Question

In 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.

  • Metallurgy
WhatsApp

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+ + 6F1-

  • Fluorspar : CaF2 ⇌ Ca2+ + 2F1-
  • Alumina : Al2O3 ⇌ 2Al3+ + 3O2-
  • At cathode : 2Al3+ + 6e– ⟶ 2Al
  • At anode : 3O2- – 6e– ⟶ 3[O] ⟶ 3O2

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

Was this answer helpful?

Didn't liked the above answer ?

Text Generation Tool

💡 Some Related Questions

Metallurgy

Give reasons for the following pertaining to Hall Herault’s process.
(i) The fusion temperature of the electrolyte has to be lowered before conducting the electrolytic reduction.
(ii) The constituents of the electrolyte in addition to one part of fused alumina contains three parts of cryolite and one part of fluorspar.
(iii) A layer of powdered coke sprinkled over the electrolytic mixture, protects the carbon electrodes.
(iv) It is preferred to use a number of graphite electrodes as anode, instead of a single graphite electrode.

Open Answer »