Question

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

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

Answer 1

Answer: (B)

$8.1 \times 10^4 g$

Answer 2

$W =\frac{ E }{96500} \times I \times t$
$W =\frac{9}{96500} \times 4.0 \times 10^{4} \times 6 \times 3600$
$=8.1 \times 10^{4}\, g$