Question

${Al}_2{O}_3$ is reduced by electrolysis at low potential and high current. If ${4.0\times 10}^4$ amperes of current is passed through molten ${Al}_2{O}_3$ for $6$ hours, what mass of aluminium is produced ? $(Assume110$

• A. $1.3\times {10}^{4}g$
• B. $9.0\times {10}^{3}g$
• C. $8.1\times {10}^{4}g$
• D. $2.4\times {10}^{5}g$

Answer 1

The correct option is C $8.1\times {10}^{4}g$

$A{l}_2{O}_3\rightleftharpoons A{l}^{+}+Al{O}_3^{-3}$

At cathode= $A{l}^{3+}+3e^{-}\to Al$

Mass of Aluminium $3F=27$

$\therefore \frac{27\times 4\times {10}^4\times 6\times 3600}{3F}=8.1\times {10}^{4}g$ .