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Standard electrode potentials: E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(I2/I⁻) = +0.54 V. For the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I2, E° equals:
A+1.31 V
B0
C-0.23 V
D+0.23 V
Answer & Solution
Correct answer: D. +0.23 V
E°_cell = E°_cathode - E°_anode = 0.77 - 0.54 = +0.23 V. Positive, so reaction spontaneous. Fe³⁺ oxidizes I⁻ to I2.
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