The reaction at the cathode in the Hall-Heroult cell is

A $Al X^{3 +} + 3 e X^{-} Al$
B $Al Al X^{3 +} + 3 e X^{-}$
C $C + O X^{2 -} CO + 2 e X^{-}$
D $2 O X^{2 -} O X_{2} + 4 e X^{-}$

Answer: A

Reduction occurs at the cathode, so aluminium ions gain three electrons to form molten aluminium. The other three are oxidation reactions that take place at the anode.

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