During an electrolysis of conc. `H_(2)SO_(4)`, peroxydisulphuric acid `(H_(2)S_(2)O_(8))` and `O_(2)` form in an equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be

Options

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

Answer

A

Explanation

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`