Question

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

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Solution

## 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.

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