Practice question
Question
What is the emf of a cell Zn(s) | Zn²⁺(0.001 M) || H⁺(1 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Zn²⁺/Zn = -0.76 V , E°H⁺/H_₂ = 0.00 V )
Explanation
E°cell = 0.00 - (-0.76) = 0.76 V . Ecell = 0.76 - (0.059/2) log (0.001/1²) = 0.76 + 0.0885 = 0.8485 V .
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