Question

The amount of electricity required to deposit $0.9 \,g$ of aluminium when the electrode reaction is $Al ^{3+}+3 e^{-} \longrightarrow Al ;$ (atomic mass of $\left. Al =27\right)$

• A. $9.65 \times 10^{3} \,C$
• B. $1.93 \times 10^{4}\, C$
• C. $9.65 \times 10^{4}\, C$
• D. $4.34 \times 10^{5}\, C$

Answer 1

Answer: (A)

$ A l^{3+}+3 e^{-} \longrightarrow A l$

$27 \,g$ of $Al$ require $=3 \times 96500\, C$

$0.9 \,g$ of $Al$ requires $=\frac{3 \times 96500}{27} \times 0.9=9650 \,C =9.65 \times 10^{3}\, C$