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

Let us suppose that the acidic solution of `Cu^(2+)` salt contains `H_(2)SO_(4)`.
In the beginning of electrolysis, Cu will be deposited at the cathode and `O_(2)` will discharged at anode.
`Cu^(2+) + 2e rarr Cu` (at cathode)
`{:(2OH^(-) rarr 2OH + 2e),(2OH rarr H_(2)O + (1)/(2)O_(2)):}}` (at anode)
`therefore` equivalent of oxygen evolved = eq. of Cu deposited
`= (0.4)/(31.8)`
`therefore` volume of `O_(2)` (at NTP) evolved `= (0.4)/(31.8) xx 5600 = 70.44 mL`.
(1 eq. of oxygen at NTP = 5600 mL)
During another even minutes (420 s) of electrolysis, `H_(2)` and `O_(2)` will evolve at cathode and anode respectively.
Now, charge `= 1.2 xx 420 = 504` coulombs `= (504)/(96500)F`.
`therefore` eq. of `H_(2)` evolved `= (504)/(96500)`. (at cathode)
Volume of `H_(2)` evolved at NTP `= (504)/(96500) xx 11200 = 58.49 mL`.
(1 eq. of `H_(2)` at NTP = 1200 mL)
Eq. of `O_(2)` evolved `= (504)/(96500)` (at anode)
`therefore` volume of `O_(2)` evolved at NTP `= (504)/(96500) xx 5600 = 29.24 mL`.
Thus, during the entire electrolysis,
`H_(2)` evolved = 58.49 mL
`O_(2)` evolved = (70.44 + 29.24) mL
= 99.68 mL.