Question

The number of Faradays(F) required to produce $4.5\text{g}$ of aluminium from molten $AlC{l}_3$ is:
(Atomic mass of $Al=27u$)

• A. $2$
• B. $3$
• C. $1$
• D. $0.5$

Answer 1

The correct option is D $0.5$

$A{l}^{3+}\stackrel{+3e^{-}}{\to }\underset{27g}{Al}$
$\because \text{27 g Al is deposited by 3 F}$
$\therefore \text{4.5 g Al is deposited by 0.5 F}$