
The chemical equivalent of copper and silver are 32 and 108 respectively. When copper and silver voltmeters are connected in series and an electric current is passed through for some time, 1.6 g of copper is deposited. Then, the mass of silver deposited will be
A. 3.5 g
B. 2.8 g
C. 5.4 g
D. None of these
Answer
162.6k+ views
Hint: When two electrical components are connected in series then the electricity in both components are same. As electricity is defined as the flow of charge, so, is the amount of the charge transferred to copper and silver. We use faraday’s first law of electrolysis to determine the amount of silver deposited.
Formula Used:\[m = Zit\], where m is the mass of the ion deposited, Z is the electrochemical equivalent, i is the electric current and t is the time and Z is the electric equivalence of the electrolyte.
Complete answer:As the copper and silver voltammeter are connected in series, so the charges to each will be the same. Let the equal electric current to copper and silver be I for the same interval of time.
As we know that the mass of the ion deposited is proportional to the current passing through the electrolyte and the time duration using Faraday’s first law of electrolysis,
\[m = Zit\]
We have given the chemical equivalence of copper and silver as 32 and 108.
\[{E_{Cu}} = 32\]
\[{E_{Ag}} = 108\]
Then the electrochemical equivalence can be determined as,
\[Z = \dfrac{E}{F}\]
For the copper and silver branches respectively,
\[{m_{Cu}} = {Z_{Cu}}it\]
And,
\[{m_{Ag}} = {Z_{Ag}}it\]
The amount of copper deposited is given as 1.6g.
On dividing the expression for the masses of the ion deposited, we get
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{Z_{Cu}}it}}{{{Z_{Ag}}it}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{\left( {\dfrac{{{E_{Cu}}}}{F}} \right)}}{{\left( {\dfrac{{{E_{Ag}}}}{F}} \right)}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{E_{Cu}}}}{{{E_{Ag}}}}\]
\[{m_{Ag}} = \left( {\dfrac{{{E_{Ag}}}}{{{E_{Cu}}}}} \right){m_{Cu}}\]
\[{m_{Ag}} = \left( {\dfrac{{108}}{{32}}} \right) \times 1.6g\]
\[{m_{Ag}} = 5.4g\]
So, the mass of silver deposited is 5.4g.
Therefore,
the correct option is (C).
Note: We should be careful about how the copper and silver voltammeter are arranged in the circuit. If the circuit is series then the current in both will be the same and if the circuit is in parallel then the current in both will be different.
Formula Used:\[m = Zit\], where m is the mass of the ion deposited, Z is the electrochemical equivalent, i is the electric current and t is the time and Z is the electric equivalence of the electrolyte.
Complete answer:As the copper and silver voltammeter are connected in series, so the charges to each will be the same. Let the equal electric current to copper and silver be I for the same interval of time.
As we know that the mass of the ion deposited is proportional to the current passing through the electrolyte and the time duration using Faraday’s first law of electrolysis,
\[m = Zit\]
We have given the chemical equivalence of copper and silver as 32 and 108.
\[{E_{Cu}} = 32\]
\[{E_{Ag}} = 108\]
Then the electrochemical equivalence can be determined as,
\[Z = \dfrac{E}{F}\]
For the copper and silver branches respectively,
\[{m_{Cu}} = {Z_{Cu}}it\]
And,
\[{m_{Ag}} = {Z_{Ag}}it\]
The amount of copper deposited is given as 1.6g.
On dividing the expression for the masses of the ion deposited, we get
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{Z_{Cu}}it}}{{{Z_{Ag}}it}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{\left( {\dfrac{{{E_{Cu}}}}{F}} \right)}}{{\left( {\dfrac{{{E_{Ag}}}}{F}} \right)}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{E_{Cu}}}}{{{E_{Ag}}}}\]
\[{m_{Ag}} = \left( {\dfrac{{{E_{Ag}}}}{{{E_{Cu}}}}} \right){m_{Cu}}\]
\[{m_{Ag}} = \left( {\dfrac{{108}}{{32}}} \right) \times 1.6g\]
\[{m_{Ag}} = 5.4g\]
So, the mass of silver deposited is 5.4g.
Therefore,
the correct option is (C).
Note: We should be careful about how the copper and silver voltammeter are arranged in the circuit. If the circuit is series then the current in both will be the same and if the circuit is in parallel then the current in both will be different.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

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

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
