Question

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C.

The initial concentrations of Ni2+ and Zn2+ are 1.50molL−1 and 0.100 molL−1, respectively.

Zn2+(aq)+2e−→Zn(s) E∘=−0.76V

Ni2+(aq)+2e−→Ni(s) E∘=−0.23V

What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.46 V?

Answer 1

oxidation : Zn --------------> Zn+2 + 2e-    Eo = - 0.76 V

reduction : Ni+2 + 2e- --------------> Ni      Eo = - 0.23

overall reaction : Zn + Ni +2 -----------> Zn +2 + Ni

Eocell = Eo red - Eo ox

          = - 0.23 - (- 0.76)

          = 0.53 V

E cell = E0 - 0.05916/n x log K

Given [Ni+2] =1.50 M

[Zn+2] = 0.100 M

Ecell = 0.53 - 0.05916 /2 log [Zn+2] / [Ni+2]

0.46 = 0.53 - 0.05916 / 2 log [0.1 + x / 1.50 - x ]

[0.1 + x / 1.50 - x ] = 232.5

0.1 + x = 348.78 - 232.5 x

x = 1.493

[Zn+2] = 0.1 + 1.493 = 1.59 M

[Zn+2] = 1.59 M

[Ni+2] = 0.00700 M