Question

A well stirred $\left(1L\right)$ solution of $0.1MCuS{O}_4$ is electrolysed at ${25}^{\circ }C$ using copper electrodes with a current of $25mA$ 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?

• A. $0.0856M$
• B. $0.092M$
• C. $0.0986M$
• D. $0.1M$

Answer 1

Answer: (C)

$0.0986M$

Using Faraday's law of electrolysis,

$W=Zit$

where $W$ is the weight deposited and $Z$ is the electrochemical equivalent and $i$ is the current.

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

$W=0.089gm$

Moles deposited $=1.398milimoles$

Initial moles of $CuS{O}_4=M\times V=1\times 0.1=0.1moles=100milimoles$

$\therefore$ Remaining moles of $C{u}^{2+}$ in the solution $=100-1.398=98.6milimoles$

Concentration of $C{u}^{2+}$ in the solution (Molarity) $=\frac{98.6}{1\times 1000}=0.0986M$