Question

The weight of silver (Atomic weight = 108g) displaced by a quantity of electricity which displaces \[5600ml\] of \[{O_2}\] at STP will be
(A) \[5.4g\]
(B) \[10.8g\]
(C) \[54.0g\]
(D) \[108.0g\]

Answer 1

Hint: To calculate the weight of silver displaced by a quantity of electricity displaces \[5600ml\] of \[{O_2}\] , we need to know the Faraday’s Second Law of Electrolysis and the equivalent weight silver and \[{O_2}\].

Complete answer:
Faraday’s Second Law of Electrolysis: - This law states that when the same quantity of electric current is passed through different electrolytes, then the mass of the substances deposited are directly proportional to their respective equivalent weights.
\[W \propto E\]
Where, \[W = \]mass of the chemical substance deposited
\[E = \]equivalent weight of the respective chemical substance
So, according to the Faraday’s second law of electrolysis
\[W \propto {\rm E}\]
Where, \[W = \]weight of the chemical substance deposited
\[E = \]the equivalent weight of the chemical substance.
Since, the question is asking us the amount of Ag displaced by the electric current which also displaces \[5600ml\] of \[{O_2}\], so, from Faraday’s second law of electrolysis, it can be represented as,
\[\dfrac{{{W_{Ag}}}}{{{W_{{O_2}}}}} = \dfrac{{{E_{Ag}}}}{{{E_{{O_2}}}}}\]
Where, \[{W_{Ag}}\]= amount of silver deposited
\[{E_{Ag}}\]= equivalent weight of silver
\[{W_{{O_2}}}\] = weight of \[{O_2}\] released during the electrolysis
\[{E_{{O_2}}}\] = equivalent weight of \[{O_2}\]
The above expression can be written as,
\[\dfrac{{{W_{Ag}}}}{{{E_{Ag}}}} = \dfrac{{{W_{{O_2}}}}}{{{E_{{O_2}}}}}\]
Now, we have to calculate the weight of \[{O_2}\]released during the electrolysis,
\[{W_{{O_2}}}\]=\[\dfrac{{{V_{{O_2}}}}}{V} \times M\]
Where, \[{V_{{O_2}}}\] = volume of \[{O_2}\] released which is \[5600ml\] as given in the question
\[V\]= volume of 1 mole of \[{O_2}\] which is \[22400ml\]
\[M\]= molar mass of \[{O_2}\]
Hence, \[{W_{{O_2}}}\] \[ = \dfrac{{5600}}{{22400}} \times 32 = 8gram\]
Finally, for calculating the amount the amount of silver displaced during the electrolysis,
From equation (i)
\[
  \dfrac{{{W_{Ag}}}}{{{E_{Ag}}}} = \dfrac{{{W_{{O_2}}}}}{{{E_{{O_2}}}}} \\
\Rightarrow \dfrac{{{W_{Ag}}}}{{108}} = \dfrac{8}{8}...(ii) \\
 \]
Here, for calculating the equivalent weight of \[{O_2}\] and Ag, we have to use the following formula,
Equivalent weight\[ = \dfrac{{atomic{\text{ }}weight}}{{Valency}}\]
Equivalent weight is defined as the weight of the substance which will combine ore displace a unit weight of hydrogen.
So, \[{E_{{O_2}}} = \dfrac{{32}}{4} = 8\]
And \[{E_{Ag}} = \dfrac{{108}}{1} = 108\]
Hence, from equation (ii), we can write
\[
  \dfrac{{{W_{Ag}}}}{{108}} = \dfrac{8}{8} \\
   \Rightarrow {W_{Ag}} = 108gram
 \]
Hence, \[108gram\] of silver will be displaced by the quantity of electric current which displaces \[5600ml\] of \[{O_2}\] at STP.

Option D is correct.

Note:
Students must know the equivalent weights of different elements as they are directly used in the questions and if the question asks for the amount of more than 2 different substances deposited during electrolysis, then for that Faraday’s Second Law will be used as
\[{W_1},{W_2},{W_3} = {E_1},{E_2},{E_3}\] where \[{W_1},{W_2},{W_3}\] are the amount of the 3 different substances deposited during electrolysis respectively and \[{E_1},{E_2},{E_3}\] are their respective equivalent weight.