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