All question types · Electrochemistry
Electrode potential away from standard conditions
UKChO Round 1 · unreviewed
Fe³⁺/Fe²⁺ half-cell at non-standard concentrations.
Worked example
Calculate the electrode potential at 298 K of a platinum electrode in a solution with [Fe³⁺] = 0.10 mol dm⁻³ and [Fe²⁺] = 0.010 mol dm⁻³. E°(Fe³⁺/Fe²⁺) = 0.77 V.
1. Divide [Fe²⁺] ÷ [Fe³⁺] → Q = [Fe²⁺]/[Fe³⁺]
= 0.1000
2. Nernst equation: solve for E
= 0.8291 V
Answer: E = 0.829 V
Method
For Fe³⁺ + e⁻ → Fe²⁺: E = E° − (RT/F) ln([Fe²⁺]/[Fe³⁺]). At 298 K, RT/F = 0.0257 V, so each tenfold change in the ratio shifts E by 59 mV.
Common mistakes
- For the reduction Fe³⁺ + e⁻ → Fe²⁺, Q = [Fe²⁺]/[Fe³⁺] (product over reactant). More Fe³⁺ makes E more positive.
Formulas used
Nernst equation