A solution of \(CuSO_4\) is electrolyzed for 10 minutes using a current of 1.5 amperes \((\text{1 mol of Cu = 63 g})\). How much copper is deposited at the cathode? \(1F = 96487 \text{ Cmol}^{-1}\)

0.2938 g

The correct answer is (1) \(0.293 g\)

Time \((t) = 10 \text{ min = }10 × 60 \text{ = 600 sec}\)

Current \((A) = 1.5 A\)

From the above question, we can form the reaction at the cathode as:

\(\underset{\text{copper ions}}{Cu^{2+} (aq)} + \underset{electrons}{2e^-} \longrightarrow \underset{Cu (s)}{Cu(s)}\)

Thus, 2 electrons are transferred here.

The mass of copper \(= 63.5 \text{ g mol}^{-1}\)

We know that,

\(\text{Charge = time } × \text{ current}\)

\( ⇒ \text{Charge = 600 }×\text{ 1.5}\)

\( ⇒ \text{Charge = 900C}\)

Now, the Mass of copper deposited

\(= \frac{\text{Molar mass }×\text{ Charge}}{\text{Electrons transferred }×\text{ Faradays constant}}\)

\(= \frac{63.5 × 900}{2 × 96500}\)

\(= \frac{57150}{193000}\)

\(≈ 0.2938 g\)