Question

100 mL of 1 M solution of $Cu{Br}_2$ was electrolyzed with a current of 0.965 ampere hour. What is the normality of the remaining $Cu{Br}_2$ solution?

• A. 1.64
• B. 3.28
• C. 0.82
• D. 4.92

Answer 1

The correct option is C 0.82

The reaction taking place would be:

$C{u}^{2+}+2e^{-}⟶Cu$

$\therefore 2\times 96500=193000C$ is used to reduce $1$ mole of $C{u}^{2+}$.

The amount of charge used= $Q=I\times t$

$=0.965A\times 3600sec$

$=3474C$

$\therefore 3474C$ will reduce= $\frac{3474}{193000}=0.018$ mole of $C{u}^{2+}$

Number of moles in initial solution= $1M\times \frac{100ml}{1000ml{L}^{-1}}$

$=0.1mol$

$\therefore$ Number of moles of $C{u}^{2+}$ left after electrolysis= $0.1-0.018=0.082$

Molarity of $CuB{r}_2$ after deposition

$=\text{moles of}C{u}^{2+}\text{left / Volume of solution}$

$=\frac{0.082}{100/1000}=0.82M$