$\mathrm{Al}_{2} \mathrm{O}_{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0…

$\mathrm{Al}_{2} \mathrm{O}_{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^{4}$ amperes of current is passed through molten $\mathrm{Al}_{2} \mathrm{O}_{3}$ for 6 hours, what mass of aluminium is produced? (Assume $100 \%$ current efficiency. At. mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ )
  1. $8.1 \times 10^{4} \mathrm{~g}$
  2. $2.4 \times 10^{5} \mathrm{~g}$
  3. $1.3 \times 10^{4} \mathrm{~g}$
  4. $9.0 \times 10^{3} \mathrm{~g}$

Solution

$\quad \because \mathrm{Q}=i \times t$
$\therefore \mathrm{Q}=4.0 \times 10^{4} \times 6 \times 60 \times 60 \mathrm{C}=8.64 \times 10^{8} \mathrm{C}$
Now since 96500 C liberates $9 \mathrm{~g}$ of $\mathrm{Al}$ $8.64 \times 10^{8} \mathrm{Cliberates}$
$\frac{9}{96500} \times 8.64 \times 10^{8} \mathrm{~g} \mathrm{Al}=8.1 \times 10^{4} \mathrm{~g}$ of $\mathrm{A}$ ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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