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.

Answer

`V(O_(2)) = 99.68mL V(H_(2)) = 58.46 mL`, Total vol `=158.1 mL`

Explanation

In beginning
At anode: `2e^(-) +Cu^(+2) rarr Cu`
At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)`
Eq. of `O_(2) = Eq. of Cu^(+2)`
`=(0.4 xx 2)/(63.5) = 0.012598`
vol of `O_(2) = (0.012598 xx 22400)/(4) = 70.55mL`
Later on:
At anode:
`2H_(2)O +2e^(-) rarr H_(2) +2OH^(-)`
At anode:
`2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)`
Faraday passed
`= (1.2 xx 7 xx 60)/(96500) = 0.00522`
moles of `O_(2) = (0.00522)/(4)`
moles of `H_(2) = (0.00522)/(2)`
Volume of `O_(2) = (0.00522)/(4) xx 22400 = 29.24mL`
Volume of `H_(2) = (0.00522)/(2) xx 22400 = 58.464 mL`
Total volume of `O_(2)`
`= 29.24 + 70.55 = 99.79 mL`
Total volume of `H_(2)`
`= 58.464 mL`