Question
Explain the electrometallurgy of aluminium.
✓
Answer
Electrochemical extraction of aluminium Hall-Herold process: In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as an anode. A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber. About 10%, calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K.
The chemical reactions involved in this process are as follows:
(i) Ionisation of alumina:
$Al _2 O _3 \longrightarrow 2 Al ^{3+}+3 O ^{2-}$
(ii) Reaction at cathode:
$Al ^{3+}($ melt $)+3 e ^{-} \longrightarrow Al _{( l )}$
(iii) Reaction at anode:
$2 O ^{2-}($ melt $) \longrightarrow O _2($ melt $)+4 e ^{-}$
Since carbon acts as an anode the following reaction also takes place on it.
$C _{( s )}+ O ^2( melt ) \longrightarrow CO +2 e ^{-}$
$C _{( s )}+2 O _2$ (melt) $\longrightarrow CO _2+4 e ^{-}$
Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom.
The net electrolysis reaction can be written as follows:
$4 Al ^{3+}($ melt $)+6 O ^{2-}( melt )+3 C _{( s )} \longrightarrow 4 Al _{( l )}+3 CO _{2( g )}$
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