Statement I: At the end of electrolysis using platinum electrodes an aqueous solution of copper sulphate turns colourless

Statement II: Copper in CuSO₄ is converted to Cu(OH)₂ during electrolysis

Statement I is correct but Statement II is false

The answer is (3) Statement I is correct but Statement II is false.

Statement I is correct because copper sulfate solution turns colorless at the end of electrolysis using platinum electrodes. This is because the copper ions in the solution are reduced to copper metal, which is a colorless substance.

Statement II is false because the electrode potential decreases on dilution of the solution. This is because the concentration of the ions in the solution decreases, which makes it more difficult for the electrons to be transferred from the electrode to the ions.

The correct answer is 3.

Here is a more detailed explanation of Statement I:

When an aqueous solution of copper sulfate is electrolyzed using platinum electrodes, the copper ions in the solution are reduced to copper metal at the cathode. The reduction reaction is:

Cu²⁺ + 2e^- → Cu

The copper metal that is produced at the cathode is deposited on the electrode, and this causes the solution to turn colorless.

Here is a more detailed explanation of Statement II:

The electrode potential is the potential difference between an electrode and a reference electrode. The electrode potential depends on the concentration of the ions in the solution, the type of electrode, and the temperature.

On dilution of the solution, the concentration of the ions in the solution decreases. This makes it more difficult for the electrons to be transferred from the electrode to the ions, and this causes the electrode potential to decrease.

Therefore, Statement I is correct but Statement II is false.