Practice question
Question
What is the emf of a cell Cd(s) | Cd²⁺(0.1 M) || Ag⁺(0.001 M) | Ag(s) at 298 K? (Given: E°Cd²⁺/Cd = -0.40 V , E°Ag⁺/Ag = 0.80 V )
Explanation
E°cell = 0.80 - (-0.40) = 1.20 V . Ecell = 1.20 - (0.059/2) log (0.1/0.001) = 1.20 - 0.059 = 1.141 V .
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