Question

What is the role of cryolite (Na₃AlF₆) in the electrolytic reduction of alumina in the Hall's process?

Answer 1

The extraction of aluminium from pure ore is done by electrolysis, called the Hall-Heroult's process. The electrolyte used in the cell consists of fused pure alumina (${\mathrm{Al}}_2{\mathrm{O}}_3$) mixed with cryolite (${\mathrm{Na}}_3{\mathrm{AlF}}_6$) and fluorspar (${\mathrm{CaF}}_2$). The addition of cryolite (Na₃AlF₆) and fluorspar (CaF₂) raises the electrical conductance of alumina by reducing its melting point from 2050 ^oC to 950 ^oC.
Cryolite (${\mathrm{Na}}_3{\mathrm{AlF}}_6$) acts as a solvent for the electrolytic mixture. It helps dissolve the fused pure alumina and fluorspar evenly, thereby, forming a molten electrolytic mixture.