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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]

Options

  1. AAg = 10.8 g, Cu = 12.7 g, Fe = 1.11 g
  2. BAg = 540 g, Cu = 367.5 g, Fe = 325 g
  3. CAg = 108 g, Cu = 63.5 g, Fe = 56 g
  4. DAg = 540 g, Cu = 158.8 g, Fe = 93.3 g

Correct answer

D. Ag = 540 g, Cu = 158.8 g, Fe = 93.3 g

Step-by-step solution

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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