Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{Al}}^{3+}+\underset{3\mathrm{F}}{3{\mathrm{e}}^{-}}⟶\underset{1\mathrm{mole}}{\mathrm{Al}}$ 1 mole of Al , i.e. 27 g of Al are deposited by charge $=3\mathrm{F}$ $\therefore 4.5\mathrm{g}$ of Al will be deposited by charge $=\frac{3}{27}\times 4.5\mathrm{F}=0.5\mathrm{F}$ $2{\mathrm{H}}^{+}+\underset{2\mathrm{F}}{2{\mathrm{e}}^{-}}⟶\underset{1\mathrm{mole}}{{\mathrm{H}}_2}$ 2 F of charge liberate ${\mathrm{H}}_2=1\mathrm{mol}=22.4$ litre at STP $\therefore 0.5\mathrm{F}$ of charge will liberate ${\mathrm{H}}_2=\frac{22.4}{2}\times 0.5$ $=5.6\mathrm{L}$ at STP