The charge (in F) required to deposites all Al from the electrolysis of 1 mol molten Al $_{2}$ O $_{3}$ is :

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Solution

$2 A l^{3 +} + 6 e^{-} \to 2 A l$
Hence we notice that 6 electrons are used in the reduction.
Hence for 1 mole of $A l_{2} O_{3}$ we need $6 F$ electrons.

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Similar questions

Δ G = - 827 k J {m o l}^{- 1}

O_{2}

{A l}_{2} O_{3}

(F = 96500 C {m o l}^{- 1})

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