Question

$100\text{mL}$ of a $0.05\text{M}CuS{O}_4\text{aqueous}$ solution was electrolysed using inert electrodes by passing current till the pH of the resulting solution was 2. The solution after electrolysis was neutralised and then treated with excess $KI$ and the formed $I_2$ was titrated with $0.04\text{M}N{a}_2{S}_2{O}_3$. Calculate the volume of $N{a}_2{S}_2{O}_3$ required in $\text{mL}$.

• A. $112.5\text{mL}$
• B. $100\text{mL}$
• C. $125\text{mL}$
• D. None of these

Answer 1

The correct option is A $112.5\text{mL}$

During electrolysis of $CuS{O}_4$
At cathode,
$C{u}^{2+}+2e^{-}\to Cu\left(s\right)$
$E^o=0.34$

At anode,
$2H_{2}O\to O_2+4H^{+}+4e^{-}$ $E^o=-1.23$

Therefore some of $C{u}^{2+}$ is converted to $Cu\left(s\right)$ and $H^{+}$ is released at anode

$pH=2$ hence, $\left[H^{+}\right]={10}^{-2}$M
$\left[S{O}_4^{2-}\right]=\frac{{10}^{-2}}{2}$M
$=5\times {10}^{-3}$M

Initial milli moles of $CuS{O}_4$ = 50
Amount of $CuS{O}_4$ left after electrolysis $=50-5$
$=45$ milli moles
$C{u}^{2+}=0.045$M
Also,
$CuS{O}_4+KI\to CuI+I_2$
$I_2+N{a}_2{S}_2{O}_3\to N{a}_2{S}_4{O}_6+NaI$
As $C{u}^{2+}$ is reacting with $KI$,
Therefore $MeqC{u}^{2+}$ = $MeqKI$
Also $MeqKI$ = $Meq{I}_2$
$Meq{I}_2$ = $MeqN{a}_2{S}_2{O}_3$
$MeqC{u}^{2+}$ = $0.045\times 1\times 100$
= 4.5

$MeqN{a}_2{S}_2{O}_3$ = $0.04\times 1\times V_{N{a}_2{S}_2{O}_3}$ $0.04\times 1\times V_{N{a}_2{S}_2{O}_3}=4.5$
$V_{N{a}_2{S}_2{O}_3}=\frac{4.5}{0.04}$
$=112.5mL$