$\mathrm{Al}_2 \mathrm{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.5 \times…
$\mathrm{Al}_2 \mathrm{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.5 \times 10^4 \mathrm{~A}$ of current is passed through molten $\mathrm{Al}_2 \mathrm{O}_3$ for $6 \mathrm{~h}$, what mass of aluminium is produced? (Assume 100\% current efficiency, at. mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$9.0 \times 10^3 \mathrm{~g}$
$8.1 \times 10^4 \mathrm{~g}$
$2.4 \times 10^5 \mathrm{~g}$
$1.3 \times 10^4 \mathrm{~g}$
Solution
$\mathrm{Al}_2 \mathrm{O}_3$ ionises as,
$\mathrm{Al}_2 \mathrm{O}_3 \rightleftharpoons \underset{\text { cathode }}{3+}+\begin{array}{r}
\mathrm{AlO}_3^{3-} \\
\text { anode }
\end{array}$
At cathode
$\mathrm{Al}^{3+}+\underset{3 \mathrm{~F}}{3 \mathrm{e}^{-}} \longrightarrow \stackrel{\mathrm{Al}}{27 \mathrm{~g}}$
$\because$ Mass of aluminium deposited by $3 \mathrm{~F}$ of electricity $=27 \mathrm{~g}$
$\therefore$ Mass of aluminium deposited by $4.0 \times 10^4 \times 6 \times 3600 \mathrm{C}$ of electricity
$\begin{aligned}
& =\frac{27 \times 4.0 \times 10^4 \times 6 \times 3600}{3 \mathrm{~F}} \mathrm{~g} \\
& =\frac{27 \times 4.0 \times 10^4 \times 6 \times 3600}{3 \times 96500} \mathrm{~g} \\
& =8.1 \times 10^4 \mathrm{~g}
\end{aligned}$