Question

In electrolysis, amount of silver deposited equivalent of 112 ml of Hydrogen is?

Answer 1

Answer:1.08gm

Explanation:don't know pls explain someone.

Answer 2

Answer:

The weight of silver deposited is $0.54g$.

Explanation:

Given data,

The volume of hydrogen = $112ml$ = $0.112L$

The amount of silver deposited =?

As we know, according to Faraday's law:

• The weight of any given substance liberated/deposited at any electrode on passing a certain amount of charge is directly proportional to its equivalent weight.

As given also, the amount of silver deposited is equivalent to the $112ml$ of hydrogen.

Thus, $\frac{W_{Ag} }{W_{H} } = \frac{Eq_{Ag} }{Eq_{H} }$      -------equation (1)

Firstly, we have to find the weight of hydrogen.

As we know, at STP ( standard temperature and pressure ) volume is $22.4L/mol$.

• The hydrogen's number of moles = $\frac{0.112}{22.4}$ = $0.005mol$
• The molecular weight of hydrogen, $M_{H}$ = $1g/mol$

Therefore,

• The weight of hydrogen = $1\times 0.005$ = $0.005g$.

As we know,

• The hydrogen's valency = 1 ( valence electrons )
• The silver's valency = 1
• The molecular weight of silver, $M_{Ag}$ = $108g/mol$

Therefore,

• The equivalent weight of hydrogen = $\frac{M_{H} }{Valency}$ = $\frac{1}{1}$ = $1g/mol$
• The equivalent weight of silver = $\frac{M_{Ag} }{Valency}$ = $\frac{108}{1}$ = $108g/mol$

After putting the calculated values of the equivalent weights and the weight of hydrogen in equation (1), we get:

• $\frac{W_{Ag} }{0.005} = \frac{108 }{1 }$
• $W_{Ag} = 0.54g$

Hence, the weight of silver deposited is $0.54g$.