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Two faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu)
- a)0 g
- b)63.5 g
- c)2 g
- d)127 g
Correct answer is option 'B'. Can you explain this answer?
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Two faraday of electricity is passed through a solution of CuSO4. The ...
Calculation of Mass of Copper Deposited at the Cathode
Given:
- Faraday constant = 96500 C/mol
- Charge passed = 2 Faraday = 2 x 96500 C
- Atomic mass of Cu = 63.5 amu
We know that during the electrolysis of CuSO4, Cu2+ ions are reduced at the cathode to form Cu metal. The reduction half-reaction is:
Cu2+ + 2e- → Cu
From the above half-reaction, we can see that 2 electrons are required for the reduction of 1 Cu2+ ion to Cu metal.
Using Faraday's law of electrolysis, we can calculate the amount of Cu deposited at the cathode as follows:
- Amount of charge passed = Current x Time
- Current = Charge/Time
- Amount of Cu deposited = (Amount of charge passed)/(2 x Number of electrons required for the reduction of Cu2+ to Cu)
Substituting the given values, we get:
- Amount of charge passed = 2 x 96500 C
- Current = Amount of charge passed/Time
- Amount of Cu deposited = (2 x 96500 C)/(2 x 2) x (1 mol Cu/2 x 96500 C) x (63.5 g Cu/1 mol Cu)
Simplifying the above expression, we get:
- Amount of Cu deposited = 31.75 g
Therefore, the mass of copper deposited at the cathode is 63.5 g.
Given:
- Faraday constant = 96500 C/mol
- Charge passed = 2 Faraday = 2 x 96500 C
- Atomic mass of Cu = 63.5 amu
We know that during the electrolysis of CuSO4, Cu2+ ions are reduced at the cathode to form Cu metal. The reduction half-reaction is:
Cu2+ + 2e- → Cu
From the above half-reaction, we can see that 2 electrons are required for the reduction of 1 Cu2+ ion to Cu metal.
Using Faraday's law of electrolysis, we can calculate the amount of Cu deposited at the cathode as follows:
- Amount of charge passed = Current x Time
- Current = Charge/Time
- Amount of Cu deposited = (Amount of charge passed)/(2 x Number of electrons required for the reduction of Cu2+ to Cu)
Substituting the given values, we get:
- Amount of charge passed = 2 x 96500 C
- Current = Amount of charge passed/Time
- Amount of Cu deposited = (2 x 96500 C)/(2 x 2) x (1 mol Cu/2 x 96500 C) x (63.5 g Cu/1 mol Cu)
Simplifying the above expression, we get:
- Amount of Cu deposited = 31.75 g
Therefore, the mass of copper deposited at the cathode is 63.5 g.
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Two faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu)a)0 gb)63.5 gc)2 gd)127 gCorrect answer is option 'B'. Can you explain this answer?
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