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Standard XII

Chemistry

Faraday's Second Law

Peroxy disulp...

Question

Peroxy disulphuric acid $(H_{2} S_{2} O_{8})$ can be prepared by electrolytic oxidation of $H_{2} S O_{4}$ as:
$2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-}$
Oxygen and hydrogen are by products. In such an electrolysis $9.72 l i t r e$ of $H_{2}$ and $2.35$ litre of $O_{2}$ were generated at NTP. What is the mass of peroxy disulphuric acid formed?

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Solution

Anodic reaction: $2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-}$
$2 H_{2} O \to 4 H^{+} + O_{2} + 2 e^{-}$
Cathodic reaction : $2 H_{2} O + 2 e^{-} \to 2 O H^{-} + H_{2}$ $

Total equivalent of $H_{2} S_{2} O_{8} +$ equivalent of oxygen $=$ Equivalent of $H_{2}$

$9.72 l i t r e H_{2} = \frac{9.72}{11.2} = 0.868 e q u i v a l e n t$

$2.35 l i t r e O_{2} = \frac{2.35}{5.6} = 0.42 e q u i v a l e n t$

Equivalent of $H_{2} S_{2} O_{8} = (0.868 - 0.420)$

$= 0.448$

Mass of $H_{2} S_{2} O_{8} = 0.448 \times \frac{194}{2} = 43.456 g$ .

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Peroxy disulphuric acid (H2S2O8) can be prepared by electrolytic oxidation of H2SO4 as:2H2SO4→H2S2O8+2H++2e−Oxygen and hydrogen are by products. In such an electrolysis 9.72 litre of H2 and 2.35 litre of O2 were generated at NTP. What is the mass of peroxy disulphuric acid formed?