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⁺/N
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 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry