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 $Al^{3+} + 3e^- \to Al^{\circ}$ 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)

$5.49 \times 10^7$ C of electricity

Answer 2

$Al^{3+} + 3e^- \to Al$ $w = zQ$ where w = amount of metal $= 5.12\, kg = 5.12 \times 10^3\,g$ z = electrochemical equivalent $=\frac{Equivalent \,weight}{96500} = \frac{Atomic \,mass}{Electrons \times 96500}$ $= \frac{27}{3\times96500}\times Q$ $5.12\times10^{3} = \frac{27}{3\times 96500}\times Q$ $Q = \frac{5.12\times10^{3}\times3\times96500}{27}C$ $= 5.49 \times 10^{7} \,C$