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Standard XII

Chemistry

Faraday's Second Law

What is the a...

Question

What is the amount of Al deposited on the electrolysis of molten ${A l}_{2} O_{3}$ when a current of 9.65 A is passed for 10.0 s.

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Solution

$W = z . i . t = \frac{E}{F} . i . t$ $= (\frac{27 \times 9.65 \times 10}{n \times 96500}) g .$
$∵ {A l}^{3 +} \to n f a c t o r = 3$
$∴ W = 9 \times 10^{- 3} g .$

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{A l}_{2} O_{3}

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A l = 27 g {m o l}^{- 1}

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{A l}_{2} O_{3}

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What is the amount of Al deposited on the electrolysis of molten Al2O3 when a current of 9.65 A is passed for 10.0 s.

W=z.i.t=EF.i.t=(27×9.65×10n×96500)g.∵Al3+→nfactor=3∴W=9×10−3g.

What is the amount of Al deposited on the electrolysis of molten ${A l}_{2} O_{3}$ when a current of 9.65 A is passed for 10.0 s.

$W = z . i . t = \frac{E}{F} . i . t$ $= (\frac{27 \times 9.65 \times 10}{n \times 96500}) g .$
$∵ {A l}^{3 +} \to n f a c t o r = 3$
$∴ W = 9 \times 10^{- 3} g .$