During an electrolysis of conc. `H_(2)SO_(4)` perdisulphuric acid `(H_(2)S_(2)O_(6))` and `O_(2)` form in equimolar amount. The amount of `H_(2)` than

Monjur Alahi

During an electrolysis of conc. `H_(2)SO_(4)` perdisulphuric acid `(H_(2)S_(2)O_(6))` and `O_(2)` form in equimolar amount. The amount of `H_(2)` than will form simultaneously at other electrode will be `(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-))`
A. thrice that of `O_(2)` in moles
B. twice that of `O_(2)` in moles
C. equal to that of `O_(2)` in moles
D. half of the of `O_(2)` in moles


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Salim Ahmed Kawsar

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Correct Answer - A
Anode `{{:(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-)),(2H_(2)OtoO_(2)+4H^(+)+4e^(-)):}}`
`underline(" Cathode")" "{2H_(2)OtoH_(2)+2OH^(-)-2e^(-)}3)`.
Net: `2H_(2)SO_(4)+8H_(2)OtoH_(2)S_(2)O_(8)+O_(2)+3H_(2)+6H^(+)+6OH^(-)`
Hence ratio of `n_(O_(2))` and `n_(H_(2))` is `1:3`

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