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WAEC Chemistry 2011 Paper

Question : 34

Total: 46

Solution:

Concept:

Electrolysis of

\mathrm{CuSO}_4

deposits copper at the cathode.

Each

\mathrm{Cu}^{2+}

ion requires 2 electrons to become copper metal:

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

Explanation:

Current is

I = 10.0 \, \text{A}

and time is

t = 1 \, \text{hour} = 3600 \, \text{s}

Total charge passed:

Q = I \times t = 10.0 \times 3600 = 36000 \, \text{C}

Moles of electrons:

n_{e^-} = \dfrac{Q}{F} = \dfrac{36000}{96500} = 0.373 \, \text{mol}

Since 2 moles of electrons deposit 1 mole of copper:

n_{\mathrm{Cu}} = \dfrac{0.373}{2} = 0.1865 \, \text{mol}

Mass of copper formed:

m = n_{\mathrm{Cu}} \times M_{\mathrm{Cu}} = 0.1865 \times 63.5 = 11.8 \, \text{g}

Answer:

The mass of copper formed is

11.8 \, \text{g}

, which corresponds to option B.

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