Practice question
Question
What is the emf of the cell Cd(s) | Cd²⁺(0.01 M) || Ni²⁺(0.1 M) | Ni(s) at 298 K? (Given: E°Cd²⁺/Cd = -0.40 V , E°Ni²⁺/Ni = -0.25 V )
Explanation
E°cell = -0.25 - (-0.40) = 0.15 V . Ecell = 0.15 - (0.059/2) log (0.01/0.1) = 0.15 + 0.0295 = 0.1795 V .
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