$$A{l_2}{O_3}$$  is reduced by electrolysis at low potentials and high currents. If $$4.0 \times {10^4}A$$   of current is passed through molten $$A{l_2}{O_3}$$  for $$6$$ $$h,$$ what mass of aluminium is produced? ( Assume $$100\% $$  current efficiency, atomic mass of $$Al = 27\,g\,mo{l^{ - 1}}$$    )

Solution :
$$\eqalign{ & A{l_2}{O_3}\,\,{\text{ionises as,}} \cr & A{l_2}{O_3} \rightleftharpoons \mathop {A{l^{3 + }}}\limits_{{\text{Cathode}}} + \mathop {AlO_3^{3 - }}\limits_{{\text{Anode}}} \cr & {\text{At cathode}} \cr & A{l^{3 + }} + \mathop {3{e^ - }}\limits_{3\,F} \to \mathop {Al}\limits_{27\,g} \cr} $$
$$\because $$  Mass of aluminium deposited by $$3$$ $$F$$ of electricity $$=27$$ $$g$$
$$\therefore $$  Mass of aluminium deposited by
$$\eqalign{ & 4.0 \times {10^4} \times 6 \times 3600\,C\,\,{\text{of electricity}} \cr & {\text{ = }}\frac{{27 \times 4.0 \times {{10}^4} \times 6 \times 3600}}{{3\,F}}g \cr & = \frac{{27 \times 4.0 \times {{10}^4} \times 6 \times 3600}}{{3 \times 96500}}g \cr & = 8.1 \times {10^4}g \cr} $$