The time required to deposit 4.5g of copper from CuSO\(_4\) solution by passing a current of 2.5 Amperes is (Cu = 64g ; 1F = 96500C/mol)

Copper is deposited from CuSO₄ solution by the reduction of Cu²⁺ ions:

\[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\]

This means each mole of copper requires 2 moles of electrons (2 faradays) to be deposited.

Step 1: Calculate the moles of copper to be deposited.

\[n_{\text{Cu}} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{64} = 0.0703125 \text{ mol}\]

Step 2: Calculate the total charge required.

Since 1 mole of Cu requires 2 faradays:

\[Q = n_{\text{Cu}} \times 2 \times F = 0.0703125 \times 2 \times 96500\]

\[Q = 0.140625 \times 96500 = 13570.3 \text{ C}\]

Step 3: Calculate the time using \(Q = It\).

\[t = \frac{Q}{I} = \frac{13570.3}{2.5} = 5428 \text{ sec}\]

The time required is 5428 seconds.