Question

Aluminium oxide may be electrolyte at 1000${}^0$C to furnish aluminium metal (Atomic mass = 27 amu, 1 Faraday = 96,500 Coulombs). The cathode reaction is $A{l}^3+3e^{-}\to A{l}^o$. 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

The correct option is D $5.49\times {10}^7$ C of electricity

Mass of Al deposited $=5.12kg=5.12\times {10}^{3}g.$

Mass of Al deposited $=\frac{Q\left(C\right)}{96500\frac{C}{mol{e}^{-}}}\times moleratio\times molarmass$.

Substitute the mass of Al deposited in the above equation.

Hence, $5.12\times {10}^{3}g=\frac{Q\left(C\right)}{96500\frac{C}{mol{e}^{-}}}\times \frac{1molAl}{3mol{e}^{-}}\times 27\frac{g}{mol}$

$Q\left(C\right)=5.49\times {10}^{7}C.$

Option D is correct.