What is the emf of a cell Ni(s) | Ni²⁺(0.1 M) || Cu²⁺(0.001 M) | Cu(s) at 298 K? (Given: E°Ni²⁺/Ni = -0.25 V , E°Cu²⁺/Cu
E°cell = 0.34 - (-0.25) = 0.59 V . Ecell = 0.59 - (0.059/2) log (0.1/0.001) = 0.59 - 0.059 = 0.531 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement