Question

Aluminium oxide may be electrolysed at 1000 ${}^{\circ }$C to furnish aluminium metal (Atomic mass = 27 amu: 1 faraday = 96,500 Coulombs). The cathode reaction is
$A{l}^{3+}+3e^{-}\to A{l}^{\circ }$
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

Answer 1

Answer: (D)

$A{l}^{3+}+3e^{-}\to Al$
$w=zQ$
where w = amount of metal
$=5.12kg=5.12\times 10^{3}g$
z = electrochemical equivalent
$=\frac{Equivalentweight}{96500}=\frac{Atomicmass}{Electrons\times 96500}$
$=\frac{27}{3\times 96500}\times Q$
$5.12\times 10^3=\frac{27}{3\times 96500}\times Q$
$Q=\frac{5.12\times 10^3\times 3\times 96500}{27}C$
$=5.49\times 10^{7}C$