The three cells with their E_cell^∘ values are given below.
Cells E_cell^∘/V
(a) Fe|Fe²⁺||Fe³⁺) |Fe 0.404
(b) Fe∣|Fe²⁺||Fe³⁺,Fe²⁺|Pt 1.211
(c) Fe∣|Fe³⁺||Fe³⁺,Fe²⁺|Pt 0.807
The standard Gibbs free energy change values for three cells are, respectively, ( F represents charge on 1 mole of electrons)

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+2.424F,+2.422F,+2.42 F

-2.424F,-2.422F,-2.421F

-0.808F,-2.422F,-2.421F

-1.212F,-1.211F,-0.807F

answer is C.

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Detailed Solution

Formula ΔG^∘=-nFE^∘_cell
(a) n=6∴ΛG^∘ =-6×F×0.404
=-2.424F.
(b) n=2∴ΔG^∘ =-2×F×1.211
=-2.422F.
(c) n=3∴ΔG^∘ =-3×F×0.807
=-2.42F.

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