Practice question
Question
What is the emf of the cell Mn(s) | Mn²⁺(0.005 M) || Hg²⁺(0.02 M) | Hg(l) at 298 K? (Given: E°Mn²⁺/Mn = -1.18 V , E°Hg²⁺/Hg = 0.85 V )
Explanation
E°cell = 0.85 - (-1.18) = 2.03 V . Ecell = 2.03 - (0.059/2) log (0.005/0.02) = 2.03 + 0.02065 = 2.05065 V .
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