Practice question
Question
The emf of a cell Ni(s) | Ni²⁺(0.001 M) || Ag⁺(0.002 M) | Ag(s) at 298 K is (Given: E°Ni²⁺/Ni = -0.25 V , E°Ag⁺/Ag = 0.80 V )?
Explanation
E°cell = 0.80 - (-0.25) = 1.05 V . Ecell = E°cell - (0.059/2) log ([Ni²⁺]/[Ag⁺]²) = 1.05 - (0.059/2) log (0.001/(0.002)²) = 0.989 V .
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