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Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited?  
  • a)
    16.20 min, Copper 0.487g, Zinc 0.437 g
  • b)
    15.10 min, Copper 0.452g, Zinc 0.437 g
  • c)
    14.40 min, Copper 0.427g, Zinc 0.437 g
  • d)
    13.10 min, Copper 0.403g, Zinc 0.437 g
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 ...
Ag has +1 charge in AgNO3
So charge transfer, n = 1 
F = 96487 Coulombs
Molar mass of Ag = 108 g 
Use formula required charge  = nF
So required charge for 1 mol or 108 g of Ag = Coulombs.
Charge required for 1.45 g of Ag = 96487 Coulombs x 1.45g/108g 
              = 1295.43 Coulombs
Given that current I = 1.5 A
Use formula Charge = I x t
Time t  = charge / I 
           = 1295.43 Coulombs/ 1.5 A 
          = 863.6 s
Divide by 60 to convert in minute 
= 864/60 min
= 14.40 min
Charge in Cu in CuSo4 is +2 
Use formula required charge for 1 mol = nF 
So charge required for 1 mol or 63.5 g  of Cu = 2 x  96487 = 192974 Coulombs 
192974 Coulombs of charge deposit = 63.5 g of Cu
1295.43 Coulombs of charge will deposit = 63.5 g x 1295.43/192974
        = 0.426 g of Cu
Similarly for Zn 
Zn has charge in ZnSO4  = +2 
Use formula required charge for 1 mol = nF 
So charge required for 1 mol or 63.5 g  of Zn = 2 x  96487 = 192974 Coulombs 
192974 Coulombs of charge deposit = 63.5 g of Zn
2 x 96487 C of charge deposit = 65.4 g of Zn
1295.43 Coulombs of charge will deposit = 65.4 g x 1295.43/192974
        = 0.439 g of Zn
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Most Upvoted Answer
Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 ...
Given information:
- Three electrolytic cells A, B, C connected in series containing solutions of ZnSO4, AgNO3 and CuSO4 respectively.
- A steady current of 1.5 amperes passed through them.
- 1.45 g of silver deposited at the cathode of cell B.

To find:
- How long did the current flow?
- What mass of copper and zinc were deposited?

Solution:
1. Calculation of time:
- We can calculate the time taken for the current to flow using the formula:
Time = Quantity of electricity / Current
- Quantity of electricity can be calculated using the formula:
Quantity of electricity = Current x Time
- We know the current and the quantity of silver deposited at the cathode of cell B.
- The molar mass of silver is 107.87 g/mol and the charge on one mole of electrons is 96500 C.
- Using these values, we can calculate the quantity of electricity required for the deposition of 1.45 g of silver as follows:
Quantity of electricity = (1.45 g / 107.87 g/mol) x (96500 C/mol) = 1297.88 C
- Now, we can calculate the time taken for the current to flow through the cells as follows:
Time = Quantity of electricity / Current = 1297.88 C / 1.5 A = 864.58 s = 14.41 min ≈ 14.4 min
- Therefore, the time taken for the current to flow is approximately 14.4 min.

2. Calculation of mass of copper and zinc:
- The mass of copper and zinc deposited can be calculated using the formula:
Mass = Current x Time x Equivalent weight / 96500
- Equivalent weight is the atomic weight divided by the valency.
- For the deposition of copper and zinc, the valency is 2. Therefore, the equivalent weight of copper and zinc can be calculated as follows:
Equivalent weight of copper = 63.55 g/mol / 2 = 31.77 g/equiv
Equivalent weight of zinc = 65.38 g/mol / 2 = 32.69 g/equiv
- Using the given current and the time taken for the current to flow, we can calculate the mass of copper and zinc deposited as follows:
Mass of copper = 1.5 A x 864.58 s x 31.77 g/equiv / 96500 C = 0.427 g
Mass of zinc = 1.5 A x 864.58 s x 32.69 g/equiv / 96500 C = 0.437 g
- Therefore, the mass of copper and zinc deposited is 0.427 g and 0.437 g, respectively.

Therefore, option (c) 14.4 min, Copper 0.427g, Zinc 0.437 g is the correct answer.
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Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited? a)16.20 min, Copper 0.487g, Zinc 0.437 gb)15.10 min, Copper 0.452g, Zinc 0.437 gc)14.40 min, Copper 0.427g, Zinc 0.437 gd)13.10 min, Copper 0.403g, Zinc 0.437 gCorrect answer is option 'C'. Can you explain this answer?
Question Description
Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited? a)16.20 min, Copper 0.487g, Zinc 0.437 gb)15.10 min, Copper 0.452g, Zinc 0.437 gc)14.40 min, Copper 0.427g, Zinc 0.437 gd)13.10 min, Copper 0.403g, Zinc 0.437 gCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited? a)16.20 min, Copper 0.487g, Zinc 0.437 gb)15.10 min, Copper 0.452g, Zinc 0.437 gc)14.40 min, Copper 0.427g, Zinc 0.437 gd)13.10 min, Copper 0.403g, Zinc 0.437 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited? a)16.20 min, Copper 0.487g, Zinc 0.437 gb)15.10 min, Copper 0.452g, Zinc 0.437 gc)14.40 min, Copper 0.427g, Zinc 0.437 gd)13.10 min, Copper 0.403g, Zinc 0.437 gCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited? a)16.20 min, Copper 0.487g, Zinc 0.437 gb)15.10 min, Copper 0.452g, Zinc 0.437 gc)14.40 min, Copper 0.427g, Zinc 0.437 gd)13.10 min, Copper 0.403g, Zinc 0.437 gCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
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