Question

How much electricity in terms of Faraday is required to produce $40.0g$ of $Al$ from molten $A{l}_2{O}_3$?

• A. 1
• B. 2.50
• C. 3.21
• D. 4.44

Answer 1

The correct option is D 4.44

Electrolysis of molten $A{l}_2{O}_3$:
At cathode : $A{l}^{3+}+3e^{-}\to Al$
At anode : $2O^{2-}+4e^{-}\to O_2$
Overall equation :
$2A{l}_2{O}_3\left(l\right)\to 4Al\left(l\right)+3O_2\left(g\right)$

$1mol\text{of}{e}^{-}=1\text{Faraday of charge}$
1 mole of $A{l}^{3+}$ react with 3 mole of $e^{-}$ to give 1 mol of $Al$
So, 3 Faraday of electricity is required to produce 1 mole of $Al$.
Now,
$\text{No. of moles in 40 g of Al}=\frac{40}{27}=1.48\text{mol of Al}$
$\therefore$
$1.48mol$ of $Al$ can be produced by 1.48 $\times 3$ Faraday
Hence,
$4.44F$ of electricity is required to produce $40g$ of $Al$ from molten $A{l}_2{O}_3$