Question 7In the electrolytic reduction of pure alumina to pure aluminium by Hall Herault's process, give the electrolytic reactions involved in the same, resulting in formation of aluminium at the cathode.

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Accepted Answer

In the electrolytic reduction of pure alumina to pure aluminium by Hall Herault's process, the constitutents of electrolyte are cryolite (Na₃AlF₆) and fluorspar (CaF₂) and alumina (Al₂O₃)

The electrolytic reactions are as follows :

Cryolite : Na₃AlF₆ ⇌ 3Na¹⁺ + Al³⁺ + 6F^1-

Fluorspar : CaF₂ ⇌ Ca²⁺ + 2F^1-

Alumina : Al₂O₃ ⇌ 2Al³⁺ + 3O^2-

At cathode : 2Al³⁺ + 6e^- ⟶ 2Al

At anode : 3O^2- - 6e^- ⟶ 3[O] ⟶ 3O₂

Al³⁺ are discharged in preference to Na¹⁺ and Ca²⁺ ions due to it's lower position in the electrochemical series.

Metallurgy — Question 16

Question 7