Calculate the standard Gibbs energy change for the following cell Cd (s) | Cd ⁺⁺(1 M ) || Sn ⁺⁺(1 M )| Sn (s) if E^0_ cell is 0.27 V .

Step-by-step solution

The cell reaction is Cd (s) + Sn ²⁺ Cd ²⁺ + Sn (s) . The number of electrons transferred in the reaction is n = 2 . The standard Gibbs free energy change is given by the relation: G^0 = -nFE^0_ cell Substituting the given values: G^0 = -2 96500 C mol ⁻¹ 0.27 V G^0 = -52110 J mol ⁻¹ G^0 = -52.11 kJ mol ⁻¹ Answer: -52.11 kJ