Question

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

• A. 1 Faraday
• B. 2 Faraday
• C. 3.2 Faraday
• D. 4.44 Faraday

Answer 1

Answer: (D)

4.44 Faraday

Cathode: $A{l}^{3+}+3e^{-}\to Al$

Anode: $2O^{2-}+4e^{-}\to O_2$

Overall equation: ${2A{l}_2{O}_3}_{\left(l\right)}\to 4A{l}_{\left(l\right)}+{3O_2}_{\left(g\right)}$

So $3$ Faraday of electricity is required to produce $1$ mole of $Al$.

Now, $40g$ of $Al=\frac{40}{27}$ mole$=1.48$ moles of $Al$

$\therefore 1.48$ moles of $Al$ can be produced by $1.48\times 3$ Faraday i.e $4.44$ Faraday of electricity.