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
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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.
A current of 0.50 ampere is passing through a CuSO4solution. How many ...