Practice question
Question
A cell Zn(s) | Zn²⁺(0.02 M) || Fe³⁺(0.05 M), Fe²⁺(0.01 M) | Pt(s) operates at 298 K. What is the cell potential? (Given: E°Zn²⁺/Zn = -0.76 V , E°Fe³⁺/Fe²⁺ = 0.77 V )
Explanation
E°cell = 0.77 - (-0.76) = 1.53 V . Ecell = 1.53 - (0.059/1) log ([Zn²⁺][Fe²⁺]/[Fe³⁺]) = 1.53 - 0.059 log (0.02 × 0.01/0.05) = 1.53 + 0.0566 = 1.5866 V .
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