Question

$A{l}_2{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0\times {10}^{4}amperes$ ofcurrent is passed through molten $A{l}_2{O}_3$ for $6hours$, what mass of aluminium is produced? (Assume $100\%$ current efficiency, at. mass of $Al=27gmo{l}^{–1}$)

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

Answer 1

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

Applying formula:
E = Z × 96500
Where, $E$ = equivalent weight of $Al=\frac{27}{3}$ for aluminium
$Z$ = electrochemical equivalent mass of one coulomb charge
⇒ $\frac{27}{3}$= $Z\times 96500$
$Z=\frac{27}{3\times 96500}$
= $\frac{9}{96500}$
Now applying the formula, $w=Z\times I\times t$
$W=\frac{9}{96500\times (4\times 104)\times (6\times 60\times 60)}$
= $8.1\times {10}^{4}g$