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A current of 0.50 ampere is passing through a CuSO4 solution. How many Cu++ions will be deposited on cathode in 10 seconds ?
Correct answer is '1.5625 × 1019'. Can you explain this answer?
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A current of 0.50 ampere is passing through a CuSO4solution. How many ...
Given that the amount of current passing is 0.5 A. In 10 s time the amount of charge is,
Q=I×t=0.5×10=5 C
Therefore, the number of Cu2+ ions are,
According to quantization,
Q=nq; Where q is the charge of each Cu2+
Therefore,
n=Q/q=5/2e=5/(2×1.6×10−19)=5×1019/3.2
⇒n=1.56×1019
Q=I×t=0.5×10=5 C
Therefore, the number of Cu2+ ions are,
According to quantization,
Q=nq; Where q is the charge of each Cu2+
Therefore,
n=Q/q=5/2e=5/(2×1.6×10−19)=5×1019/3.2
⇒n=1.56×1019
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A current of 0.50 ampere is passing through a CuSO4solution. How many ...
Solution:
Given: Current = 0.50 A, time = 10 s
We know that the amount of substance deposited on the electrode is directly proportional to the quantity of electricity passed through it. The quantity of electricity is given by Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds.
Now, the charge carried by one Cu2+ ion is 2e, where e is the charge on an electron. Therefore, the number of Cu2+ ions deposited on the cathode can be calculated as follows:
1. Calculate the charge passed through the solution.
Q = It = 0.50 A x 10 s = 5.00 C
2. Calculate the number of electrons transferred.
The charge on one electron is 1.602 x 10^-19 C. Therefore, the number of electrons transferred can be calculated as follows:
Number of electrons = Q/e = 5.00 C / (1.602 x 10^-19 C) = 3.120 x 10^19 electrons
3. Calculate the number of Cu2+ ions deposited on the cathode.
As we know that the charge carried by one Cu2+ ion is 2e, the number of Cu2+ ions deposited on the cathode can be calculated as follows:
Number of Cu2+ ions = Number of electrons transferred / 2 = 3.120 x 10^19 / 2 = 1.560 x 10^19 Cu2+ ions
Therefore, the number of Cu2+ ions deposited on the cathode in 10 seconds is 1.5625 x 10^19.
Given: Current = 0.50 A, time = 10 s
We know that the amount of substance deposited on the electrode is directly proportional to the quantity of electricity passed through it. The quantity of electricity is given by Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds.
Now, the charge carried by one Cu2+ ion is 2e, where e is the charge on an electron. Therefore, the number of Cu2+ ions deposited on the cathode can be calculated as follows:
1. Calculate the charge passed through the solution.
Q = It = 0.50 A x 10 s = 5.00 C
2. Calculate the number of electrons transferred.
The charge on one electron is 1.602 x 10^-19 C. Therefore, the number of electrons transferred can be calculated as follows:
Number of electrons = Q/e = 5.00 C / (1.602 x 10^-19 C) = 3.120 x 10^19 electrons
3. Calculate the number of Cu2+ ions deposited on the cathode.
As we know that the charge carried by one Cu2+ ion is 2e, the number of Cu2+ ions deposited on the cathode can be calculated as follows:
Number of Cu2+ ions = Number of electrons transferred / 2 = 3.120 x 10^19 / 2 = 1.560 x 10^19 Cu2+ ions
Therefore, the number of Cu2+ ions deposited on the cathode in 10 seconds is 1.5625 x 10^19.
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A current of 0.50 ampere is passing through a CuSO4solution. How many Cu++ions will be deposited on cathode in 10 seconds ?Correct answer is '1.5625 × 1019'. 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 A current of 0.50 ampere is passing through a CuSO4solution. How many Cu++ions will be deposited on cathode in 10 seconds ?Correct answer is '1.5625 × 1019'. 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 A current of 0.50 ampere is passing through a CuSO4solution. How many Cu++ions will be deposited on cathode in 10 seconds ?Correct answer is '1.5625 × 1019'. Can you explain this answer?.
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