The G ^o for the reaction, Cd ²⁺(aq) + Zn (s) Zn ²⁺(aq) + Cd (s) is: [E^o_ Cd ²⁺/ Cd = -0.403, E^o_ Zn ²⁺/ Zn = -0.763 V ]

Step-by-step solution

For the reaction Cd ²⁺(aq) + Zn (s) Zn ²⁺(aq) + Cd (s) , Zn is oxidized (anode) and Cd ²⁺ is reduced (cathode). E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = -0.403 - (-0.763) = 0.360 V The number of electrons transferred in the reaction is n = 2 . The standard Gibbs free energy change is given by: G^ = -nFE^ _ cell G^ = -2 96500 0.360 G^ = -69480 J = -69.48 kJ Rounding to one decimal place, G^ -69.5 kJ . Answer: -69.5 kJ