An acidic solution of `Cu^(2+0` 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.

Answer

`158.14mL`

Explanation

For first of electolytes `:`
At anode `:2H_(2)O rarr 4H^(o+)+O_(2)+4e^(-)`
At cathode`:Cu^(2+)+2e^(-) rarr Cu`
`:.` Equivalent of `O_(2)` formed `=` equivalent of `Cu`
`=(0.4xx2)/(63.5)=12.58xx10^(-3)`
For second part of electrolysis `:`
Since `Cu^(2+)` ions are discharged completely and thus further passage of current through solution will lead to following changes `:`
At anode `:2H_(2)O rarr 4H^(o+) +O_(2)+4e^(-)`
At cathode `:2H_(2)O+2e^(-) rarr H_(2)+2overset(c-)(O)H`
Thus, equivalent of `H_(2)=` equivalent of `O_(2)`
`=(It)/(96500)`
`=(1.2xx7xx60)/(96500)=5.22xx10^(-3)`
`:. ` Total equivalent of `O_(2)=` Equivalent of `O_(2)` for first part `+` Equivalent of `O_(2)` for second part of electrolysis
`=5.22xx10^(-3)+12.58xx10^(-3)`
`=17.8xx10^(-3)`
`:.` 4 equivalents of `O_(2)` at `STP=22.4L` ltbr. `:. 17.8xx10^(-3)` equivalent of `O_(2)` at `STP`
`=(22.4xx17.8xx10^(-3))/(4)=99.68mL`
Now , equivalent of `H_(2)=5.22xx10^(-3)`
`:' 2` equivalent of `H_(2)` at `STP`
`=(22.4xx5.22xx10^(-3))/(2)=58.46mL`
`:. ` Total volume of `O_(2)+H_(2)=99.68+58.46`
`=158.14mL`