Problem: 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 V2 IO3-(aq) + 10 e- <=> I2(aq), E° = 1.10 Va. +0.33 Vb. -1.87 Vc. +1.87 Vd. -0.33 Ve. 0 V

FREE Expert Solution

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). 


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Problem Details

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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