We’re asked to determine the Ecell of the reaction given the half-reactions:
Fe3+(aq) + e- ⇋ Fe2+(aq) , E° = 0.77 V
2 IO3-(aq) + 10 e- ⇋ I2(aq) E° = 1.10 V
In this case, we need to determine the oxidation/reduction reactions:
Loss of Electron is an Oxidation process undergone by the Reducing Agent (LEORA). In the Anode, Oxidation happens (An Ox)
Gain of Electron is a Reduction process undergone by the Oxidizing Agent (GEROA). Reduction happens in the Cathode (Red Cat)
Upon inspection, notice that the overall reaction needs 2 IO3-(aq) in the product side.
This means, the half-reaction 2 IO3-(aq) + 10 e- ⇋ I2(aq) must be the oxidation reaction, thus the anode in this reaction (disregarding which has greater or less than Ecell).
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
a. +0.33 V
b. -1.87 V
c. +1.87 V
d. -0.33 V
e. 0 V
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