Question

Draw labelled diagram of electrolytic cell for the extraction of aluminium and the overall reaction taking place in it.

Answer 1

Oxidation at anode:
$6O^{2-}-12e^{-}\to 3O_2$
Reduction at anode:
$4A{l}^{3+}+12e^{-}\to 4Al$
Overall reaction:
$2A{l}_2{O}_3\to 4Al+3O_2$

Following two concepts are there for the reaction taking place during electrolysis:
1. According to the first concept which is most famous, ionisation of alumina occurs firstly:
At cathode $\left\{\begin{array}{cc}\underset{Molten}{A{l}_2{O}_3}& \stackrel{Electrolysis}{\to }2A{l}^{3+}+3O^{2-}\\ A{l}^3+3e^{-}& \to Al\left(l\right)\end{array}\right\}$
At anode $\left\{\begin{array}{ccc}{O}^{2-}& \to & O+2e\\ O+O& \to & O_2\\ 4C+3O_2& \to & 2CO+2C{O}_2\end{array}\right\}$
$2C+O_2\to 2CO$
$2CO+O_2\to 2C{O}_2$
Hence, corrosion of graphite rods (anode) occurs and they have to be replaced periodically. For $1kg$ of aluminium produced about $0.5kg$ of anode (graphite) is consumed.
2. According to the second concept, ionisation of 'Cryolite' occurs firstly on passing electric current.
$\underset{Moltencryolite}{N{a}_{3}Al{F}_6}\stackrel{Electrolysis}{\rightleftharpoons }3NaF+Al{F}_3$
$Al{F}_3\stackrel{Electrolysis}{\rightleftharpoons }A{l}^{3+}+3F^{-}$
At cathode : $A{l}^{3+}+3e^{-}\to A{l}_{\left(l\right)}$
At anode : $F^{-}\to F+e^{-}$
$3C+2A{l}_2{O}_3\to 4Al+3C{O}_2$
At anode, $A{l}_2{O}_3$ is changed into $Al{F}_3$ and the process of