Given the standard potential for the following half-cell reaction at 298 K, C u + a q + e - → C u s ; E ° = 0.52 V C u 2 + a q + e - → C u + a q ; E ° = 0.16 V Calculate the Δ G ° (kJ) for the reaction, 2 C u + a q → C u s + C u 2 +

Step-by-step solution

C u + + e - → C u ; E ° = 0.52 V Δ G 1 = - n F E ° = - 1 × 96500 × 0.52 ...(i) C u 2 + + e - → C u + ; E ° = 0.16 V or C u + → C u 2 + + e - ; E ° = - 0.16 V Δ G 2 = - 1 × 96500 × - 0.16 ...(ii) On adding equations (i) and (ii), we get 2 C u + → C u + C u 2 + , Δ G = Δ G 1 + Δ G 2 = - 96500 × 0.52 + 96500 × 0.16 = 96500 - 0.52 + 0.16 = - 96500 × 0.36 = − 34740 J = − 34 .740 kJ