Practice question
Question
A cell Fe(s) | Fe²⁺(0.02 M) || I₂(s) | I⁻(0.1 M) | Pt(s) operates at 298 K. What is the cell potential? (Given: E°Fe²⁺/Fe = -0.44 V , E°I_₂/I⁻ = 0.54 V )
Explanation
E°cell = 0.54 - (-0.44) = 0.98 V . Ecell = 0.98 - (0.059/2) log ([Fe²⁺][I⁻]²/1) = 0.98 - 0.0295 log (0.02 × 0.01) = 0.98 + 0.068 = 1.048 V .
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