ElectrochemistryHard
Question
During an electrolysis of conc. H2SO4, perdisulphuric acid (H2S2O8) and O2 form in equimolar amount. The amount of H2 that will form simultaneously at other electrode will be (2H2SO4 → H2S2O8 + 2H+ + 2e-)
Options
A.thrice that of O2 in moles
B.twice that of O2 in moles
C.equal to that of O2 in moles
D.half of that of O2 in moles
Solution
Anode 
Cathode {2H2O → H2 + 2OH- - 2e-} × 3.
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Net : 2H2SO4 + 8H2O → H2S2O8 + O2 + 3H2 + 6H+ + 6OH-
Hence ratio of and is 1 : 3.
Cathode {2H2O → H2 + 2OH- - 2e-} × 3.
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Net : 2H2SO4 + 8H2O → H2S2O8 + O2 + 3H2 + 6H+ + 6OH-
Hence ratio of and is 1 : 3.
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