The standard electrode potential (E^°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is: [Given E^°(Fe³⁺/Fe)=-0.04 V and E^°(Fe²⁺/Fe)=-0.44 V]
- A.+0.40 V
- B.+0.76 V✓
- C.-0.48 V
- D.+0.92 V
Correct Answer
(B) +0.76 V
Solution & Explanation
Concept: standard potentials are NOT additive, but Gibbs energies ΔG° = -nFE° ARE additive. Write the target half-reaction as a combination: Fe³⁺ + 3e⁻ → Fe (n=3, E° = -0.04 V) Fe²⁺ + 2e⁻ → Fe (n=2, E° = -0.44 V) The desired step is Fe³⁺ + e⁻ → Fe²⁺, obtained as (Fe³⁺→Fe) minus (Fe²⁺→Fe). Using ΔG° = -nFE°: ΔG°(Fe³⁺/Fe) = ΔG°(Fe³⁺/Fe²⁺) + ΔG°(Fe²⁺/Fe) -3F(-0.04) = -(1)F·E° + [-2F(-0.44)] Divide by -F: 3(-0.04) = E° + 2(-0.44) -0.12 = E° - 0.88 E° = 0.88 - 0.12 = +0.76 V. Hence E°(Fe³⁺/Fe²⁺) = +0.76 V — answer (B). Trap: never simply average the two given potentials; weight by the number of electrons via ΔG°.
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