A well stirred (1 L) solution of 0.1 M CuSO₄ is electrolysed at 25°C using copper electrodes with a current of 25 mA for 6 hours. If current efficiency is 50%. At the end of the duration what would be the concentration of copper ions in the solution?

According to the Faraday's law of electrolysis, 

w = $\frac{\mathrm{Eit}}{\mathrm{F}}$

where w is the weight deposited, E is the chemical equivalent and i is the current.

w = $\frac{63.5}{2\times 96500}\times 25\times {10}^{-3}\times 6\times 3600\times \frac{1}{2}$

⇒ w = 0.089 g

So, Moles deposited =1.398 milimoles

Initial moles of CuSO₄ ​= M$\times$V = 1$\times$0.1 = 0.1 moles = 100 millimoles
∴ Remaining moles of Cu²⁺ = 98.6 milimoles

So, The Concentration of Cu²⁺ = $\frac{98.6}{1\times 1000}$ = 0.0986 M