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For the reaction taking place in the cell: Pt s H 2 g H + aq Ag + aq Ag s E cell ° = + 0 . 5332 V . The value of Δ f G ⊖ is kJmol - 1 . (in nearest integer)

Correct answer

0

Step-by-step solution

Anode   ⇒ 1 2 H 2 g ⟶ H + aq + e - Cathode   ⇒ Ag + + e - ⟶ Ag ( s ) 1 2 H 2 g + Ag + aq ⟶ n = 1 H + aq + Ag s Relation between Gibbs Free Energy and EMF of a Cell is given by ΔG ° = - nFE cell ° = - 1 × 96500 × 0 . 5332 = 51453 . 8   J   mol - 1 = 51 . 4538   kJ   mol - 1

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