For the galvanic (voltaic) cell Cd²⁺(aq) + Fe(s) ⟶ Cd(s) + Fe²⁺(aq) (E° = 0.0400 V), what is the ratio [Fe²⁺]/[Cd²⁺] when E = 0.006 V? Assume T is 298 K

Answer 1

Looking up standard reduction potentials for the 2 elements involved, I find...

Fe²⁺ ==> Fe Eº = -0.44

Cd²⁺ ==> Cd Eº = -0.40

So, Redox equation is Cd²⁺ + Fe ==> Cd + Fe²⁺ Eºcell = -0.40 - (-0.44) = 0.04 V

Use the Nernst Equation:

Ecell = Eºcell - RT/nF ln Q which @ 298K converts to...

Ecell = Eºcell - 0.0592/n log Q and we are to solve for Q which is the ratio of Fe²⁺ / Cd²⁺

n = # electrons transferred = 2

0.006 = 0.0400 - 0.0592/2 log Q

-0.034 = -0.0296 log Q

log Q = 1.149

Q = 14.1

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