Question

In the electrolysis of alumina using Hall Heroult’s process the electrolyte is covered with powdered coke. Explain why.

Answer 1

Hint: Aluminium extraction includes mainly two phases: Bayer' process and the Hall-Heroult process. Bayer's process is the refining of bauxite ore to obtain aluminum oxide and Hall-Heroult process is the industrial method for smelting of aluminum, which gives pure aluminum from aluminum oxide.

Complete step by step answer:
The Hall-Heroult process is the electrolytic process of alumina to obtain molten aluminum. In the Hall-Heroult process, powdered coke is used to reduce heat loss by radiation and prevent the burning of anode.
 In Hall-Heroult process, pure alumina (\[A{{l}_{2}}{{O}_{3}}\]) is mixed with cryolite (\[N{{a}_{3}}Al{{F}_{6}}\]). This results in a lowering of the melting point of the mixture, so its ability to conduct electricity increases. A steel vessel with a lining of carbon and graphite is used in the process.
The carbon lining acts as cathode and carbon lining acts as anode. After passing electricity through the electrolyte consists of a carbon electrode, oxygen is formed at anode. This leads to the formation of carbon monoxide and carbon dioxide by reacting the formed oxygen with the carbon anode. In this process, for every 2 units of aluminum is formed 1 unit of carbon anode is burned. To prevent this burning of anode we use powdered coke as a covering for the electrolyte.

Note:
The Bayer process includes the refining of bauxite ore to obtain aluminum oxide(alumina). Here sodium hydroxide is used as a leaching agent. The ore of aluminum usually contains impurities like silica, iron oxide and titanium oxide. Hence the concentration of bauxite ore is modified by mixing the powdered ore with a concentrated solution of sodium hydroxide at \[473-523K\] and \[35-36bar\] pressure. By this procedure alumina is leached out as sodium aluminate leaving behind the impurities. The insoluble impurities are called red mud.
\[\Rightarrow A{{l}_{2}}{{O}_{3}}.2{{H}_{2}}O+2NaOH\xrightarrow{473-523K}2NaAl{{O}_{2}}+3{{H}_{2}}O\]
\[\Rightarrow NaAl{{O}_{2}}+2{{H}_{2}}0\to NaOH+Al{{(OH)}_{3}}\]
\[\Rightarrow 2Al{{(OH)}_{3}}\to A{{l}_{2}}{{O}_{3}}+3{{H}_{2}}O\]