An electric charge of 5 Faradays is passed through three electrolytes $A g N O_{3}, C u S O_{4}$ and $F e C l_{3}$ solution. The grams of each metal liberated at cathode will be
[Atomic weight; Fe - 56 g/mol, Cu - 63.5 g/mol, Ag - 108 g/mol]

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A g + + e - 1 mol ≡ 1F → A g C u 2 + + 2 e - 2 mol ≡ 2F → C u F e 3 + + 3 e - 3 mol ≡ 3F → F e Grams of Ag liberated = 108 1 × 5 = 540 g Grams of Cu liberated = 63.5 2 × 5 = 158.75 g Grams of Fe liberated = 56 3 × 5 = 93.3 g

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