Question

Why the Aluminum is extracted through an electrolytic process and not by using a reducing agent? Write the reaction taking place at anode and cathode during the electrolytic reduction of Alumina.

Answer 1

Extraction of Aluminum

Aluminum is an active metal, extracted from the electrolytic process, since using reducing agents such as Carbon, Sodium, Magnesium, etc. is difficult and economically unfeasible.

Any metal in the activity series above Carbon must be extracted via electrolysis.

Electrolytic process of fused Alumina

Aluminum$\left(\mathrm{Al}\right)$ is extracted in an electrolytic cell by electrolysis of molten Aluminum oxide or Alumina$\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$.

In this cell, there is a rectangular iron tank with a sloping bottom that aids in the removal of molten Aluminum.

Carbon(graphite) is used to make the anode of the cell since it oxidizes to Carbon dioxide$\left({\mathrm{CO}}_2\right)$ during the process.

The surface of the electrolyte is strewn with powdered coke.

The addition of Cryolite$\left({\mathrm{Na}}_3{\mathrm{AlF}}_6\right)$ lowers the fusion temperature from${205}^0$to ${950}^{o}C$ and increases conductivity.

Fluorspar$\left({\mathrm{CaF}}_2\right)$ serves as a solvent for the electrolytic mixture, increasing its conductivity.

Powdered coke reduces heat loss due to radiation and prevents anode burning.

There are several types of ion solutions, including ${\mathrm{Na}}^{+},{\mathrm{Ca}}^{2+},{\mathrm{Al}}^{3+}$and${\mathrm{F}}^{-},{\mathrm{O}}^{2-}$.

Aluminum$\left(\mathrm{Al}\right)$ is present at a lower position in the activity series, it is preferred to be discharged at the cathode while Oxygen$\left({\mathrm{O}}_2\right)$ is discharged at the anode.

Electrolysis of Alumina is as follows:

${\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)\rightleftharpoons 2{\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{O}}^{2-}\left(\mathrm{aq}\right)$

At Anode: $4{\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+12{\mathrm{e}}^{-}\to 4\mathrm{Al}\left(\mathrm{s}\right)$

At Cathode: $6{\mathrm{O}}^{2-}\left(\mathrm{aq}\right)\to 3{\mathrm{O}}_2\left(\mathrm{g}\right)+12{\mathrm{e}}^{-}$

Therefore, Aluminum$\left(\mathrm{Al}\right)$ is extracted through an electrolytic process and is not reduced using reducing agents.