A solution of \(CuSO_4\) is electrolyzed for \(10\, \ min\) with a currecnt of \(1.5 A\). What is the mass of copper deposited at the cathode (molar mass of \(Cu = 63 g\)) (\(1F = 96487C\))

0.2938 g

The correct answer is option 4. 0.2938 g.

Given,

Time \((t) = 10\, \ min = 600 s\)

Current \((I) = 1.5 A\)

The reaction of cathode can be formulated as:

\(Cu^{2+} + 2e^- \longrightarrow Cu(s)\)

Thus, 2 electrons are transferred here.

Mass of \(Cu = 63\, \ g mol^{-1}\)

We know that,

\(Charge\, \  =\, \  time \, \  × \, \  current\)

\(⇒ Charge\, \ = 600 × 1.5 = 900C\)

Now, mass of copper deposited

\(= \frac{\text{Molar mass } × \text{ Charge}}{\text{electrons transferred } × \text{ Faraday's constant}}\)

\(= \frac{63 × 900}{2 × 96487}\)

\(= \frac{56700}{192974}\)

\(= 0.2938 g\)