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Standard XII

Chemistry

Faraday's Second Law

One of the me...

Question

One of the methods of preparation of per disulphuric acid, $H_{2} S_{2} O_{8}$ , involve electrolytic oxidation of $H_{2} S O_{4}$ at node $(2 H_{2} S O_{4} \to H_{2} S O_{4} + 2 H^{+} + 2 e^{-})$ with oxygen and hydrogen as by -products. In such an electrolysis, 9.722 L of $H_{2}$ and 2.35 L of $O_{2}$ were generated at STP. What is the weight of $H_{2} S_{2} O_{8}$ formed in gms? (write the value to the nearest integer).

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Solution

At anode, $(2 H_{2} S O_{4} \to H_{2} S O_{4} + 2 H^{+} + 2 e^{-})$
$4 O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$
At cathode, $H_{2} + 2 e^{-} ⇌ 2 H^{+}$
The number of moles of hydrogen liberated $= \frac{9.722}{22.4} = 0.434$ mol.
The number of moles of electrons required for liberation of hydrogen $= 2 \times 0.434 m o l e^{-} = 0.868 m o l e^{-}$
The number of moles of oxygen liberated $= \frac{2.35}{22.4} = 0.1049$ mol.
The number of moles of electrons released during liberation of oxygen $= 4 \times 0.1049 m o l e^{-} = 0.4196 m o l e^{-}$
The number of moles of electrons released during formation of $H_{2} S_{2} O_{8} = 0.868 m o l e^{-} - 0.4196 m o l e^{-} = 0.448 m o l e^{-}$
The number of moles of $H_{2} S_{2} O_{8}$ formed $= \frac{0.448}{2} = 0.224 m o l$
Mass of $H_{2} S_{2} O_{8}$ formed $= 0.224 m o l \times 194.1 g / m o l = 43.45 g$ .

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