Practicing Success
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 2.938 g 29.38 g 0.02938 g |
0.2938 g |
The correct answer is option (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 \(= \frac{\text{Molar mass }×\text{ Charge}}{\text{Electrons transferred }×\text{ Faradays constant}}\) \(= \frac{63.5 × 900}{2 × 96500}\) \(= \frac{57150}{193000}\) \(≈ 0.2938 g\) |