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Calculate the mass of copper deposited when a current of 0.5 ampere was passed through...

Chemistry
JAMB 2017

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
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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

There is an explanation video available below.


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Explanation Video

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