Calculate E°(cell) for the reaction, 10 Fe3+(aq) + I2(aq) <=> 2 IO3-(aq) + 10 Fe2+(aq) given the reduction potentials:

Question

Fe³⁺(aq) + e^- <=> Fe²⁺(aq) , E° = 0.77 V

2 IO₃^-(aq) + 10 e^- <=> I₂(aq), E° = 1.10 V

answer: -0.33 V

Looking for an explanation on how to work this problem because I thought 1.10 V represented electrode at the cathode, but this problem seems to take a different approach.

Answer 1

Take a look at the overall reaction which shows Fe³⁺ going to Fe²⁺ and I₂ going to IO^3-.

This indicates the following:

Fe³⁺ + electrons ==> Fe²⁺ REDUCTION (cathode)

I₂ ==> IO₃^- + electrons OXIDATION (anode)

So, while you would be correct that 1.10 V would normally represent the cathode, in this case, it is opposite and this is the oxidation reaction, so it is the anode. Clearly the reaction will not be spontaneous since the Eº is negative.

Eº = cathode - anode

Eº = 0.77 V - 1.10 V = -0.33 V