The minimum voltage required to electrolyse alumina in the Hall-Heroult process is

Given, $\Delta {G}_f^{∘} \left({Al}_{2} {O}_{3}\right) = - {1520 kJ mol}^{-1}$

$\Delta {G}_f^{∘} \left({CO}_{2}\right) = - {394 kJ mol}^{-1}$

Question

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

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Correct answer: B. 1.60 V

2 Al _ 2 O _ 3 (s) + 3 C 1 2 e^ - → 4 Al (s) +3 CO _ 2 (g) &Delta; G^ ∘ = 3 &Delta; G _f^ ∘ ( CO _ 2 ) - 2 &Delta; G^ ∘ ( Al _ 2 O _ 3 ) = - 3 × 3 9 4 - 2 (-1 5 2 0) = - 1 1 8 2 + 3 0 4 0 = + 1858 kJ &Delta; G^ ∘ = - nFE ^ ∘ 1 8 5 8 × 1 0 0 0 = - 1 2 × 9 6 5 0 0 × E^ ∘ &there4; E^ ∘ = - 1.60 V Thus, voltage requires = 1.60 V

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