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Standard XII

Chemistry

Faraday's Second Law

100 mL of 0...

Question

$100 m L$ of $0.05 M C u S O_{4} (a q)$ solution was electrolyzed using inert electrodes by passing current till the $p H$ of the resulting solution was $2$ . The solution after electrolysis was neutralized and then treated with excess $K I$ and formed $I_{2}$ titrated with $0.04 M N a_{2} S_{2} O_{3}$ . Calculate the required volume (in mL) of $N a_{2} S_{2} O_{3}$ .

112.5 m L

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90 m L

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125 m L

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None of these

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Solution

The correct option is A $112.5 m L$
When the copper sulphate solution is electrolyzed, till the pH of solution changes to $2$ , the number of milliequivalents of copper sulphate changes from $5$ to $4.5$ .
The, balanced chemical equations are;
$C u^{2 +} + 2 I^{-} \to C u + I_{2}$
$2 S_{2} O_{3}^{2 -} + I_{2} \to 2 I^{-} + S_{4} O_{6}^{2 -}$
Hence, $C u^{2 +} \equiv I_{2} \equiv 2 S_{2} O_{3}^{2 -}$
The volume of $N a_{2} S_{2} O_{3}$ required is $\frac{112.5 \times 0.04}{0.05} = 90 m L$ .

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