Practice question
Question
What is the emf of a cell Fe(s) | Fe²⁺(0.1 M) || H⁺(0.001 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Fe²⁺/Fe = -0.44 V , E°H⁺/H_₂ = 0.00 V )
Explanation
E°cell = 0.00 - (-0.44) = 0.44 V . Ecell = 0.44 - (0.059/2) log (0.1/0.001²) = 0.44 - 0.0885 = 0.3515 V .
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