Question

In the electrolysis of alumina using cryolite, the reaction that takes place at cathode is:

• A. $4H_{2}O+4e^{-}⟶2H_2+4O{H}^{-}$
• B. $4A{l}^{3+}+12e^{-}⟶4Al$
• C. $6F_2+2A{l}_2{O}_3⟶4Al{F}_3+3O_2$
• D. $12F^{-}\to 6F_2+12e^{-}$

Answer 1

The correct option is B $4A{l}^{3+}+12e^{-}⟶4Al$

$4A{l}^{3+}+12e-\to 4Al$ (aluminium metal at the (-)cathode) reduction.
Aluminium oxide $\left(A{l}_2{O}_3\right)$ is an ionic compound. When it is melted the $A{l}^{3+}$ and $O^{2-}$ ions are free to move and conduct electricity.
Electrolysis of the alumina/cryolite solution gives aluminium at the cathode and oxygen at the anode.
$4A{l}^{3+}+12e-\to 4Al$ (aluminium metal at the (-)cathode) reduction.
$6O^{2-}-12e-\to 3O_2$ (oxygen gas at the (+)anode) oxidation.
Aluminium is more dense than the alumina/cryolite solution and so it falls to the bottom of the cell where it can be tapped off as pure liquid metal.
$2A{l}_2{O}_3\left(l\right)\to 4Al\left(l\right)+3O_2\left(g\right)$
Oxygen is given off at the positive carbon anode. Carbon dioxide is also given off at the carbon anode because hot oxygen reacts with the carbon anode to form carbon dioxide gas.
$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$
The carbon anodes slowly disappear because each molecule of carbon dioxide which is given off takes a little piece of carbon away with it. The carbon anodes need to be replaced when they become too small.