Question

In the Hall's process for extraction of aluminium,
(a) give the formula and purpose of fluorspar and cryolite.
(b) state the location of cathode and anode and explain what occurs at each electrode.

Answer 1

(a) The formula for cryolite and fluorspar is Na₃AlF₆ and CaF_2, respectively. The purpose of adding cryolite and fluorspar to fused pure alumina is to raise the electrical conductance of alumina and reduces its melting point from 2050 ^oC to 950 ^oC.

(b) During the extraction of aluminium from pure ore by electrolysis; the electrolytic cell is made of steel with an inner layer of carbon. The inner gas carbon layer is taken as the cathode. The thick carbon rods dipped in fused electrolyte act as the anode. Fused pure alumina (${\mathrm{Al}}_2{\mathrm{O}}_3$) mixed with cryolite (${\mathrm{Na}}_3{\mathrm{AlF}}_6$) and fluorspar (${\mathrm{CaF}}_2$) is taken as the electrolyte. On passing electricity through the electrodes, the following reactions occur:

Dissociation of alumina: ${\mathrm{Al}}_2{\mathrm{O}}_3\leftrightarrow 2{\mathrm{Al}}^{3+}+3{\mathrm{O}}_{}^{2-}$

Reaction at the cathode: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Al}$

Aluminium is formed in a molten state at the cathode and settles at the bottom of the cell. This is collected from time to time. The aluminium so obtained is 99% pure.

Reaction at anode: $$\begin{gathered} {\mathrm{O}}_{}^{2-}\stackrel{}{\to }\mathrm{O}+2{\mathrm{e}}^{-} \\ \mathrm{O}+\mathrm{O}\stackrel{}{\to }{\mathrm{O}}_2 \end{gathered}$$
The oxygen produced in the process reacts with the carbon anode and forms CO and CO₂. Due to oxidation at the carbon anode, it needs to be replaced regularly.