Question

State the reactions that take place at the cathode and anode during the process of electrolytic reduction of Alumina (Al₂O₃).

Answer 1

Extraction of aluminium:- Aluminium is obtained from Bauxite ore. This ore contain ferric oxide and silica as impurities.

Extraction of Aluminium involved in the following steps:-

• The concentration of the ore: If the ore contains ferric oxide is an impurity, it is concentrated by Baeyer's process

Baeyer's process:

Reactions involved in the production of Alumina:

a) Bauxite treated with Sodium hydroxide to give Sodium meta aluminate,

${\mathrm{Al}}_2{\mathrm{O}}_3.2{\mathrm{H}}_2\mathrm{O}+2\mathrm{NaOH}\to 2{\mathrm{NaAlO}}_2+3{\mathrm{H}}_2\mathrm{O}$

b) Sodium meta aluminate passing through water produce aluminium hydroxide,

$2{\mathrm{NaAlO}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

c) Aluminium hydroxide, which is heated at 1500^oC produces Alumina.

$2\mathrm{Al}{\left(\mathrm{OH}\right)}_3\stackrel{{1500}^{\mathrm{o}}\mathrm{C}}{\to }{\mathrm{Al}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}$

Electrolytic reduction of pure Alumina:

The electrolysis is carried out in an iron tank lined inside with carbon which acts as the cathode.

The anode consists of a carbon rod dipped in the fused mixture of electrolytes, it is covered with a layer of coke.

On passing current, Aluminium is discharged at the cathode.

Aluminium being heavier than electrolyte settles down in the bottom, and oxygen is liberated at the anode.

It attacks carbon rods forming CO and CO₂.

To increase the electrical conductivity and to reduce fusion temperature Cryolite(Na₃AlF₆) and Fluorspar (CaF₂) are added to Alumina.

Mechanism:-

AlF₃ from Cryolite ionises as

${\mathrm{AlF}}_3(\mathrm{alluminium}\mathrm{tryflouride})\rightleftharpoons {\mathrm{Al}}^{3+}+3{\mathrm{F}}^{-}$

At cathode: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

At anode: $2{\mathrm{F}}^{-}\to {\mathrm{F}}_2+2{\mathrm{e}}^{-}$

The liberated fluorine react with alumina to form AlF₃ and O₂

${\mathrm{Al}}_2{\mathrm{O}}_{3\to }+3{\mathrm{F}}_2\to 2{\mathrm{AlF}}_3+\frac{3}{2}{\mathrm{O}}_2$

$2\mathrm{C}+{\mathrm{O}}_2\to 2\mathrm{CO}$

$\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2$

Therefore in the electrolytic method aluminium acts as a cathode and impure aluminium acts as anode.