JAMB Chemistry 2012 Paper

Concept:Copper(II) ions are reduced to copper metal by gaining two electrons.So depositing one mole of copper requires two moles of electrons.Explanation:The cathode half-reaction is:$$\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$$From this equation, $1$ mole of $\mathrm{Cu}$ needs $2$ moles of electrons.One mole of electrons carries $96500\ \mathrm{C}$, so the charge needed for $1$ mole of $\mathrm{Cu}$ is:$$2 \times 96500\ \mathrm{C} = 193000\ \mathrm{C}$$Now, calculate the moles deposited by $360\ \mathrm{C}$:$$n = \frac{360\ \mathrm{C}}{193000\ \mathrm{C\ mol^{-1}}} = 0.001865\ \mathrm{mol}$$Expressed in standard scientific notation:$$n = 1.87 \times 10^{-3}\ \mathrm{mol}$$Answer:The correct option is B. $1.87 \times 10^{-3}$ mole.