(a) Name the ore from which aluminium is extracted.
(b) How is ore purified (give equations also)?
(c) How is aluminium obtained from pure ore?
(d) Which method of refining is used to get 99.8% pure aluminium?

(a) Aluminium is mainly extracted from bauxite ore (${\mathrm{Al}}_2{\mathrm{O}}_3.x{\mathrm{H}}_2\mathrm{O}$).
(b) Bauxite ore is purified by Baeyer's process. In this process, the bauxite ore is powdered and mixed with sodium hydroxide solution and heated strongly. This results in the formation of sodium aluminate.

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{NaOH}\stackrel{}{\to }2{\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{From}\mathrm{Bauxite}\mathrm{ore}\mathrm{Sodium}\mathrm{hydroxide}\mathrm{Sodium}\mathrm{aluminate} \end{gathered}$$

The sodium aluminate solution is filtered and cooled. Then, it is passed through tanks called precipitators where hydrolysis of sodium aluminate takes place and insoluble aluminium hydroxide is formed.

$$\begin{gathered} {\mathrm{NaAlO}}_2+2{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\mathrm{NaOH}+\mathrm{Al}(\mathrm{OH}{)}_3 \\ \mathrm{Sodium}\mathrm{aluminate}\mathrm{Sodium}\mathrm{hydroxide}\mathrm{Aluminium}\mathrm{hydroxide} \end{gathered}$$

The aluminium hydroxide recovered through filtration is dried and heated strongly to around 1100 ^oC to 1200 ^oC. This results in the decomposition of aluminium hydroxide to aluminium oxide (alumina).

$$\begin{gathered} 2\mathrm{Al}(\mathrm{OH}{)}_3\stackrel{}{\to }{\mathrm{Al}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Aluminium}\mathrm{hydroxide}\mathrm{Aluminium}\mathrm{oxide} \end{gathered}$$

(c) Aluminium is extracted from pure ore by electrolysis. This is called the Hall-Heroult's process. In this method, an electrolytic cell is made of steel with an inner layer of carbon. The inner gas carbon layer with thick carbon rods dipped in fused electrolyte is taken as the anode. Fused pure alumina (${\mathrm{Al}}_2{\mathrm{O}}_3$) mixed with cryolite (${\mathrm{Na}}_3{\mathrm{AlF}}_6$) and fluorspar (${\mathrm{CaF}}_2$) is taken as the electrolyte. This mixture raises the electrical conductance of alumina and reduces its melting point from 2050 ^oC to 950 ^oC. ​When electricity is passed through the electrodes, the following reactions occur at the cathode and anode:

Dissociation of alumina: ${\mathrm{Al}}_2{\mathrm{O}}_3\leftrightarrow 2{\mathrm{Al}}^{3+}+3{\mathrm{O}}_{}^{2-}$

Reaction at cathode: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Al}$

Reaction at anode: $$\begin{gathered} {\mathrm{O}}_{}^{2-}\stackrel{}{\to }\mathrm{O}+2{\mathrm{e}}^{-} \\ \mathrm{O}+\mathrm{O}\stackrel{}{\to }{\mathrm{O}}_2 \end{gathered}$$
The oxygen so produced reacts with the carbon anode and forms CO and CO₂. Due to oxidation, the carbon anode needs to be replaced regularly.

(d) Hoope's electrolytic refining method is used to get 99.8% pure aluminium.