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NTA Abhyas JEE Main2020ChemistryGeneral Principles and Processes of Isolation of MetalsPractice

The minimum voltage required to electrolyse alumina in the Hall-Heroult process is Given, Δ G f ∘ Al 2 O 3 = - 1520 kJ mol -1 Δ G f ∘ CO 2 = - 394 kJ mol -1

Options

  1. A1.575 V
  2. B1.60 V
  3. C1.312 V
  4. D- 2.62 V

Correct answer

B. 1.60 V

Step-by-step solution

2 Al 2 O 3 s + 3 C → 1 2 e - 4 Al s + 3 CO 2 g Δ G ∘ = 3 Δ G f ∘ CO 2 - 2 Δ G ∘ Al 2 O 3 = - 3 × 3 9 4 - 2 - 1 5 2 0 = - 1 1 8 2 + 3 0 4 0 = + 1858 kJ Δ G ∘ = - nFE ∘ 1 8 5 8 × 1 0 0 0 = - 1 2 × 9 6 5 0 0 × E ∘ ∴ E ∘ = - 1.60 V Thus, voltage requires = 1.60 V

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