Question

When same quantity of electricity is passed through aqueous $AgNO_3$ and $H_2SO_4$ solutions connected in series, $ 5.04 \times 10^{-2} \, g $ of $H_2$ is liberated. What is the mass of silver (in grams) deposited? (Eq. wts. of hydrogen = $ 1.008,$ silver = $108$)

• A. 54
• B. 0.54
• C. 5.4
• D. 10.8

Answer 1

Answer: (C)

Given, weight of hydrogen liberated
$=5.04 \times 10^{-2} g$
Eq. wt. of hydrogen $=1.008$
Eq. wt. of silver $=108$
Weight of silver deposited, $w =?$
According to Faraday's second law of electrolysis
$\frac{\text { weight of silver deposited }}{\text { weight of hydrogen liberated }}$
$=\frac{\text { eq. wt .of silver }}{\text { eq. wt. of hydrogeon }}$
$\frac{w}{5.04 \times 10^{-2}}=\frac{108}{1.008}$
$W=\frac{108 \times 5.04 \times 10^{-2}}{1.008}$
$=5.4\, g$