ElectrochemistryHard
Question
Perdisulphuric acid, H2S2O8 can be prepared by electrolytic oxidation of H2SO4, oxygen and hydrogen gases are by products. In such an electrolysis, 9.08 L of H2 and 2.27 L of O2 were generated at STP. What is the mass of H2S2O8 formed?
Options
A.0
B.77.6 g
C.38.8 g
D.19.4 g
Solution
$n_{eq}H_{2}\left( \text{at Cathode} \right) = n_{eq}O_{2} + n_{eq}H_{2}S_{2}O_{8}\left( \text{at Anode} \right)$
$\text{Or, }\frac{9.08}{22.7} \times 2 = \frac{2.27}{22.7} \times 4 + \frac{w}{194} \times 2\left\lbrack 2SO_{4}^{2 -} \rightarrow S_{2}O_{8}^{2 -} + 2e^{-} \right\rbrack $$$\therefore w = 38.8\text{ gm}$$
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