100-mL of a 0-05-M-CuSO-4-aqueous nbsp-solution was electrolysed using inert electrodes by passing current till the pH of the resulting solution was 2- The solution after electrolysis nbsp-was neutralised and then treated with excess KI and the formed I-2 was titrated with 0-04-M-Na-2S-2O-3- Calculate the volume of Na-2S-2O-3 required in mL-112-5-mL100-mL125-mLNone of these

The correct option is A 112.5 mLDuring electrolysis of CuSO_4 At cathode, Cu^2+ + 2e^ → Cu(s) nbsp; nbsp; E^o = 0.34 At anode, 2H_2O → O_2 + 4H^+ + 4e^ ...

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Standard XII

Chemistry

Application of Electrolysis

100 mL of a 0...

Question

$100 mL$ of a $0.05 M C u S O_{4} aqueous$ 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 $K I$ and the formed $I_{2}$ was titrated with $0.04 M N a_{2} S_{2} O_{3}$ . Calculate the volume of $N a_{2} S_{2} O_{3}$ required in $mL$ .

112.5 mL

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100 mL

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125 mL

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None of these

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Solution

The correct option is A $112.5 mL$
During electrolysis of $C u S O_{4}$
At cathode,
$C u^{2 +} + 2 e^{-} \to C u (s)$
$E^{o} = 0.34$

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

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

$p H = 2$ hence, $[H^{+}] = 10^{- 2}$ M
$[S O_{4}^{2 -}] = \frac{10^{- 2}}{2}$ M
$= 5 \times 10^{- 3}$ M

Initial milli moles of $C u S O_{4}$ = 50
Amount of $C u S O_{4}$ left after electrolysis $= 50 - 5$
$= 45$ milli moles
$C u^{2 +} = 0.045$ M
Also,
$C u S O_{4} + K I \to C u I + I_{2}$
$I_{2} + N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + N a I$
As $C u^{2 +}$ is reacting with $K I$ ,
Therefore $M e q C u^{2 +}$ = $M e q K I$
Also $M e q K I$ = $M e q I_{2}$
$M e q I_{2}$ = $M e q N a_{2} S_{2} O_{3}$
$M e q C u^{2 +}$ = $0.045 \times 1 \times 100$
= 4.5

$M e q N a_{2} S_{2} O_{3}$ = $0.04 \times 1 \times V_{N a_{2} S_{2} O_{3}}$ $0.04 \times 1 \times V_{N a_{2} S_{2} O_{3}} = 4.5$
$V_{N a_{2} S_{2} O_{3}} = \frac{4.5}{0.04}$
$= 112.5 m L$

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