Question

On passing $1.5F$ charge, the number of moles of aluminum deposited at the cathode is:
[Molar mass of $Al=27$ gram mol${}^{-1}$]

• A. $1.0$
• B. $13.5$
• C. $0.50$
• D. $0.75$

Answer 1

The correct option is C $0.50$

The balanced chemical equation for the deposition of aluminum is shown below.

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

One mole of $A{l}^{3+}$ ions requires 3 moles of electrons.

On passing $3F$ charge ($3$ moles of electrons), the number of moles of aluminium deposited at the cathode is $1$.

On passing $1.5F$ charge, the number of moles of aluminium deposited at cathode are $1\times \frac{1.5}{3}=0.5$

Hence, option (C) is the correct answer.