Question

During the high temperature non-aqueous electrolytic extraction of Aluminum from purified $A{l}_2{O}_3$, the oxygen evolved at the anode reacts with the electrode (graphite). As a result the electrodes are periodically replaced. – True or False?

• A. True
• B. False

Answer 1

The correct option is A

True

In the electrolytic extraction of Aluminium, purified alumina is mixed with $N{a}_{3}Al{F}_6$ and $Ca{F}_2$. Cryolite $\left(N{a}_{3}Al{F}_6\right)$ helps dissolve Alumina and lowers the melting point of this mix to about 1050 ${}^{0}C$. Pure Alumina melts at about 2100 ${}^{0}C$. $Ca{F}_2$ is added to enhance the conductivity.

As shown in the above image, the surface on the inside of the steel case is coated with a graphite lining, which acts as the cathode. The electrolyte is a molten mixture of cryolite, molten Alumina and $Ca{F}_2$. Fluorspar $Ca{F}_2$ is added to this to enhance the conductivity of the electrolyte.

Although not completely accurate, the half-cell reactions “maybe” represented as

$A{l}^{3+}+3e^{-}\to Al$ (Cathode : Reduction)
$2O^{2-}\to O_2+4e^{e-}$ (Anode : Oxidation)

In reality, there seems to be considerable debate regarding what actually goes on inside the cell. The above reactions are at best only simplistic approximations. At our current level, these equations may be taken to be true.

Do note that the whole electrolysis process is carried out at close to 950 ${}^{0}C$ - 1050 ${}^{0}C$. The oxygen evolved at the cathode does react with the graphite electrode producing CO and $C{O}_2$. Also, the solid-liquid-gas interface where molten electrolyte is in touch with the electrode and air is very vulnerable to rapid corrosion. To mitigate the rate of corrosion, the top surface of the fused electrolyte is protected by a layer of powdered coke.