Question
Aluminium oxide may be electrolysed at $${1000^ \circ }C$$ to furnish aluminium metal ( At. Mass = 27 amu; 1 Faraday= 96,500 Coulombs ). The cathode reaction is $$ - A{l^{3 + }} + 3{e^ - } \to A{l^ \circ }.$$ To prepare 5.12 $$kg$$ of aluminium metal by this method we require
A.
$$5.49 \times {10^1}C\,{\text{of}}\,{\text{electricty}}$$
B.
$$5.49 \times {10^4}C\,{\text{of}}\,{\text{electricty}}$$
C.
$$1.83 \times {10^7}C\,{\text{of}}\,{\text{electricty}}$$
D.
$$5.49 \times {10^7}C\,{\text{of}}\,{\text{electricty}}$$
Answer :
$$5.49 \times {10^7}C\,{\text{of}}\,{\text{electricty}}$$
Solution :
$$\eqalign{
& 1\,mole\,{\text{of}}\,{e^ - } = 1F = 96500C \cr
& 27g\,{\text{of}}\,Al\,{\text{is deposited by}}\,3 \times 96500\,C \cr
& 5120g\,{\text{of}}\,Al\,{\text{will be deposited by}} \cr
& = \frac{{3 \times 96500 \times 5120}}{{27}} \cr
& = 5.49 \times {10^7}C \cr} $$