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Question

Marshall's acid or perdisulphuric acid (H2S2O8) can be prepared by electrolytic oxidation of H2SO4 as :
2H2SO4H2S2O8+2H+2e
O2 and H2 are byproducts. In such an electrolysis 9.72 L of H2 and 2.35 L of O2 were produced at S.T.P. What is the weight of H2S2O8 formed ?

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Solution

H2SO42H+SO24
Or H2SO4H+HSO4
Two reactions are competing at anode :
2HSO4H2S2O8+2e
and H2O2H+12O2+2e
At cathode : 2H+2eH2(g)
Since one reaction at cathode and two reactions at anode are taking place, therefore, equivalent of H2(g) produced at cathode should be equal to the equivalent of O2 produced and equivalent of H2S2O8 formed at anode.
Eq of H2= Eq of O2+ Eq of H2S2O8

9.7222.4/2=2.3522.4/4+W(H2S2O8)194/2
WH2S2O8=43.456g
Note :-
1 Eq of H2=22.4L211.2L at STP (n factor = 2)
1 Eq of O2=22.4L4=5.6L at STP (n factor = 4)
Ew of H2S2O8=Mw2 (n factor = 2)
Mw of H2S2O8=194 g mol1

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