
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
Answer
169.5k+ views
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.
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.
Recently Updated Pages
Molarity vs Molality: Definitions, Formulas & Key Differences

Preparation of Hydrogen Gas: Methods & Uses Explained

Polymers in Chemistry: Definition, Types, Examples & Uses

P Block Elements: Definition, Groups, Trends & Properties for JEE/NEET

Order of Reaction in Chemistry: Definition, Formula & Examples

Hydrocarbons: Types, Formula, Structure & Examples Explained

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Displacement-Time Graph and Velocity-Time Graph for JEE

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Instantaneous Velocity - Formula based Examples for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions Exercise - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes - 2025-26

NCERT Solution for Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
