An acidic solution of `Cu^(2+)` salt containing `0.4g` of `Cu^(2+)` is electrolyzed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at `100mL` and the current at `1.2A`. Calculate the volume of gases evolved at `STP` during the entire electrolysis.

Explanation

For I part of electrolysis
Anode: `2H_(2)O rarr 4H^(+)+O_(2)+4e`
Cathode: `Cu^(2+)+2e rarr Cu`
`:.` Eq. of `O_(2)` formed = Eq. of `Cu`
`= (0.4 xx 2)/(63.6) = 12.58 xx 10^(-3)`
For II part of electrolysis: Since `Cu^(+)` ions are discharged completely and thus further passage of current through solution will lead the following changes:
Anode : `2H_(2)O rarr 4H^(+)+O_(2)+4e`
Cathode: `2H_(2)O+2e rarr H_(2)+2OH^(-)`
`{:( :. "Total Eq. of "O_(2)),(=5.22xx10^(-3)+12.58xx10^(-3)),(=17.8xx10^(-3)),(because 4"Eq." O_(2) at N.T.P.),(=22.4 litre),( :. 17.8xx10^(-3) Eq. O_(2)at N.T.P.),(=(22.4xx17.8xx10^(-3))/4 litre),(=99.68 mL):}:|{:(Eq. of H_(2)=5.22xx10^(-3)),( :. 2 Eq. of H_(2) at N.T.P.),(=22.4 litre),( :. 5.22xx10^(-3) Eq. at),(N.T.P),(=(22.4xx5.22xx10^(-3))/2 litre),(=58.46 mL):}`
`:.` Total volume of `O_(2)+H_(2) = 99.68+58.46` = `158.14 mL`
Note: If `Cu^(2+)` is as `CuCl_(2)`, then `Cl_(@)` will come out in I step and `H_(2)` and `O_(2)` in II step. Calculate volumes.