# An acidic solution of Cu+2 salt containing 0.4g of Cu+2 is electrolysed untill all the copper is deposited. The electrolysis is continued for 7 more minutes with the volume of solution kept at 100ml and the current at 1.2 ampere. Calculate volume of gases evolved at NTP during the entire electrolysis. (Atomic mass of Cu = 63.5) .

Assuming Cu2+ salt to be copper sulphate, the reactions occurring at the electrodes would be:

Anode  H2O  ------> 2H+  + ½ O2  + 2e  Cathode Cu2+ + 2e  ----> Cu

After the complete deposition of copper, the reactions would be

Anode  H2O ---> 2H+  + ½ O2  + 2e  Cathode  2H2O  + 2e ----> H2  + 2OH-

Now Amount of Cu deposited = 0.4 g/63.6 g/mol = 0.00629 mol

Amount of Oxygen liberated = ½ x 0.00629 mol = 0.003145 mol

Quantity of electricity passed in seven minutes after the deposition of the entire

copper  = (1.2A)(7x60 s)  = 504 C

Amount of electrons carrying this much of electricity = 504 C/96500C/mol

= 0.00522 mol

From the electrode reactions which occur after the deposition of copper, we conclude

Amount of oxygen liberated = ¼ x 0.00522 mol = 0.001305 mol

Amount of hydrogen liberated = ½ x 0.00522 mol = 0.00261 mol

Total amount of gases liberated in the entire electrolysis

= (0.003145+0.001305+0.00261) mol

= 0.00706 mol

Volume of the gases evolved at STP durig the entire electrolysis

= (0.00706 mol) (22414 ml/mol)

= 158.2 ml

If the salt of Cu2+ ions is CuCl2, then the anodic reaction during the deposition of copper would be 2Cl-  ------> Cl2  + 2e

In this case, Total amount of gases liberated would be

= (0.00629 + 0.001305 + 0.00261) mol

= 0.010205 mol

Volume of the gases evolved would be = (0.010205 mol)(22414 ml/mol) = 228.7 ml

• -15
What are you looking for?