During an electrolysis of conc `H_(2)SO_(4)` , perdisulphuric acid `(H_(2)S_(2)_(8))` and `O_(2)` are formed in equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be
`a. `Thrice that of `O_(2)" "b.` Twice that of `O_(2)`
`c.` Equal to that of `O_(2)." "d.` Half of that of `O_(2)`

Explanation

`a.` This is a special case of electrolysis where two products are being obtained at anode .
At anode `:`
`4 overset(-)(O)H rarrO_(2)+2H_(2)O+4e^(c-) ,,,,,,,...(i)`
`2SO_(4)^(2-) rarr S_(2)O_(8)^(2-)+2e^(-)" "....(ii)`
`1 mol O_(2)` requires `4F` electricity and `1 mol S_(2)O_(8)^(2-)(-=H_(2)S_(2)O_(8))` requires `2F` electricity.
So, if `x mol `of `O_(2)` are being produced, electricity being passed at anode is `:`
`4x( f o r O_(2))+2x(3 f o r S_(2)O_(8)^(2-))=6xF`
At cathode `:`
`2H^(o+)+2e^(-) rarr H_(2)" "......(iii)`
`2F` electricity `-=1 mol H_(2)` is produced
`implies 6xF` electricity `-=1 mol H_(2)` is produced
`implies ` Moles of `H_(2)` produced at cathode `=3 mol `of `O_(2)` produced at anode.