Calculate the number of coulombs required to deposit 6.75 g of $A l$ when the electrode reaction is:
$A l^{3 +} + 3 e^{-} \to A l$

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Solution

Given,
weight = 6.75 g.
n = 3.
According to Faraday's first law,
$W α Q$
$W = Z Q$
$6.75 = \frac{A t . m a s s}{n \times 96500} \times Q$
$\Rightarrow 6.75 = \frac{27}{3 \times 96500}$
$\Rightarrow \frac{6.75 \times 96500}{9} = Q$
$∴ Q = 72, 375 C$

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