When CuSO4 is electrolyzed using platinum electrodes, copper is liberated at the anode and oxygen is liberated at the cathode. This can be explained by the process of electrolysis and the reactions occurring at each electrode.

Electrolysis is the process in which an electric current is passed through an electrolyte, causing a chemical reaction to occur. In this case, the electrolyte is CuSO4, which is an aqueous solution of copper sulfate.

1. Anode reaction:
At the anode, oxidation takes place. The copper ions (Cu2+) in the CuSO4 solution are attracted to the positive electrode (anode). When they reach the anode, they lose two electrons each to form copper atoms (Cu). This reaction can be represented as:

Cu2+ + 2e- → Cu

As a result, copper metal is deposited on the platinum anode.

2. Cathode reaction:
At the cathode, reduction takes place. The water molecules (H2O) in the CuSO4 solution are attracted to the negative electrode (cathode). When they reach the cathode, they gain electrons and are reduced to form hydrogen gas (H2). This reaction can be represented as:

2H2O + 2e- → H2 + 2OH-

The hydroxide ions (OH-) formed during this reaction combine with the copper ions (Cu2+) in the solution to form copper hydroxide (Cu(OH)2). However, this copper hydroxide is unstable and quickly decomposes to form copper oxide (CuO) and water (H2O):

Cu(OH)2 → CuO + H2O

Overall, the electrolysis of CuSO4 using platinum electrodes results in the liberation of copper at the anode and oxygen at the cathode. The copper ions are reduced at the anode to form copper metal, while the water molecules are reduced at the cathode to form hydrogen gas.