Question

Explain briefly the extraction of aluminium from bauxite.

Answer 1

Hint – You can start by describing the Bayer process that converts bauxite to alumina (bauxite is processed to make sodium aluminate, which is then processed to aluminium hydroxide and then finally to alumina). Then follow up by describing the electrolytic process which converts alumina to aluminium.

Complete step by step solution:

The extraction of aluminium from bauxite takes place in two major steps –
Step 1 – Refining bauxite to produce alumina (aluminium oxide)
- The process of refining bauxite to produce alumina ($A{l_2}{O_3}$). Bauxite, the most important ore of aluminium, contains $30 - 60\%$aluminium oxide, and a mixture of silica, iron oxides and titanium dioxide.
- The bauxite is heated ($150^\circ to200^\circ C$) with sodium hydroxide ($NaOH$) solution. The aluminium is separated out as sodium aluminate ($NaAl{O_2}$) and silicon separates out as sodium silicate$N{a_2}Si{O_3}$.
$A{l_2}{O_3} + 2NaOH\xrightarrow{{}}2NaAl{O_2} + {H_2}O$
$2NaOH + Si{O_2}\xrightarrow{{}}N{a_2}Si{O_3} + {H_2}O$
- Then the solution is cooled and then carbon dioxide ($C{O_2}$) is bubbled through it. This coverts sodium aluminate to aluminium hydroxide.
$2NaAl{O_2} + 3{H_2}O + C{O_2}\xrightarrow{{}}2Al{(OH)_3} + N{a_2}C{O_3}$
- Aluminium hydroxide when heated at $1470K$ in rotary kilns produces alumina (aluminium oxide)
$2Al{(OH)_3}\xrightarrow{{}}A{l_2}{O_3} + 3{H_2}O$
Step 2 – The reduction of Alumina by electrolysis
The reduction of alumina by electrolysis is carried out in an iron box and is supported by fluorspar ($Ca{F_2}$) and cryolite ($N{a_3}Al{F_6}$). Cryolite and fluorspar is used to lower the melting point and increase the conductivity of the electrolyte. We use graphite rods as anode (positive) and cathode (negative).
In electrolysis aluminium ions being positive move towards cathode and oxygen towards anode.

1. At anode the oxygen liberated reacts with carbon from graphite rods to produce $CO$ and $C{O_2}$.
   $C + {O^{2 - }}\xrightarrow{{}}CO + 2{e^ - }$
   $C + 2{O^{2 - }}\xrightarrow{{}}C{O_2} + 4{e^ - }$
   At cathode –
   $A{l^{3 + }} + 3{e^ - }\xrightarrow{{}}Al$
2. At cathode $98.8\%$ pure aluminium is obtained.
   Hence, Bayer processes and electrolysis convert bauxite to alumina.

Note - The process of refining bauxite to produce alumina is called the Bayer process. This process can be heavily affected by the presence of organic impurities. The Bayer process is considered uneconomic when bauxites have more than $10\%$silica, because sodium aluminium silicate is formed, reducing yield and in this case some other process is used.