In the electrolysis of CuCl₂ solution using Cu electrodes, the weight of Cu anode increased by 2 g at the cathode. In the anode

2 g of copper goes into solution as Cu²⁺

The correct answer is option 4. 2 g of copper goes into solution as \(Cu^{2+}\).

Let us analyze the electrolysis of \(CuCl_2\) solution using \(Cu\) electrodes based on the information provided:

Given Data:

Electrolysis using \(Cu\) electrodes.

Weight of \(Cu\) anode increased by \(2 g\).

\(CuCl_2\) solution is used.

Analysis:

Weight of \(Cu\) anode increased by 2 g:

This indicates that copper ions \((Cu^{2+})\) from the \(CuCl_2\) solution are being deposited onto the anode (\(Cu\) electrode), causing its weight to increase.

In the anode:

The anode is where oxidation occurs during electrolysis. Here, \(Cu(s)\) from the electrode is oxidized to \(Cu^{2+}\) ions that enter the solution.

Options Analysis:

1. \(0.2\) mole of \(Cu^{2+}\) will go into solution: To determine this, we can convert the increase in weight of the anode (2 g) into moles of \(Cu^{2+}\) ions. The molar mass of \(Cu\) is approximately \(63.55 g/mol\).

2. 560 mL \(O_2\) liberated: This is typically associated with the cathode during electrolysis when water is present, but \(CuCl_2\) electrolysis primarily involves \(Cu^{2+}\) and \(Cu^{2+}\) ions.

3. No loss in weight: This contradicts the given increase in weight of the anode (2 g).

4. 2 g of copper goes into solution as \(Cu^{2+}\): This directly correlates with the 2 g increase in weight of the anode, where Cu(s) is oxidized to \(Cu^{2+}\) ions.

Based on the information provided and the principles of electrolysis: 2 g of copper goes into solution as \(Cu^{2+}\).

This statement correctly explains the increase in weight of the anode during the electrolysis of \(CuCl_2\) solution using \(Cu\) electrodes. It aligns with the oxidation process occurring at the anode, where \(Cu(s)\) loses electrons to form \(Cu^{2+}\) ions in solution. Therefore, the correct answer is that 2 g of copper goes into solution as \(Cu^{2+}\).