Find the number of faradays of electricity required to produce $45 \mathrm{~g}$ of $\mathrm{Al}$ from molten…
Find the number of faradays of electricity required to produce $45 \mathrm{~g}$ of $\mathrm{Al}$ from molten $\mathrm{Al}_2 \mathrm{O}_3$.
$1 \mathrm{~F}$
$3 \mathrm{~F}$
$5\mathrm{~F}$
$7 \mathrm{~F}$
Solution
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
The equation shows that 3 moles of electrons are required to produce 1 mole of $\mathrm{Al}$ (i.e., $27 \mathrm{~g}$ of $\mathrm{Al}$ ).
$\therefore \quad 3 \mathrm{~F}$ of electricity is required to produce $27 \mathrm{~g}$ of $\mathrm{Al}$ from molten $\mathrm{Al}_2 \mathrm{O}_3$.
$\therefore \quad$ Faradays of electricity required to produce $45 \mathrm{~g}$ of $\mathrm{Al}=\frac{3}{27} \times 45=5 \mathrm{~F}$