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Question

A 5 ampere current is passed through a solution of zinc sulphate for 40 minutes. The amount of zinc deposited at the cathode is:

A
0.4065 g
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B
65.04 g
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C
40.65 g
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D
4.065 g
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Solution

The correct option is D 4.065 g
Current (I) = 5 ampere and time (t) = 40 minutes=2400 seconds.
Amount of electricity passed (Q) = I×t
= 5×2400=12000 C
Now, Zn2+ + 2e Zn (1 mole=65.39 g)
Since, two mole electronic charges (i.e., 2×96500 C) deposits 65.39 g of zinc, therefore 12000 C will deposit
= (65.39×120002×96500)
= 4.065 g of zinc

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