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]
- 0.300g
- 0.250g
- 0.2242g
-
0.448g
The correct answer is: 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 = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol-1
no of charge = + 2
Mass = \(\frac{64 \times 1350}{96500 \times 2}\)
Mass = 0.448g
There is an explanation video available .