Question

In electrolysis of \[A{l_2}{O_3}\] by Hall - Heroult process, .
A.Cryolite \[N{a_3}[Al{F_6}]\] lowers the M.P. of \[A{l_2}{O_3}\] and increases its electrical conductivity
B.Al is obtained at the cathode and probably \[C{O_2}\] at anode.
C.Both (a) and (b) are correct
D.None of the above

Answer 1

Hint: Hall-Heroult process is a major industrial process that is used for the smelting of Al. In the Hall-Heroult process, alumina is dissolved in molten cryolite \[N{a_3}[Al{F_6}]\] by using Bauxite that is the chief ore of Al. This molten salt bath is then electrolysed in a purpose-built cell.

Complete step by step solution: -
In Hall-Heroult process, pure aluminium oxide or alumina \[(A{l_2}{O_3})\] is combined with \[Na{F_2}\] or \[N{a_3}[Al{F_6}]\] which results in the lowering of melting point of the mixture and the ability to conduct electricity increases. A steel vessel that is lined with carbon and graphite rods is used in this process.
The carbon lining is used as the cathode whereas the graphite rods are used as anodes. Upon passage of electricity through the electrolytic cell, oxygen is produced at the anode. This oxygen reacts with the carbon of the anode, to for carbon monoxide and carbon dioxide.
Aluminium ions are produced at the cathode from \[(A{l_2}{O_3})\] and sink to the bottom as these ions are heavier than the cryolite solution. The oxygen of \[(A{l_2}{O_3})\] reacts with carbon dioxide with the graphite rods.
Reaction at cathode is:
 \[A{l^{3 + }} + 3{e^ - } \to Al(l)\]
Reaction at the anode:
 \[
  C(s) + {O^{2 - }} \to CO(g) + 2{e^ - } \\
  C(s) + 2{O^{2 - }} \to C{O_2}(g) + 4{e^ - } \\
 \]
The overall reaction is:
 \[2A{l_2}{O_3} + 3C \to 4Al + 3C{O_2}\]
Hence, the correct answer is C.

Note: Smelting is the process of obtaining a metal by heating its ore above the melting point in the presence of oxidizing agents such as air or reducing agents such as coke. Most ores are oxides and this process removes the oxygen from the oxide, leaving behind the metal. Even then, if the metal is still not pure, so then another process is needed.