The standard Gibbs free energy change (∆Gº in kJ mol–1), in a Dan

1.	The standard Gibbs free energy change (∆Gº in kJ mol–1), in a Daniel cell ( V Eºcell =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» 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

The standard Gibbs free energy change (∆Gº in kJ mol–1), in a Daniel cell ( V Eºcell =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»

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

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The standard Gibbs free energy change (∆Gº in kJ mol–1), in a Daniel cell ( V Eºcell =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

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