Question

The number of coulombs required to deposit $5.4\text{g}$ of Aluminium when the given electrode reaction is
$A{l}^{3+}+3e^{-}\to Al$, is:

• A. 57900
• B. 57900.00
• C. 57900.0

Answer 1

Answer: (A), (B), (C)

According to Faraday's first law of electrolysis
$W=\frac{Z\times Q}{F}$
$W=\text{Mass deposited}=5.4g$
$Z=\text{Equivalent mass}=\frac{\text{Molar mass}}{\text{n-factor}}$
$F=\text{Faraday constant}=96500C$
$Q=\text{Charge}$
For the given reaction,
$A{l}^{3+}+3e^{-}\to Al$
n-factor is 3
$\therefore$
$5.4=\left(\frac{27}{3}\right)\times \frac{Q}{96500}$

$Q=57900C$

Thus, $57900C$ is required to deposit $5.4g$ of $Al$