Question

The electrolytic reduction of alumina to aluminium through the Hall-Heroult process is carried out in the presence of:
a.) NaCl
b.) Fluoride
c.) Cryolite which forms a melt with lower melting temperature
d.) Cryolite which forms a melt with higher melting temperature

Answer 1

Hint: Recall that the chemical formula of Alumina is $A{{l}_{2}}{{O}_{3}}$ and that of Cryolite is $N{{a}_{3}}Al{{F}_{6}}$. Also remember that the Hall-Heroult process is an industrial procedure for the large-scale smelting of Aluminium and that Alumina and Cryolite are both ores of Aluminium.

Complete step-by-step answer:
Let us start by first discussing the Hall-Heroult process for the smelting of Aluminium in detail.
The Hall–Héroult process involves dissolving aluminium oxide (alumina) in molten cryolite, and electrolysing the molten salt bath, typically in a purpose-built cell.
The Hall–Héroult process applied at industrial scale happens at 940–980°C and produces 99.5–99.8% pure aluminium.
Recycled aluminium requires no electrolysis; thus, it does not end up in this process.
In Hall-Heroult’s process, pure $A{{l}_{2}}{{O}_{3}}$ is mixed with $N{{a}_{3}}Al{{F}_{6}}$.This results in lowering of the melting point of the mixture and increases its ability to conduct electricity.
The overall equation of the reaction is:
$\mathbf{2A}{{\mathbf{l}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}~+\text{ }\mathbf{3C}\text{ }\to ~\mathbf{4Al}\text{ }+\text{ }\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}$
With the Cathode equation being:
$\mathbf{Al}{{~}^{\mathbf{3}+}}~+\text{ }\mathbf{3}{{\mathbf{e}}^{}}~\to ~\mathbf{Al}\text{ }$
And that at the anode being:

\mathbf{C}\text{ }+\text{ }{{\mathbf{O}}^{\mathbf{2}-~}}\to ~\mathbf{CO}\text{ }+\text{ }\mathbf{2}{{\mathbf{e}}^{}} \\\ \mathbf{C}\text{ }+\text{ }\mathbf{2}{{\mathbf{O}}^{\mathbf{2}-~}}\to ~\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}~\text{ }+\text{ }\mathbf{4}{{\mathbf{e}}^{}} \\\ \end{array}$$ Therefore, through this detailed analysis of the reaction that takes place and our consistent observations throughout the process, we can conclude that the melting point of the overall melt decreases. Therefore, the answer to this question is c). NOTE: Be mindful of the chemical compositions of alumina and cryolite and the particulars of the Hall-Heroult process, as any slip up in the aforementioned concepts will result in incorrect answers.