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Question

During the electrolysis of conc H2SO4, it was found that H2S2O8 and O2 were liberated in a molar ratio of 3:1. How many moles of H2 were found is terms of moles of H2S2O8 ?
[Express your answer as : 3× moles of H2 ; integer answer is between 0 and 15].

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Solution

Let x mol of O2 is liberated and 3x mol of H2S2O8 is formed.
Reactions at cathode (reduction) :
2H2O+2eH2+2OH
Reactions at anode (oxidation) :
(i) 2H2OO2+4H+4e [1molO24FxmolO2=4xF]
(II) 2SO24S2O28+2e [1mol S2O28=2F3x mol S2O28=6xF]
Total faradays at anode = (4x + 6x)F = 10x F.
Total Faradays at acthode = 2F = 1 mol H2
10x F Total Faradays at cathode = Total faradays at anode
2Fcathode1mol ofH2.
10x F at cathode 12F×10xF=5xmolofH2
Ratio=Molesof ,H2 at,cathodeMolesofH2S2O8at anode=5x3x=53
Number of moles of H2=3×53=5

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