An acidic solution of Cu^(2+) salt containing 0.4 g of Cu^(2+) is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of the solution kept at 100 mL and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.

An acidic solution of Cu^(2+) salt containing 0.4 g of Cu^(2+) is elec

1.	An acidic solution of Cu^(2+) salt containing 0.4 g of Cu^(2+) is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of the solution kept at 100 mL and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.
Answer» Solution :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.