Question

Consider an electrochemical cell based on the following reaction:

Zn(s) + 2H2O(l) + 2OH-(aq) ⇄ [Zn(OH)4]2-(aq) + H2(g)

All dissolved species are at 2.5 x 10-3 M; pressure of H2 is 1 bar (standard state) Use the Nernst equation to calculate E.

Answer 1

Oxidation half cell ( Anode)

Zn(s) + 2H2O(l) + 2OH-(aq) -------> [Zn(OH)4]2- + 2e

E°red = -1.199V

Reduction Half Cell( Cathode)

2H+ + 2e --------> H2(g)

E°red = 0.000V

Overall eeaction

Zn(s) + 2H2O(l) + 2OH- (aq) ------> [Zn(OH)4]2-(aq) + H2(g)

E°cell = E°red,cathode - E°red,anode

= 0.000V - ( - 1.199V)

= + 1.199V

Nernst equation is

Ecell = E°cell - ( 0.0592V/n)logQ

number of electron transfer ,n = 2

Q = [Zn(OH)4]2- × p(H2)/[OH-]^2

= 0.0025/(0.0025)^2

= 400

Therefore,

Ecell = 1.199V - (0.0592V/2)log(400)

= 1.199V - 0.077V

= 1.122V