Calculate the number of coulombs required to deposit $5.4 g$ of $A l$ when the electrode reaction is:
${A l}^{3 +} + 3 e^{-} \to A l$ [Atomic weight of $A l = 27$ $g / m o l$ ]

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Solution

${A l}^{3 +} + 3 e^{-} \to A l$
For $1$ mol $(27 g)$ of $A l$ requires $= 3 \times 96500 C$
$1 g$ of $A l$ requires $= \frac{3 \times 96500}{27}$
$5.4 g$ of $A l$ requires $= \frac{3 \times 96500}{27} \times 5.4 = 57900 C$

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