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

If the salt is `CuSO_4`
During deposition of Cu at cathode, O2(g) will evolve at anode. Gram-equivalent of Cu deposited
`=(0.4xx2)/(63.5)=0.126`
Volume of O2 liberated at NTP at anode `=0.0126xx5600 mL=70.56 m L`
In the next 7 min `H_2` at cathode and `O_2` at anode would be produced
Faraday’s passed `=(1.2xx7xx60)/(96500)=5.22x10^(-3)`
`implies` Volume of `O_2` (at NTP) `=5.22xx10^(-3)xx11200 mL=58.46 mL`
volume of `O_2` (at NTP) `5.22 xx10^(-3)xx5600 mL=29.23 mL`
Therefore , `O_2 (g)` at NTP `=70.56+29.33=99.89`
`H_2(g)` at NTP `=58.46 mL`