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On the basis of information available for the reaction. 4 3 Al + O 2 → 2 3 Al 2 O 3 ; ΔG = –827 kJ/mol of O 2 The minimum emf required to carry out an electrolysis of A l 2 O 3 is: Given: One Faraday = 96500 C

Options

  1. A2.14 V
  2. B4.28 V
  3. C6.42 V
  4. D8.56 V

Correct answer

A. 2.14 V

Step-by-step solution

A l → A l 3 + + 3 e – 4 3 mol   Al ≡ 4 3   ×   3   mol   e − ≡ 4   mol   e – n = 4 Δ G = – n F E – 827 × 1000 = – 4 × 96500 × E E = 2.14   V

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