Calculate the volume of gases liberated at anode and cathode at `NTP` from the electrolysis of `Na_(2)SO_(4)(aq.)` solution by a current of `2` ampere passed for `10` minute.

Explanation

At cathode: `2H_(2)O+2e rarr H_(2)+2OH^(-)`
At anode: `2H_(2)O rarr 4H^(+)+4e+O_(2)`
`:.` At anode `E_(O_(2)) = (32)/(4) = 8`
`:. W_(O_(2)) = (E.i.t)/(96500) = (32 xx 2 xx 10 xx 60)/(4 xx 96500) = 0.0995g`
At NTP : Volume of `O_(2) = (0.0995 xx 22.4)/(32)`
`= 0.0696` litre
Similarly at cathode `w_(H_(2))`
`= (E.i.T)/(96500) = (2 xx 2 xx 10 xx 60)/(2 xx 96500) = 0.0124g`
At NTP : Volume of `H_(2) = (0.0124 xx 22.4)/(2)`
`= 0.139` litre