Question

In electrtolysis of $A{l}_2{O}_3$ by Hall-Heroult process:

• A. Cryolite $N{a}_3\left[AI{F}_6\right]$ lowers the melting point of $A{l}_2{O}_3$ and increases its electrical conductivity.
• B. $Al$ is obtained at cathode and probably $C{O}_2$ at anode
• C. both (a) and (b) are correct
• D. none of the above is correct

Answer 1

Answer: (C)

both (a) and (b) are correct

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina and cryolite along with minor quantities of aluminum fluoride and fluorspar. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature. Thus, the option A is correct. Also at anode, alumina reacts with fluorine to give oxygen. The liberated oxygen reacts with carbon to form $CO$ and $C{O}_2$. These gases are liberated at anode.