Chemistry - Electrochemistry Question with Solution | TestHub

${\text{Al}}^{3+}+3{\text{e}}^{-}\to \text{Al }$

1 mole Al requires 3 mole e^-(3 F charge) to dischage on electrode.

Also, 1 F = 96500 C.

Thus,$\text{27 g of Al require}=3\times 96500 C$

$\text{0.9 g of Al requires}=\frac{3\times 96500 C}{27}\times \text{0.9}=\text{9650 C}$.

$=9.65\times {10}^3\text{ C}$