Question

Electrolytic reduction of alumina to aluminium by the Hall-Heroult process is carried out _________

• A. In the presence of NaCl
• B. In the presence of fluorite
• C. In the presence of cryolite which forms a melt with a lower melting temperature
• D. In the presence of cryolite which forms a melt with a higher melting temperature

Answer 1

The correct option is C

In the presence of cryolite which forms a melt with a lower melting temperature

Explanation:

Hall-Heroult: the process is widely used in the extraction of aluminium. In the Hall-Heroults process, pure ${\mathrm{Al}}_2{\mathrm{O}}_3$ mixed with ${\mathrm{CaF}}_2\mathrm{and}{\mathrm{Na}}_3{\mathrm{AlF}}_6$. This results in a lowering of the melting point of the mixture and increases its ability to conduct electricity. A steel vessel with a lining of carbon and graphite rods is used.

The mechanism in the Hall-Heroult process

• The carbon lining acts as a cathode and graphite acts as an anode. When electricity is passed through the electrolytic cell which consists of carbon electrodes oxygen is formed at the anode.
• This oxygen formed reacts with the carbon of the anode to form carbon monoxide and carbon dioxide. In this method of production of aluminium for every 1 kg of Al produced.

The explanation for the correct answer-

• Aluminium ions are created at the adverse cathode from the aluminium oxide and then sink down because they are heavier than the cryolite solution. Then, the liquid shape of the aluminium sunk to the bottom. On the other side, at the positive anode, the oxygen from the aluminium oxide forms and responds to carbon dioxide CO2 with the graphite carbon.

The overall reaction is:

$2{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{aq}\right)+3\mathrm{C}\left(\mathrm{s}\right)\to 4\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{CO}}_2\left(\mathrm{g}\right)$

The electrolytic reactions are:

At the cathode:

${\mathrm{Al}}^{3+}+3\mathrm{e}\to \mathrm{Al}\left(\mathrm{l}\right)$

At the anode:

$$\begin{gathered} \mathrm{C}\left(\mathrm{s}\right)+{\mathrm{O}}^{2-}\to \mathrm{CO}\left(\mathrm{g}\right)+2{\mathrm{e}}^{-} \\ \mathrm{C}\left(\mathrm{s}\right)+2{\mathrm{O}}^{2-}\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{e}}^{-} \end{gathered}$$

During the process of electrolysis,

1. Aluminium ions that are positively loaded gain electrons from the cathode and form molten aluminium.
2. Oxide ions lose anode electrons and form molecules of oxygen.

The explanation for the incorrect option:

• In Hall Heroult's process aluminium is produced by alumina by electrolysis.
• As alumina has a high melting point around$2000°C$ so in order to decrease the melting point we use cryolite and fluorspar which are mixed with alumina it forms melt

Therefore electrolytic reduction of alumina to aluminium by the Hall-Heroult process is carried out in the presence of cryolite which forms a melt with a lower melting temperature. Hence, option (c) is correct.