What is the emf of the cell Sn(s) | Sn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Sn²⁺/Sn = -0.14
E°cell = 0.00 - (-0.14) = 0.14 V . Ecell = 0.14 - (0.059/2) log (0.05/0.1²) = 0.14 + 0.039 = 0.179 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant