Electrolytic reduction of alumina to aluminiu | vedclass

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Question

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 (

Vedclass · Accepted Answer

Cryolite which forms a melt with lower melting temperature

Vedclass · Answer

Hall and Heroult's process:

$2 Al _2 O _3+3 C ightarrow 4 Al +3 O _2$

Cathode $: Al ^{3+}($ melt $)+3 e ^{-} ightarrow Al (l)$

Anode $: C ( s )+ O ^{2-}( g )($ melt $) ightarrow CO ( g )+2 e ^{-}$

$C ( s )+2 O ^{2-}( g )(	ext { melt }) ightarrow CO _2( g )+4 e ^{-}$

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina $\left( Al _2 O _3ight)$ and cryolite $\left( Na _3 AlF _6ight.$, sodium hexafluoroaluminate) along with minor quantities of aluminum fluoride $\left( AlF _3ight)$ and fluorspar $\left( CaF _2ight)$. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature.

Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out in the presence of

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