What is the cell potential (in V) at 298 K and 1 bar for the following cell?

Zn(s)|ZnBr₂(aq, 0.20 mol/kg) ||AgBr(s)|Ag(s)|Cu

(given \(E^0_{zn^{+2}/zn}\) = -0.762V, \(E^0_{AgBr/Ag}\) = +0.730V, and assuming γ± of ZnBr₂ solution = 0.462)?

Concept:-

• Nerst equation gives a relationship between the electrode potential and ionic concentration of the electrolyte solution.
• For a reduction occurring at an electrode Mn+|M

Mn+ + ne- → M (s)

E = Eº - \(\frac{RT}{nF}ln\frac{[M]}{[M^{n+}]}\)

E = Eº + \(\frac{RT}{nF}ln{[M^{n+}]}\) (molar concentration of pure solid and liquid is taken as unity)

E = Eº + \(\frac{0.0591}{n}log{[M^{n+}]}\) at 25ºC

where, n is the number of electrons exchanged in the reaction.

• The relation between the activity of the individual ions and the concentration of the solute is,

\(a_+ = \gamma _+(C_m)_+\).............(ii)​

where a+ is the activity of the cation,

\(\gamma _+\) is the activity coefficient of the cation,

\((C_m)_+\) is the concentration of the cation.

Explanation:-

• The given electrochemical cell is,

Zn(s)|ZnBr2(aq, 0.20 mol/kg) ||AgBr(s)|Ag(s)|Cu

• The corresponding cell reaction is

2 AgBr(s) + Zn(s) → 2Ag(s) + ZnBr2 (aq)

​Reaction at Anode:

Zn → Zn+2 + 2e-

​Reaction at Cathode:​

AgBr(s) + e- → Ag(s) + Br-

• For the cell,

\(E^{o}\) = \(E^0_{AgBr/Ag}\) - \(E^0_{zn^{+2}/zn}\)

=  +0.730V - ( -0.762V)

= 1.492 V

• The EMF of the cell will be,

E = \(E^{o}\) - \(\frac{RT}{nF}ln\frac{a_{Ag}\times a_{ZnBr_2}}{a_{AgBr}^2 \times a_{Ag}}\)

E = 1.492 V - \(\frac{0.059}{2}ln\frac{a_{ZnBr_2}}{a_{AgBr}^2}\)

E = 1.492 V - (-0.074) (γ± of ZnBr2 solution = 0.462)

or, E = 1.566 V

Conclusion:-

Hence, the cell potential (in V) at 298 K and 1 bar for the following cell is 1.566