\(\text{Given:}\) Mass of Cu = 2.4 g, Molar mass of Cu = 63.5 g mol\(^{-1}\), 1 hour = 1 \(\times\) 60 \(\times\) 60 s = 3600 s. \(\text{To find: Current strength (in amperes)}\) \(\text{Formulae:}\) 1) Mole ratio = \(\frac{\text{Moles of product formed in half reaction}}{\text{Moles of electrons required in half reaction}}\) 2) \(W = \frac{I \cdot t}{96500 \, \frac{\text{C}}{\text{mol} \, e^{-1}}} \cdot \text{mole ratio} \cdot \text{molar mass}\). \(\text{Calculation:}\) 1) Stoichiometry for the formation of Cu is \(\text{Cu}^{2+}_{(s)} + 2e^{-} \rightarrow \text{Cu}_{(s)}\). Using formula (1), Mole ratio = \(\frac{1 \, \text{mole}}{2 \, \text{mole}}\). 2) Using formula (2), \[2.4 \, \text{g} = \frac{I \cdot t}{96500 \, \frac{\text{C}}{\text{mol} \, e^{-1}}} \cdot \frac{1 \, \text{mole}}{2 \, \text{mole} \, e^{-1}} \cdot 63.5 \, \text{g mol}^{-1}.\] Solving for \(I\): \[I = \frac{2.4 \cdot 96500 \cdot 2}{63.5 \cdot 3600} = 2.03 \, \text{A}.\] The current strength in amperes required to produce 2.4 g of Cu from CuSO\(_4\) is 2.03 A.