A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4.

Question

The Zn wire is attached to the positive terminal and the Ag/AgCl electrode is attached to the negative terminal of the potentiometer. Calculate the [Zn2+] in the solution if the cell potential, 𝐸cell, is −1.084 V . The standard reduction potential of the Zn2+/Zn half-reaction is −0.762 V.

Answer 1

Ag⁺ + e- ==> Ag(s) .. Eº = 0.197

Zn²⁺ + 2e- ==> Zn(s) .. Eº = -0.762

Since the Ag is the negative (anode) and Zn is the positive (cathode), this is a reversal of a spontaneous situation. This is also evident from the negative value for Ecell. What we have is the following:

Zn²⁺ + 2Ag(s) ==> Zn(s) + 2Ag⁺ Eº = -0.762 - 0.197 = -0.959

Using the Arrhenius equation, Ecell = Eºcell - RT / nF ln Q, we can find the [Zn²⁺].

Ecell = -1.084

Eºcell = -0.959

R = 8.314 J / Kmol

T = 298 K

n = 2 mols electrons

F = Faraday constant

Q = [Ag+]² / [Zn²⁺]

At standard conditions, and using log₁₀ this reduces to

Ecell = Eºcell - 0.0592 / n log Q

-1.084 = -0.959 - (0.0592 / 2)( log (1² / [Zn²⁺]) since under standard condition [Ag⁺] = 1 M

-1.084 = -0.959 - (0-.0296)(log 1²/ Zn²⁺)

-0.125 / -0.0296 = log 1^/Zn²⁺

4.22 = log 1/Zn²⁺

1/Zn²⁺ = 16710

Zn²⁺ = 5.98x10^-5 M

(be sure to check all of the math)