Question

When $96.5$ coulomb of electricity is passed throught a solution of $AgN{O}_3$, the amount of $Ag$ deposited is:

• A. $108mg$
• B. $5.4g$
• C. $16.2mg$
• D. $0.108mg$

Answer 1

The correct option is A $108mg$

Given:
$\text{Quantity of electricity passed, Q}=96.5\text{coulombs}$

$It=96.5\text{coulombs}$
$\text{Molar mass of Silver, M}=108g/mol$
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$
$n-factor=1$

According to Faradays first law of electrolysis,
$W=Z\times I\times t$
$W=\frac{E}{96500}\times I\times t$

$W=\frac{\left(\frac{\text{Molar mass}}{n-factor}\right)}{96500}\times I\times t$

$W=\frac{\left(\frac{\text{108}}{1}\right)}{96500}\times 96.5$

$W=0.108g$
$W=108mg$