A cell Zn(s) | Zn²⁺(0.02 M) || Fe³⁺(0.05 M), Fe²⁺(0.01 M) | Pt(s) operates at 298 K. What is the cell potential? (Given:
E°cell = 0.77 - (-0.76) = 1.53 V . Ecell = 1.53 - (0.059/1) log ([Zn²⁺][Fe²⁺]/[Fe³⁺]) = 1.53 - 0.059 log (0.02 × 0.01/0.05) = 1.53 + 0.0566 = 1.5866 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
