3+/Fe2+ systems are both in the same vessel, they must both experience the same half-cell potential.

 0.0591  E = 1.51 –    5 

  [Mn 2 ]  0.0591   log     8  = 1.51 –  [MnO4 ] [H ]   5  

  (0.09)  log  = 1.39 V (0.01) (0.08)8  

Example 15 : Estimate the cell potential of a Daniel cell having 1M Zn 2+ and originally having 1M Cu2+ after sufficient NH3 has been added to the cathode compartment to make NH3 concentration 2M, Kf for [Cu(NH3)4]2+ = 1 × 1012, Eº for the reaction Zn + Cn2+  Cu + Zn2+ is 1.1 V. 2+ Sol. Cu + 4NH3 [Cu(NH3)4]2+, Kf = 1 × 1012 x 2 1 At equilibrium Kf =

[Cu(NH3 )42  ] [x] [NH3 ]4

= 1 × 1012

x = 6 × 10–14 Zn + Cu2+  Cu + Zn2+ [Zn 2 ]  0.059  E = Eº –   log [Cu 2  ]  2 

Substituting the values we get, E = 0.71 V

22

Gyaan Sankalp

Electrochemistry Example 16 : Calculate the reduction potential for the following half cells at 25ºC. (i) Mg | Mg2+ (1.5 × 10–5 M) ;

Eo

(ii) Cl2 | Cl– (2 × 10–5 M);

Eo

(iii)Pt | Fe2+ (0.1 M), Fe3+ (0.01 M)

Eo

Sol. (i) Mg2+ + 2e

Eo

Mg 2  , Mg

= Eo

Mg 2  , Mg

(ii) ½Cl2 + e Cl2 , Cl 

Cl2 , Cl

(iii)Fe3+ + e Fe3 , Fe 2

= + 2.36 V

= + 1.36 V

Fe3 , Fe2 

= 0.77 V

(reduction) +

0.0591 log (1.5 × 10–5) 2

4  0.0591 = – 2.4782 V.. 2 Cl– (reduction)

o = E

= 1.36 –

Eo

Cl 2 , Cl

Mg

= – 2.36 –

E

Mg, Mg 2

–

0.0591 log [Cl–] 1

0.0591 log (2 × 10–5) = 1.6377 V 1 Fe2+ (reduction)

[Fe 2 ] o 0.0591 = E Fe3 , Fe2 – log [Fe3 ] n

= 0.77 –

0.1 0.0591 log = 0.7109 V. 0.01 1

Example 17 : A fuel cell uses CH4(g) and forms CO32– at the anode. It is used to power a car with 80 Amp. for 0.96 hr. How many litres of CH4(g) (STP) would be required ? (Vm = 22.4 L/mol) (F = 96500). Assume 100% efficiency.  CO32– + 7H2O + 8e– Sol. CH4 + 10OH– 

No. of Fradays required =

80  3600  0.96 96500

Hence mol. of CH4 required =

VCH4 =

1 80  3600  0.96 × 8 96500

1 80  3600  0.96 × × 22.4 L = 8.356 × 0.96 = 8.02 L 8 96500

Example 18 : The redox reaction involved in the rusting of iron is believed to be ++ 2Fe(s) + O2(g) + 4H+ (aq)   2Fe (aq) + 2H2O () Calculate the standard EMF of the cell in which this reaction occurs and what is the equilibrium constant of this reaction ? E°Fe|Fe++ = –0.44 V, E° for O2 + 4H+ + 4e = 2H2O () = 1.23 V ++ Sol. 2Fe(s) + O2(g) + 4H+ (aq)   2Fe (aq) + 2H2O () ++ + Cell : Fe|Fe (C = 1.0 m) || H (C = 1.0 m), H2O|H2(g), Pt. E° = 1.23 – (–0.44) = 1.67 V

E° =

RT 1.67 E ln K; ln K = = = 56.42 nF 0.0591 / 2 0.0296 K = 2.63 × 1056

Gyaan Sankalp

23

Electrochemistry Example 19 : The following two cells with initial concentration as given are connected in parallel with each other. (1) Fe(s) | Fe(NO3)2(aq.) (1M)||SnCl2(aq.) (1M)|Sn(s) (2) Zn(s)|ZnSO4(aq.)(1M)||Fe(NO3)2(aq.)(1M)|Fe(s) After sufficient time equilibrium is established in the circuit. What will be the concentrations (in mmoles/L) of Fe2+ ions in first and second cells respectively. 0 0 0 [Take ESn 2 / Sn = – 0.14V, E Zn 2  / Zn = – 0.76V,, E Fe 2 / Fe = – 0.44 V, 2.3 × RT = 6433, log 2 = 0.3]

Sol. Fe(s) + Sn2+ (aq.)

Fe2+(aq) + Sn(s) ; E 0cell = 0.30 V

Zn(s) + Fe2+ (aq.)

Zn2+ (aq.) + Fe(s) ; E 0cell = 0.32 V If above cells are connected in parallel then first cell will get charged up and second cell will get discharged so net cell reaction will be. Fe22+(aq.) + Fe12+ (aq.) t=0 t(aq.)

1M (1 – x)

1M (1 – x)

Zn2+ (aq.) + Sn2+ (aq.) ; E 0net = 0.02 1M 1M (1 + x) (1 + x)

 G 0net = – nF E 0net = – 2 × 96500 × 0.02 J/mole So, –2.30 RT log K. = – 2 × 96500 × 0.02 log K = (1  x) 2 (1  x)

2

2  96500  0.02 = 0.6 ; K= 4 6433

(1  x) 1 = 4  (1  x) = 2  x = 3

So, concentration of Fe2+ ions in the first cell = concentration of Fe2+ ions in the second cell =

2 M = 667 mmoles/L 3

Example 20 : An acidic solution of Cu+2 salt containing 0.4 g of Cu+2 is electrolysed until all the Cu is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the current at 1.2 ampere. Calculate volume of gases evolved at N.T.P. during entire electrolysis At. wt. of Cu = 63.6. Sol. For I part of electrolysis Anode : 2H2O  4H+ + O2 + 4e Cathode : Cu+2 + 2e  Cu Eq. of O formed = Eq. of Cu  2 0.4  2 = 12.58 × 10–3 63.6 For II part of electrolysis : Since Cu+ ions are discharged completely and thus further passage of current through solution will lead the following changes. Anode : 2H2O  4H+ + O2 + 4e Cathode : 2H2O + 2e  H2 + 2OH–

=

i.t 1.2  7  60 = = 5.22 × 10–3 96500 96500  Total eq. of O2 = 5.22 × 10–3 + 12.58 × 10–3 = 17.8 × 10–3 or 4 eq. O2 at NTP = 22.4 litre  2 Eq. of H2 at NTP = 22.4 litre

Eq. of H2 = Eq. of O2 =

24

 5.22 × 10–3 eq. at NTP =

22.4  5.22  103 L = 58.46 ml (Hydrogen) 2

and 17.8 × 10–3 eq. at NTP =

22.4 17.8 103 L = 99.68 ml (Oxygen) 4

Gyaan Sankalp

Electrochemistry Example 21 : The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 ×  /S m–1 1.237 11.85 23.15 55.53 106.74 Calculate  for all concentrations and draw a plot between  and C1/2. Find the value of  º. Sol. The data are rewritten as follows : C(mol L–1) 0.001 0.01 0.02 0.05 0.1 0.032 1.237 1.237 123.7

C  (S m–1) 4 10  (S cm–1) º 102

0.14 23.15 23.15 115.7

0.22 55.53 55.53 111.1

1000 (S cm1 ) C (mol / L)

Plot of  against

C for sodium chloride solution at 298 K

–1

The  – C plot is shown in the figure. From the plot (by graphical extrapolation),  º = 125.5 S cm2 mol–1

130

 /S cm2 mol

 º=

0.1 11.85 11.85 118.5

0.32 106.74 106.74 106.7 125.5

120

110

100

03

C

o The Ecell for Ag(s) | AgI (satd) || Ag+ (0.10M) | Ag(s) is + 0.417 V. Calculate Ksp for AgI. E

Ag  / Ag

Sol. The cell reaction occuring in this concentration cell is Oxidation : Ag(s)  Ag+ (satd AgI) + e– Reduction : Ag+ (0.10 M) + e–  Ag(s) Cell reaction : Ag+ (0.10 M)  Ag+ (satd) AgI) Now Ecell = Eºcell –

02

01

Example 22:

= 0.80 V..

Eº = – 0.80 V Eº = + 0.80 V Eºcell = 0.0 V

[Ag  from AgI] 0.059 log 0.10 M n

0.417 0.059 [Ag  ] [Ag  ] log ; log =– = – 0.07 0.059 1 0.10 0.10  [Ag+] = 8.5 × 10–8 × 0.10 = 8.5 × 10–9M In a saturated AgI solution, [Ag+] [I–] = Ksp  Ksp = [Ag+] [I–] = (8.5 × 10–9)2 = 7.2 × 10–17. Example 23 : Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity and if  º for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant ? Sol. Given :  = 7.896 × 10–5 S cm–1 Conc. of acetic acid solution, C = 0.00241 mol L–1

0.417 = 0 –

1000   1000  7.896 105 = S cm2 mol–1 C 0.00241  = 32.76 S cm2 mol–1 We have  º = 390.5 S cm2 mol–1

So,  =

32.76S cm 2 mol1  –2 Then  = = 2 1 = 8.39 × 10 = 0.0839  º 390.5 S cm mol

Gyaan Sankalp

25

Electrochemistry Acetic acid ionises as follows : CH3COOH CH3COO– + H+ C(1 –  ) C C C.C C 2 Ka = C (1  ) =  C2 1  So, Ka = 0.00241 × (0.0839)2 = 1.7 × 10–5.

Example 24 : The standard EMF of the cell Pt | H2 (g) | H+ (aq) || OH– (aq) | O2 (g) | Pt is 0.3900 V at 27ºC and 0.4000 V at 7ºC. If entropy change during formation of H2O() from H2(g) and O2(g) is – 193 J/K, then find the Sº (magnitude only, in mJ/K) for  H2O (). H+ (aq) + OH– (aq) 

Sol. anode :

1  H+ + 1e– H  2 2

cathode :

1 1  OH– H2O + O2 + 1e–  2 4

1 1 1 1e  H2 + O2 + H2O   H+ + OH– 2 4 2

net :

Sº =

1  96500 (0.3900  0.4000) J = – 48.25  20 K

Now,

 193 1 1 1  H2O ; Sº = H2 + O2  J/K = –96.5 J/K 2 4 2 2

 H+ + OH– 

1 1 1 H2 + O2 + H2 ; Sº = + 48.25 J/K 2 4 2

 H2O () ; Sº = – 48.25 J/K = – 48250 mJ/K add : H+ (aq) + OH– (aq) 

Example 25 : The conductivity of a solution may be taken to be directly proportional to the total concentration of the charge carries (ions) present in it in many cases. Find the percent decreases in conductivity (k) of a solution of a weak monoacidic base BOH when its 0.1 M solution is diluted to double its original volume. (Kb = 10–5 for BOH) (take 50 = 7.07) Sol. Initially [OH–] = 10 5  0.1 = 10–3 [ions]total = 2 × 10–3 M 5 later [OH–] = 10 

26

1 = 20

–4 50 × 10 M



[ions]total = 2 50 × 10–4 M



% change on [ions]total =

Gyaan Sankalp

2 50  20 × 100 = – 29.29% 20

Electrochemistry Example 26: Calculate the standard cell potentials of the galvanic cells in which the following reactions take place : (i) 2Cr(s) + 3Cd2+ (aq)  2Cr3+(aq) + 3Cd(s) (ii) Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s) Calculate the Gº and equilibrium constant of the reactions.

2Cr(s) + 3Cd2+(aq)

3Cd(s) + 2Cr 3+(aq) Reduction

Sol. (i)

Oxidation

Thus, Cr(s) | Cr3+(aq) is anode, and Cd2+(aq) | Cd(s) is cathode. Eºcell = Eºcathode – Eºanode = Eo

Then,

(Cd 2  |Cd) –

Eo

Cr|Cr3

Taking the values of the standard electrode potentials Eºcell = – 0.40 V – (– 0.74V) = + 0.34 V and Gº = – nF Eºcell = – 6 × 96500 × 0.34 J mol–1 = – 196860 J mol–1 Gº = – 196.86 kJ mol–1 The equilibrium constant of the cell reaction is obtained from the equation, Gº = – 2.303 RT log K – 196860 J mol–1 = – 2.303 × 8.314 J K–1 mol–1 × 298 K log K 196860 J mol1

So,log K =

2.303  8.314 J K 1mol1  298K

= 34.5

K = 3.17 × 1034 Reduction 2+

+

3+ (ii) Fe (aq) + Ag (aq) ––––– Fe (aq) + Ag(s)

Oxidation

Thus, Pt | Fe2+, Fe3+ electrode is anode and Ag+(aq) | Ag is cathode in this cell. o o Then, Eºcell = Eºcathode – Eºanode = EAg / Ag – EFe3 , Fe2

Substituting the Eº values, one gets Eºcell = 0.80 V – 0.77 V = 0.03 V and Gº = – nFEºcell = – 1 × 96500 × 0.03 J mol–1 = – 2895 J mol–1 = – 2.895 kJ mol–1 and log K =

G ocell 2.303 RT

=

2895 J mol 1 2.303  8.314 J K 1 mol 1  298K

= 0.507 or K = 3.2

Example 27 : At 0.04 M concentration the molar conductivity of a solution of electrolyte is 5000 –1 cm2 mol–1 while at 0.01 M concentration the value is 5100 W–1 cm2 mol–1. Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and also determine the degree of dissociation of strong electrolyte at 0.04 M. Sol. For strong electrolyte graph between  m and

C is a straight line . Plot st. line from given data.

From the graph we can see the  0m value of 5200 –1 cm2 mol–1. Hence 5000 = = 0.9615  0.96 5200

0 m

m

Gyaan Sankalp

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