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]
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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 = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
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
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