Question

In the Hall-Heroult process, aluminum is formed at the cathode. The cathode is made out of:
a) Platinum
b) Carbon
c) Pure aluminium
d) Copper

Answer 1

Hint: Pure alumina, $A{l}_2{O}_3$ , is extracted from the mineral bauxite using hot concentrated sodium hydroxide solution in the Bayer Process. Hall – Héroult process is used to obtain Aluminium from the ore aluminium oxide.

Complete step by step answer:
A Hall – Héroult Cell is a carbon lined reaction vessel which acts as the cathode and a row of graphite electrodes inserted into the bath act as anodes. Now let us discuss the steps involved in the Hall – Heroult process of extracting aluminium from aluminium oxide.

Steps involved in Hall-Héroult process are-
- The aluminium oxide obtained in the Bayer’s process has a very high melting point of around 2045 degree Celsius. To lower the melting point, it is dissolved in a synthetic cryolite, $N{a}_{3}Al{F}_6$. It reduces the melting point to around 970 degree Celsius and hence makes electrolysis easier.
- A nearly pure alumina is dissolved at 970 degree in a molten electrolyte composed of aluminium and cryolite. This undergoes electrolysis to give aluminium metal at the cathode and oxygen gas at the anode and this is the Hall- Héroult process.
We can write the half-cell reactions at the cathode and anode as-
At cathode: $A{l}^{3+}(melt)+3e^{-}\to 3Al$
At anode: $C(s)+2O^{2-}\to C{O}_2(g)+4e^{-}$

Thus we can write the overall reaction as $$2A{l}_2{O}_3+3C\to 4Al+3C{O}_2$$
We can understand from the above discussion that the cathode is made up of carbon lining.
So, the correct answer is “Option B”.

Note: Natural cryolite is very rare to be used for electrolysis, so a synthetic version is created from fluorite, a far more common material.
- Impurities which are stronger oxidants than such as water, silica, and iron oxides must be removed before the electrolysis of alumina because these would be reduced.