Question

How is electrolysis used to extract aluminium?

Answer 1

The aluminium is extracted by the process of electrolysis of the alumina which is mixed with the molten cryolite. The process is termed as Hoope’s process.
Since it is a process of electrolysis, it is certain that it would have an anode and a cathode component.

Complete step-by-step answer: The extraction of aluminium is done by the electrolysis of the aluminium oxide. Before the beginning of the extraction process the aluminium oxide is melted so that the electricity could pass through it. As we know that aluminium has a very high melting point $2000{}^\circ C$ and so it has a high cost of melting it. So, as an alternative, the aluminium oxide is dissolved in molten cryolite which is nothing but an aluminium compound having a much lower melting point than that of aluminium oxide.
For the process of electrolysis, we use an electrolytic cell which contains a steel container which is lined with carbon, which acts as the cathode of the electrolytic cell. And some carbon rods are suspended in the electrolyte or dipped in the electrolyte which acts as the anode of the cell. A diagram showing the electrolytic cell is shown below.

Now we will write the individual reactions which are taking place in anode and cathode.
At cathode the aluminium ion is getting reduced to aluminium, by gaining three electrons. The reaction can be represented as,
$A{{l}^{+3}}+3{{e}^{-}}\to Al$
Now if we consider the next step, which is the reaction taking place at the anode, we would notice that the oxygen is getting oxidised by loss of two electrons from each oxygen ion.
$2{{O}^{2-}}\to {{O}_{2}}+4{{e}^{-}}$
As we can see, the oxygen gas is formed at the anode. Now, in order to balance the number of electrons in both the half reactions, we will multiply the equation of aluminium with $4$ and the reaction which takes place at anode with $3$ in order to get same number of electrons on both the reactions, which is $12$ electrons.
$4A{{l}^{+3}}+12{{e}^{-}}\to 4Al$
$6{{O}^{2-}}\to 3{{O}_{2}}+12{{e}^{-}}$
Now, after writing both the equations we can see that the number of electrons are equal and so if we add both these equations, the electron number would cancel out. And the overall equation becomes,
$2{{O}^{2-}}+4A{{l}^{+3}}\to 4Al+{{O}_{2}}$
Now, this molten aluminium is collected at the bottom of the electrolytic cell as shown in the diagram. Therefore, the extraction process gets completed.

Note: The extraction of aluminium from the aluminium oxide requires melting of the oxide at higher temperature which is very expensive and so another molten aluminium compound, cryolite, is used as a solvent for carrying out the extraction.
The aluminium gets reduced at the cathode and the oxidation of the oxygen ion takes place at the anode, which then contributes to the overall reaction equation of the process.