The following cell has a potential of 0.10 V at 25oC. What is the pH of the solution in the cathode compartment?

Question

Ni_(s) | Ni²⁺_(aq) (1.0 M) || H⁺ (? M) | H_2(g) (1.0 atm) | Pt_(s)  

Answer 1

2H⁺ Ni(s) ==> H₂(g) + Ni²⁺(aq)

Nernst equation:

Ecell = Eºcell - RT/nF ln Q and @25ºC this can be written as...

Ecell = Ecell - 0.0592/n log Q

First, we'll find Eºcell: 0 - (-0.23) = +0.23 V (I used -0.23 as std reduction potential for Ni²⁺)

0.1 V = 0.23 V - (0.0592/2)(log Q) = 0.23 V - 0.0296 log Q

log Q = 4.39

Q = 24,547 = [Ni²⁺] / [H+]² = 1/x²

x² = 1/24547

x² = 4.07x10^-5

x = 6.4x10^-3 M H⁺

(please check the math)