How is alumina obtained from bauxite?

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Solution

Hall and Heroult, in 1885, developed the process for the extraction of aluminium. Aluminium is extracted from its bauxite ore $(A l_{2} O_{3} .2 H_{2} O)$ . Bauxite contains 60 % $A l _{2} O_{3}$ , the rest of the composition is sand, ferric oxide and titanium oxide.

Extraction of aluminium from bauxite involves two stages:

1. Conversion of Bauxite into Alumina by Bayer's Process

The conversion of bauxite into alumina involves the following steps:

1. Bauxite ore is finely divided and heated under pressure with concentrated caustic soda solution at 150°C to obtain sodium meta-aluminate.

$A l_{2} O_{3} .2 H_{2} O + 2 N a O H \to 2 N a A l O_{2} + 3 H_{2} O$

Bauxite Sodium meta

Aluminate

2. On diluting sodium meta-aluminate with water, aluminium hydroxide precipitate is formed.

$N a A l O_{2} + 2 H_{2} O \to N a O H + A l (O H)_{3}$

3. The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.

$2 A l (O H)_{3} \to A l_{2} O_{3} + 3 H_{2} O$

2. Electrolytic reduction of Alumina by Hall’s process

Aluminium is produced by the electrolytic reduction of fused alumina $(A l_{2} O_{3})$ in the electrolytic cell.

Cathode: Iron tank lined with graphite.

Anode: A bunch of graphite rods suspended in the molten electrolyte.

Electrolyte: Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)

Temperature: 900-950°C

Voltage used: 5-6V

The overall equation for aluminium extraction is

$2 A l_{2} O_{3} \to 4 A l + 3 O_{2}$

Aluminium is deposited at the cathode and oxygen gas is liberated at the anode. Oxygen combines with graphite to form $C O_{2}$ .

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