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)

$ W = z \times I \times t $
$ z =\frac{\text { molar mass }}{ nf \times 96500} $
$ z =\frac{27}{3 \times 96500} $
$ W =\frac{27}{3 \times 96500} \times 4 \times 10^{4} \times 6 \times 60 \times 60 $
$=8.1 \times 10^{4}\, g $