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An electric current of 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:
- a)0.005 mol
- b)0.01 mol
- c)0.02 mol
- d)0.04 mol
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An electric currentof 0.965 ampere is passed for 2000 seconds through ...
Given:
Electric current (I) = 0.965 A
Time (t) = 2000 s
[Cu(CH3CN)4] = unknown
We need to calculate the amount of Cu deposited at the cathode.
Formula used:
The amount of substance deposited at the electrode = (I * t * M) / (n * F)
where,
I = electric current
t = time
M = molar mass of the substance
n = number of electrons transferred in the reaction
F = Faraday constant
Calculation:
1. Finding the number of electrons transferred in the reaction:
[Cu(CH3CN)4] + e- → Cu + 4 CH3CN
The number of electrons transferred = 1
2. Finding the molar mass of [Cu(CH3CN)4]:
Let x be the molar mass of [Cu(CH3CN)4]
Molecular weight of Cu = 63.55 g/mol
Molecular weight of CH3CN = 41.05 g/mol
4 CH3CN molecules present in [Cu(CH3CN)4]
Molecular weight of [Cu(CH3CN)4] = 63.55 + (4 * 41.05) = 219.75 g/mol
x / 219.75 = 1 / 1
x = 219.75 g/mol
3. Finding the amount of Cu deposited:
Using the formula,
Amount of Cu deposited = (I * t * M) / (n * F)
= (0.965 A * 2000 s * 219.75 g/mol) / (1 * 96485 C/mol)
= 0.02 mol
Therefore, the amount of Cu deposited at the cathode is 0.02 mol. Hence, the correct answer is option C.
Electric current (I) = 0.965 A
Time (t) = 2000 s
[Cu(CH3CN)4] = unknown
We need to calculate the amount of Cu deposited at the cathode.
Formula used:
The amount of substance deposited at the electrode = (I * t * M) / (n * F)
where,
I = electric current
t = time
M = molar mass of the substance
n = number of electrons transferred in the reaction
F = Faraday constant
Calculation:
1. Finding the number of electrons transferred in the reaction:
[Cu(CH3CN)4] + e- → Cu + 4 CH3CN
The number of electrons transferred = 1
2. Finding the molar mass of [Cu(CH3CN)4]:
Let x be the molar mass of [Cu(CH3CN)4]
Molecular weight of Cu = 63.55 g/mol
Molecular weight of CH3CN = 41.05 g/mol
4 CH3CN molecules present in [Cu(CH3CN)4]
Molecular weight of [Cu(CH3CN)4] = 63.55 + (4 * 41.05) = 219.75 g/mol
x / 219.75 = 1 / 1
x = 219.75 g/mol
3. Finding the amount of Cu deposited:
Using the formula,
Amount of Cu deposited = (I * t * M) / (n * F)
= (0.965 A * 2000 s * 219.75 g/mol) / (1 * 96485 C/mol)
= 0.02 mol
Therefore, the amount of Cu deposited at the cathode is 0.02 mol. Hence, the correct answer is option C.
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Community Answer
An electric currentof 0.965 ampere is passed for 2000 seconds through ...
Equivalent weight of [Cu(CH3CN)4]+ =227/1 =227
mass deposited on cathode = E. W. (I×t) /96500
=227×0.965×2000/96500=4.45 gm
in moles = 4.45/227= 0.01960=0.02 mole
mass deposited on cathode = E. W. (I×t) /96500
=227×0.965×2000/96500=4.45 gm
in moles = 4.45/227= 0.01960=0.02 mole
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An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer?
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An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer?.
An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is:a)0.005 molb)0.01 molc)0.02 mold)0.04 molCorrect answer is option 'C'. Can you explain this answer?.
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