An acidic solution of `Cu^(2+)` salt contaning `0.4` of `Cu^(2+)` is electrolyzed untill 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 `1.2` amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.

Explanation

For I part of electrolysis
Anode : `2H_(2)Orarr 4H^(-)+O_(2)+4e`
Cathode : `Cu^(2+)+2e rarr Cu`
`therefore` Eq. of `O_(2)` formed = Eq. of Cu
`=(0.4xx2)/(63.6)=12.58xx10^(-3)`
For II part of electrolysis : Since `Cu^(+)` ions are discharged completely and thus further passage of current throufgh solution will lead the following changes :
Anode : `2H_(2)O rarr 4H^(+)+O_(2)+4e^(-)`
Cathod : `2H_(2)O+2e^(-)rarr H_(2)+2OH^(-)`
Eq. of `H_(2) `= Eq. of `O_(2) = (i.t)/(96500)`
`=(1.2xx7xx60)/(96500)=5.22xx10^(-3)`
`therefore` Total Eq. of `O_(2)`
`=5.22xx10^(-3)+12.58xx10^(-3)=17.8xx10^(-3)`
`because 4Eq. O_(2)` at N.T.P = 22.4 litre
`therefore 17.8xx10^(-3)Eq. O_(2)` at N.T.P
`=(22.4xx17.8xx10^(-3))/(4)` litre = 9968 mL
Eq. of `H_(2) = 5.22xx10^(-3)`
`because 2` Eq. of `H_(2)` at N.T.P `= 22.4` litre
`therefore 5.22xx10^(-3) "Eq. at N.T.P"
`=(22.4xx5.22xx10^(-3))/(2)` litre `=58.46mL`
`therefore` Total volume of `O_(2)+H_(2)=99.68+58.46=158.14mL`
Note: If `Cu^(2+)` is as `CuCl_(2)`, then `Cl_(2)` will come out in I step and `H_(2)` and `O_(2)` in II step. Calculate volumes.