Anodic product in the electrolytic reduction of alumina is:

C O_{2}, O_{2}

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O_{2}

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F_{2}

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O_{2}, 4 F_{2}

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Solution

The correct option is B $O_{2}$
Alumina is $A l_{2} O_{3}$ .

Ionisation of alumina: $A l_{2} O_{3} \to 6 O_{2}^{-} + 4 A l^{3 +}$ .

Reaction at cathode: $4 A l^{3 +} + 12 e^{-} \to 4 A l$ .

Reaction at anode: $6 O_{2}^{-} \to 3 O_{2} + 12 e^{-}$ .

Thus, the anodic product in electrolytic reduction of alumina is $O_{2}$ .

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Anodic product in the electrolytic reduction of alumina is:

The correct option is B O2Alumina is Al2O3.Ionisation of alumina: Al2O3→6O−2+4Al3+.Reaction at cathode: 4Al3++12e−→4Al.Reaction at anode: 6O−2→3O2+12e−.Thus, the anodic product in electrolytic reduction of alumina is O2.