Q. Aluminium oxide may be electrolysed at 1000℃ to furnish aluminium metal (atomic mass = 27 u; 1 F = 96500 C). The cathode reaction is ${Al}^{3 +} + {3e}^{-} \longrightarrow {Al}^{0}$ To prepare 5.12 kg of aluminium metal by this method would require
A) $5.49\ \times \ 10^{1}\$C of electricity
B) $5.49\ \times \ 10^{4}\$C of electricity
C) $1.83\ \times \ 10^{7}\$C of electricity
D) $5.49\ \times \ 10^{7}\$C of electricity
Explanation
(d)
${Al}^{3 +} + \ {3e}^{-}\ \rightarrow Al$
$w\ = ZQ$
Where, $w$ = amount of metal
$w\ = \ 5.12\ kg$
$= \ 5.12\ \times \ 10^{3}\ g\$
Z = electrochemical equivalent
$Z\ = \ \frac{equivalent\ weight}{96500} = \ \frac{atomic\ mass}{electrons\ \times 96500}$
$Z\ = \ \frac{27}{3\ \times 96500}$
$5.12\ \times \ 10^{3}\ = \ \frac{27}{3\ \times 96500}\ \ \times \ Q$
$Q\ = \ \frac{5.12\ \times \ 10^{3}\ \times 3\ \times 96500}{27}\ C$
$= \ 5.49\ \times \ 10^{7}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.