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  4. How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten Al2O3?

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Q. How much electricity in terms of Faraday is required to produce $40.0 \,g$ of $Al$ from molten $Al_2O_3$?

Electrochemistry

Solution:

$Al_2O_3 \to 2Al^{3+} + 3O^{2-}$ or, $Al^{3+} \underset{3 \,mole}{+ 3e^-} \to \underset{1\,mole(27\,g)}{Al}$
$\because 27\, g$ of aluminium needs $= 3$ moles of electrons
$= 3 \times 96500$ coulombs
$\therefore 40.0 \,g$ of aluminium will need
$ = \frac{3 \times 96500 \times 40.0}{27}$ coulombs
$= 4.28888 \times 10^5$ coulombs $(4.44 \,F)$