When same quantity of electricity is passed through aqueous $A{gNO}_{3}$ and $H_{2}{SO}_{4}$ solutions connected in series, ${5.04\ imes 10}^{- 2}\ g\ of\ H_{2}$ is liberated. What is the mass of silver (in grams) deposited? (Eq. wts.of hydrogen = 1.008, silver = 108)

Question

When same quantity of electricity is passed through aqueous $A{gNO}_{3}$ and $H_{2}{SO}_{4}$ solutions connected in series, ${5.04\ 	imes 10}^{- 2}\ g\ of\ H_{2}$ is liberated. What is the mass of silver (in grams) deposited? (Eq. wts.of hydrogen = 1.008, silver = 108)

Dr. Abubacker · Accepted Answer

(c)
Given, weight of hydrogen liberated
$= \ 5.04\  	imes \ 10^{- 2}g$
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{weight\ of\ silver\ deposited}{weight\ of\ hydrogen\ liberated} = \ \frac{eq.wt.of\ silver}{eq.wt.of\ hydrogen}\$
$\frac{w}{5.04\  	imes \ 10^{- 2}} = \ \frac{108}{1.008}$
$w\  = \ \frac{108\  	imes 5.04\  	imes \ 10^{- 2}}{1.008}\ \  = \ 5.4\ g$

Q. When same quantity of electricity is passed through aqueous $A{gNO}_{3}$ and $H_{2}{SO}_{4}$ 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)

Explanation

Reviewed by Dr. Abubacker

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