Question

The standard Gibbs free energy change (∆Gº in kJ mol^–1), in a Daniel cell ( V E^º_cell =1.1 V  ), when 2 moles of Zn(s) is oxidized at 298 K, is closest to 

(A) – 212.3 

(B) – 106.2 

(C) – 424.6 

(D) – 53.1 

Answer 1

Correct Option :- (C) – 424.6 

Explanation :-

Zn + Cu⁺² → Zn⁺² + Cu 

2Zn + 2Cu⁺² → 2Zn⁺² + 2Cu  

For 2 moles of Zn, n = 4 

∆Gº = –nFEº_Cell = – 4 × 96500 × 1.1 = – 424.6 kJ