A cell Cd(s) | Cd²⁺(0.005 M) || Br₂(l) | Br⁻(0.01 M) | Pt(s) operates at 298 K. What is the cell potential? (Given: E°Cd
E°cell = 1.07 - (-0.40) = 1.47 V . Ecell = 1.47 - (0.059/2) log ([Cd²⁺][Br⁻]²/1) = 1.47 - 0.0295 log (0.005 × 0.0001) = 1.47 + 0.103 = 1.573 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant