Question

${Al}_2{O}_3$ is reduced by electrolysis at low potentials and high currents.if $4.0\times {10}^4$ amperes current is passed through molten ${Al}_2{O}_3$ for 6 hours.What mass of aluminium is produced?

(Assume 100% current efficiency, atomic mass of $Al=27g{mol}^{-1}$)

• A. $1.3\times {10}^{4}g$
• B. $9.0\times {10}^{3}g$
• C. $8.05\times {10}^{5}g$
• D. $2.4\times {10}^5$

Answer 1

Answer: (A)

$1.3\times {10}^{4}g$

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

$Faraday=\frac{change}{96500}=\frac{4\times {10}^4\times 6\times 60\times 60}{96500}$

$=1492.227mol$

3 mol $\to$ 27g of $Al$

$1492.227\to \frac{1}{3}\times 27\times 1492.227$

$=1.3\times {10}^{4}g$ of $Al$