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On the basis of the information available from the reaction 4/3 Al + O2 → 2/3 Al2O3, ΔG = –827 kJ mol–1 of O2. The minimum e.m.f required to carry out an electrolysis of Al2O3 is (F = 96500 C mol–1)

1. 3.56 V
2. 2.14 V
3. 4.28 V
4. 6.42 V

Correct answer: 3.56 V

Solution

The minimum e.m.f. required for electrolysis is calculated using ΔG = -nFE. Here, ΔG = -827 kJ/mol = -827,000 J/mol, n = 4 (as 4 electrons are involved per O2 molecule), and F = 96500 C/mol. Solving for E gives E = ΔG / (-nF) = 827,000 / (4 × 96500) = 3.56 V.

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