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Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out in the presence of
$NaCl$
Fluorite
Cryolite which forms a melt with lower melting temperature
Cryolite which forms a melt with higher melting temperature
Solution
Hall and Heroult's process:
$2 Al _2 O _3+3 C \rightarrow 4 Al +3 O _2$
Cathode $: Al ^{3+}($ melt $)+3 e ^{-} \rightarrow Al (l)$
Anode $: C ( s )+ O ^{2-}( g )($ melt $) \rightarrow CO ( g )+2 e ^{-}$
$C ( s )+2 O ^{2-}( g )(\text { melt }) \rightarrow CO _2( g )+4 e ^{-}$
The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina $\left( Al _2 O _3\right)$ and cryolite $\left( Na _3 AlF _6\right.$, sodium hexafluoroaluminate) along with minor quantities of aluminum fluoride $\left( AlF _3\right)$ and fluorspar $\left( CaF _2\right)$. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature.
