Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]
A.
0.300g
B.
0.250g
C.
0.2242g
D.
0.448g
Correct Answer: Option D
Explanation
M = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
= \(\frac{MmIT}{96500n}\)
Copper II Chloride = CuCl2
CuCl2 → Cu2+ + 2Cl2
Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
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