Question

Alumina is a common term used for aluminium oxide $A{l_2}{O_3}\;$ . The alumina $A{l_2}{O_3}\;$ is soluble in a mixture of molten fluorspar and cryolite that lowers the melting point. The aluminium is extracted from bauxite. The bauxite is the aluminium ore. So, the process of extracting aluminium is from the aluminium oxide $A{l_2}{O_3}\;$ through the Hall- Heroult process.

Answer 1

Alumina is a common term used for aluminium oxide . The alumina is soluble in a mixture of molten fluorspar and cryolite that lowers the melting point. The aluminium is extracted from bauxite. The bauxite is the aluminium ore. So, the process of extracting aluminium is from the aluminium oxide through the Hall- Heroult process.

Complete step by step solution:
The extraction of aluminium from aluminium oxide $A{l_2}{O_3}\;$ has two steps. The first step is Bayer’s process, where the purification of aluminium oxide $A{l_2}{O_3}\;$ in bauxite takes place and the second step is Hall-Heroult process, in which the extraction of aluminium takes place from aluminium oxide $A{l_2}{O_3}\;$ .
In Hall- Heroult process the electrolytic reduction takes place which means, the separation of ions by using the electricity in which the aluminium oxide $A{l_2}{O_3}\;$ is carried out at very high temperature $\sim 980^\circ C$ to extract the aluminium from aluminium oxide $A{l_2}{O_3}\;$ .
The melting point of aluminium oxide is very high (i.e., $2000^\circ C$) . It's very difficult to attain such a high temperature.
The compound cryolite $N{a_3}Al{F_6}$ is added to aluminium oxide to lower down its temperature. The cryolite has another purpose too since it also increases the conductivity of molten aluminium oxide and this helps in the process of electrolytic reduction.
Therefore, the best way to extract the aluminium from aluminium oxide $A{l_2}{O_3}\;$ is by doing the electrolysis method. In this process fused fluorspar and cryolite are added to aluminium oxide and the mixture melts at high temperature and aluminium oxide becomes a good conductor of electricity. In this process, the aluminium is liberated at cathode.
Cathode:
$A{l^{3 + }}\, + \,3{e^ - }\, \to \,Al$
Anode:
$2{O^{2 - }}\, \to \,{O_2}\, + \,4{e^ - }$
Overall reaction:
$2A{l_2}{O_3}\, \to \,4Al\, + \,3{O_2}$

So, the correct option is A. Electrolytic reduction of $A{l_2}{O_3}\;$ .

Note: The process of removing the impurities and concentrating ores are called leaching. The process of electrolytic reduction at high temperature is known as smelting. The cryolite is added during the smelting of aluminium, because it lowers down the melting point and it also increases the conductivity.