Voltaic Equilibrium?

Question

A voltaic cell consists of a Ag/Ag+ electrode (E° = 0.80 V) and a Fe2+/Fe3+ electrode (E° = 0.77 V) with the following initial molar concentrations: [Fe2+] = 0.30 M; [Fe3+] = 0.10 M; [Ag+] = 0.30 M. What is the equilibrium concentration of Fe3+? (Assume the anode and cathode solutions are of equal volume, and a temperature of 25°C.)

Answer 1

By definition

1. Ag/Ag+ half reaction

E = 0.8 + 0.0591log [Ag+]

2. Fe2+/Fe3+ half reaction

E = 0.77 - 0.0591log([Fe2+]/[Fe3+])

The overall reaction is Fe2+ + Ag+ ===> Fe3+ + Ag

and K = [Fe3+]/([Fe2+] [Ag+])

At equilibrium the half cell voltages are equal and so

0.80 - 0.77 = 0.0591log[Fe3+]/([Fe2+][Ag+]) or log K = 0.5076. Therefore K = 3.218

The equilibrium concs will be

[Fe3+] = 0.1 + x
[Fe2+] = 0.3 - x
[Ag+] = 0.3 - x

where x is the reaction product. Substituting into K yields

(0.1 + x)/(0.3 - x)^2 = 3.218. On expansion yields the quadratic 3.3218x^2 - 2.9308x + 0.18962 = 0

Applying the quadratic formula, gives x = 0.0701 mol/l. Thus

[Fe3+] = 0.170 mol/l
[Fe2+] =0.230 mol/l
[Ag+] = 0.230 mol/l

Answer 2

emf is 0.03 V

Find K by Nernst eqn.