Aluminium oxide may be electrolysed at $1000^{\circ} C$ to furnish aluminium metal (Atomic mass $= 27 a m u; 1 f a r a d a y = 96500 c o u l o m b$ ). The cathode reaction is:
$A l^{3 +} + 3 e^{-} \to A l$
To prepare $5.12 k g$ of aluminium metal by this method we would require.

5.49 \times 10^{7} C

of electricity

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1.83 \times 10^{7} C

of electricity

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5.49 \times 10^{4} C

of electricity

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5.49 \times 10^{10} C

of electricity

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Solution

The correct option is A $5.49 \times 10^{7} C$ of electricity
Half Cell Reaction:-
$A l^{3 +} + 3 e \to A l_{(s)}$ .
(i) Mole Ratio $= \frac{m o l e s o f A l f o r m e d}{m o l e s o f e l e c t r o n s} = \frac{1}{3}$
(ii) Moles of $A l$ formed $= \frac{C h a r g e (Q)}{96500 C m o l^{- 1} e^{-}} \times \frac{1}{3}$ [Moles of A $= \frac{W e i g h t}{m o l a r m a s s} = \frac{W}{M M}$ ]
$∴ \frac{5.12 \times 10^{3} g}{27 g m o l^{- 1}} = \frac{Q (C)}{96500 C m o l^{- 1} e^{-}} \times \frac{1}{3}$
$∴ Q (C) = \frac{5.12 \times 10^{3} \times 96500 C m o l^{- 1} e^{-} \times 3}{27 g m o l^{- 1}}$
$∴ Q = 54897 \times 10^{3} C$
$= 5.4897 \times 10^{7} C$

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Aluminium oxide may be electrolysed at 1000∘C to furnish aluminium metal (Atomic mass =27 amu;1 faraday=96500 coulomb). The cathode reaction is:Al3++3e−→AlTo prepare 5.12 kg of aluminium metal by this method we would require.

Aluminium oxide may be electrolysed at $1000^{\circ} C$ to furnish aluminium metal (Atomic mass $= 27 a m u; 1 f a r a d a y = 96500 c o u l o m b$ ). The cathode reaction is:
$A l^{3 +} + 3 e^{-} \to A l$
To prepare $5.12 k g$ of aluminium metal by this method we would require.