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On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)
  • a)
    0.01 N
  • b)
    0.01 M
  • c)
    0.02 M
  • d)
    0.2 N
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
On passing a current of 1.0 ampere for 16 min and 5 sec through one li...
By Faraday's Ist Law, W/e = q/96500
(where q = it = charge of ion)
we know that no of equivalent

(where i = 1 A, t = 16×60+5 = 965 sec.) Since, we know that
Normality = no. of equivalent/Volume (in litre)
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On passing a current of 1.0 ampere for 16 min and 5 sec through one li...
To find the strength of the CuCl2 solution, we need to first find the amount of copper deposited at the cathode.

The amount of copper deposited can be calculated using Faraday's law of electrolysis, which states:

Amount of substance = (Current × Time) / (n × F)

Where:
- Current is the electric current in amperes (A)
- Time is the time in seconds (s)
- n is the number of moles of electrons transferred in the balanced chemical equation
- F is the Faraday constant, which is 96,500 C/mol

In this case, the balanced chemical equation for the deposition of copper is:

Cu2+ + 2e- → Cu

From the equation, we can see that 2 moles of electrons are required for the deposition of 1 mole of copper.

Given:
- Current = 1.0 A
- Time = 16 min and 5 sec = 16 × 60 + 5 = 965 seconds
- Molar mass of Cu = 63.5 g/mol
- Faraday constant = 96,500 C/mol

n = 2 moles of electrons / 1 mole of copper = 2

Amount of copper deposited = (1.0 A × 965 s) / (2 × 96,500 C/mol) = 0.005 mol

Now, we can find the strength of the CuCl2 solution using the equation:

Strength (Molarity) = Amount of solute (in moles) / Volume of solution (in liters)

The volume of the solution is given as 1 liter.

Strength = 0.005 mol / 1 L = 0.005 M

Therefore, the strength of the CuCl2 solution is 0.005 M.
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On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)a)0.01 Nb)0.01 Mc)0.02 Md)0.2 NCorrect answer is option 'A'. Can you explain this answer?
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On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)a)0.01 Nb)0.01 Mc)0.02 Md)0.2 NCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)a)0.01 Nb)0.01 Mc)0.02 Md)0.2 NCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)a)0.01 Nb)0.01 Mc)0.02 Md)0.2 NCorrect answer is option 'A'. Can you explain this answer?.
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