Question

Aluminium is obtained by:
A) Reducing $\mathrm{Al}_{2} \mathrm{O}_{3}$ with coke
B) Electrolysing $\mathrm{Al}_{2} \mathrm{O}_{3}$ dissolved in $\mathrm{Na}_{3} \mathrm{AlF}_{6}$
C) Reducing $\mathrm{Al}_{2} \mathrm{O}_{3}$ with Chromium
D) Heating Alumina with cryolite

Answer 1

In chemistry, electrolysis is a technique that uses direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction. Electrolysis is commercially important as a stage in the separation of elements from naturally occurring sources such as ores using an electrolytic cell.

Complete step by step answer:
Coke is a porous fuel with very high carbon content.Aluminium oxide ( $\mathrm{Al}_{2} \mathrm{O}_{3}$ )cannot be reduced by coke because it is a highly stable oxide and aluminium has a strong affinity for oxygen.
Certain oxides of Mn, Cr do not reduce by carbon due to very high melting point. They are reduced by Al. Thus, Al cannot be formed from reduction with chromium.
The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity. The fusion temperature of pure alumina is about $2000^{\circ} \mathrm{C}$
and at this temperature when electrolysis is carried out on the fused mass, the metal formed vaporizes, as the boiling point of aluminium is $1800^{\circ} \mathrm{C}$
These difficulties are overcome by using a mixture containing alumina, cryolite and fluorspar. But the addition of cryolite does not minimize the anode effect and does not remove the impurities from alumina.

Now, coming to the question given, The reactions involved are:
$4 \mathrm{Na}_{3} \mathrm{AlF}_{6} \rightleftarrows 12 \mathrm{Na}^{+}+4 \mathrm{Al}^{3+}+12 \mathrm{F}^{-}$
At cathode: $4 \mathrm{Al}^{3+}+12 \mathrm{e}^{-} \rightarrow 4 \mathrm{Al}$
At anode: $12 \mathrm{F}^{-} \rightarrow 6 \mathrm{F}_{2}+12 \mathrm{e}^{-}$

Thus, the net reaction is: $2 \mathrm{Al}_{2} \mathrm{O}_{3}+6 \mathrm{F}_{2} \rightarrow \mathrm{AlF}_{3}+\mathrm{O}_{2}$
So, the correct answer is “Option B”.

Note: The nature of electrolyte and electrodes is also an important point in electrolysis. Thus, it is important to note down the standard potentials needed in the cell for calculations.