Practice question
Question
What is the emf of the cell Mn(s) | Mn²⁺(0.05 M) || H⁺(0.005 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Mn²⁺/Mn = -1.18 V , E°H⁺/H_₂ = 0.00 V )
Explanation
E°cell = 0.00 - (-1.18) = 1.18 V . Ecell = 1.18 - (0.059/2) log (0.05/0.005²) = 1.18 - 0.07375 = 1.10625 V .
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