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A certain metal salt solution is electrolysed in series with a silver coulometer. The weights of silver and the metal deposited are 0.5094 g and 0.2653 g. Calculate the valency of the metal if its atomic weight is nearly that of silver.
A. 1
B. 2
C. 3
D. 4

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Last updated date: 26th Feb 2024
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Hint: Faraday’s law of electrolysis was given by Michael Faraday which is used to determine magnitudes of electrolytic effects and chemical changes. This law is very important to detect the amount of substance deposit at electrodes.

Formula used: \[\dfrac{{{w_1}}}{{{w_2}}} = \dfrac{{({E_{Ag}})}}{{({E_{metal}})}} = \dfrac{{{{\left( {\dfrac{{atomic\,weight}}{{valency\,factor}}} \right)}_{Ag}}}}{{{{\left( {\dfrac{{atomic\,weight}}{{valency\,factor}}} \right)}_{metal}}}}\]

Complete step-by-step answer:
Faraday’s second law of electrolysis states that the masses of different ions liberated at the electrodes, when the same amount of electricity is passed through different electrolytes are directly proportional to their chemical equivalents. We can also say that electro-chemical equivalent(Z) of an element is directly proportional to their equivalent weight(E).
Let us suppose \[{w_1}\] and \[{w_2}\] are the deposited amount of any substance and \[{E_1}\] and \[{E_2}\] are their respective equivalent weights. As per this law, we can get it for silver as shown below-
 \[\dfrac{{{w_1}}}{{{w_2}}} = \dfrac{{({E_{Ag}})}}{{({E_{metal}})}} = \dfrac{{{{\left( {\dfrac{{atomic\,weight}}{{valency\,factor}}} \right)}_{Ag}}}}{{{{\left( {\dfrac{{atomic\,weight}}{{valency\,factor}}} \right)}_{metal}}}}\]
We use this above equation to solve numerical problems. Since the atomic weight is the same for both metal and silver, we cancel them. Putting the values of \[{w_1}\] and \[{w_2}\] as 0.5094 g and 0.2653 g, we get
 \[\dfrac{{0.5094}}{{0.2653}} = \dfrac{{({E_{Ag}})}}{{({E_{metal}})}} = \dfrac{{{{\left( {valency\,factor} \right)}_{metal}}}}{{{{\left( {valency\,factor} \right)}_{Ag}}}}\]
Silver has a valency factor of +1. So, now it becomes
 \[\dfrac{{0.5094}}{{0.2653}} = \dfrac{{{{\left( {valency\,factor} \right)}_{metal}}}}{1}\]
 \[\therefore {\left( {valency\,factor} \right)_{metal}} = 1.92 \approx 2\]
One Faraday is defined as the quantity of electricity flowing through an electrolyte which will release one-gram equivalent of any substance at any electrode. Faraday’s constant(F) is charged in a unit of coulombs(C) for one mole of electrons.

Hence, the correct option is (B).

Note: Faraday’s second law is also used to determine chemical equivalents of different electrolytes. The value of Faraday’s constant is equal to 96485 \[Cmo{l^{ - 1}}\]. The current of 96500 coulombs is known as Faraday charge.