Aluminium oxide may be electrolysed at $1000^{\circ} \mathrm{C}$ to furnish aluminium metal (Atomic mass $=27 \mathrm{amu} ; 1$ Faraday $=96,500$ Coulombs). The cathode reaction is $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}^{\circ}$ To prepare 5.12 kg of aluminium metal by this method would require

(A) $5.49 \times 10^{7} \mathrm{C}$ of electricity

(B) $1.83 \times 10^{7} \mathrm{C}$ of electricity

(C) $5.49 \times 10^{4} \mathrm{C}$ of electricity

(D) $5.49 \times 10^{1} \mathrm{C}$ of electricity

JEE Main · 2005 · 08 May 2005 · Offline · Redox Reactions and Electrochemistry

Aluminium oxide may be electrolysed at

100 0^{\circ} C

to furnish aluminium metal (Atomic mass

= 27 amu; 1

Faraday

= 96, 500

Coulombs). The cathode reaction is

Al^{3 +} + 3 e^{-} ⟶ Al^{\circ}

To prepare 5.12 kg of aluminium metal by this method would require

Answer: (A)

This is a straight application of Faraday's laws, where the charge required is

Q = \frac{m n F}{M}

. Depositing aluminium takes three electrons per atom through

Al X^{3 +} + 3 e X^{-} Al

, so

n = 3

, while

m = 5.12 \times 1 0^{3}

M = 27

g mol

^{- 1}

and

F = 96500

C. Substituting,

Q = \frac{5.12 \times 1 0 ^{3} \times 3 \times 96500}{27} \approx 5.49 \times 1 0^{7}

C of electricity.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option