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How long would it take to deposit 50g of Al from an electrolytic cell containing $A l_{2} O_{3}$ using a current of 105 ampere?

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1.32 h

1.54 h

2.15 h

1.42 h

answer is B.

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Detailed Solution

$A l^{3 +} + 3 e^{-} \to A l : E q . w t . o f A l = \frac{27}{3} = 9$

$W = Z \times I \times t \Rightarrow Z = \frac{E q . w t .}{96500}$

$t = \frac{W \times 96500}{E q . w t . \times I} = \frac{50 \times 96500}{9 \times 105} = 5105.82 s$ (or) 1.42 h

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