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_2O + 2e rarr H_2+2OH^-`
At anode : `2H_2O rarr 4H^++4e+O_2`
`:.`At anode `E_(O_2)=32/4=8`
`"w"_(O_2)=(E."i"."t")/96500=(32xx2xx10xx60)/(4xx96500)`
=0.0995g
At NTP : Volume of `O_2` = `(0.0995xx22.4)/32`= 0.0696 litre
Similarly at cathode `"w"_(H_2)=(E."i"."t")/96500`
`(2xx2xx10xx60)/(2xx96500)`= 0.0124g At NTP : Volume Of `H_2` =`(0.0124xx22.4)/2`= 0.139 litre