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WAEC Chemistry 2017 Paper

Question : 41

Total: 50

\mathrm{CuSO}_{4}

Solution:

Concept:

Electrolysis: the mass of copper deposited is found using Faraday’s laws and the charge passed.

Explanation:

The cathode reaction is:

\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \to \mathrm{Cu}

Thus,

2

moles of electrons deposit

1

mole of copper.

Charge required to deposit

64\,\text{g}

of copper:

Q = 2 \times 96500\,\text{C} = 193000\,\text{C}

Time must be in seconds:

t = 1\,\text{hr}\,15\,\text{min} = 75 \times 60 = 4500\,\text{s}

Charge passed:

Q = I \times t = 0.45 \times 4500 = 2025\,\text{C}

Mass deposited:

m = \frac{64}{193000} \times 2025 = 0.6715\,\text{g}

This is approximately

0.67\,\text{g}

Answer:

0.67\,\text{g}

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