When CuSO_4 solution is electrolysed using platinum electrodes, which…

When $CuSO_4$ solution is electrolysed using platinum electrodes, which of the following is true?

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → Copper is liberated at cathode and oxygen at anode

When $CuSO_4$ (copper(II) sulfate) solution is electrolyzed using platinum electrodes, the electrolysis involves both oxidation and reduction reactions occurring at different electrodes. Here is a detailed breakdown of the process:

Electrolysis is a process where electrical energy is used to drive a non-spontaneous chemical reaction. In an electrolytic cell, two electrodes—cathode (negative) and anode (positive)—are immersed in an electrolyte solution. The electrolyte contains ions that participate in the redox (reduction-oxidation) reactions.

Composition of the Electrolyte

In $CuSO_4$ solution, the main ions present are:

$Cu^{2+}$ ions (copper ions)

$SO_4^{2-}$ ions (sulfate ions)

Water molecules ($H_2O$) which can also dissociate into $H^+$ and $OH^-$ ions.

Reactions at the Electrodes

At the Cathode (Reduction Reaction)

Reduction of Copper Ions:

At the cathode, where reduction occurs, $Cu^{2+}$ ions gain electrons to form solid copper.

The half-reaction can be written as:

$Cu^{2+} + 2e^- \rightarrow Cu(s)$

This reaction results in the deposition of copper metal on the cathode.

At the Anode (Oxidation Reaction)

Oxidation of Water:

At the anode, where oxidation occurs, water molecules are oxidized to produce oxygen gas and protons ($H^+$).

The half-reaction can be written as:

$2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$

This results in the release of oxygen gas at the anode.

Summary of Products

At the Cathode: Copper metal is deposited.

At the Anode: Oxygen gas is evolved.

Conclusion

In summary, during the electrolysis of $CuSO_4$ solution using platinum electrodes:

Copper is liberated at the cathode, where $Cu^{2+}$ ions are reduced to solid copper.

Oxygen is liberated at the anode, where water is oxidized.

Thus, the correct statement is: Copper is liberated at the cathode and oxygen at the anode.

This process is important in electrochemistry and has practical applications in electroplating and the purification of metals

When $CuSO_4$ solution is electrolysed using platinum electrodes, which of the following is true?

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → Copper is liberated at cathode and oxygen at anode

Composition of the Electrolyte

In $CuSO_4$ solution, the main ions present are:

$Cu^{2+}$ ions (copper ions)

$SO_4^{2-}$ ions (sulfate ions)

Water molecules ($H_2O$) which can also dissociate into $H^+$ and $OH^-$ ions.

Reactions at the Electrodes

At the Cathode (Reduction Reaction)

Reduction of Copper Ions:

At the cathode, where reduction occurs, $Cu^{2+}$ ions gain electrons to form solid copper.

The half-reaction can be written as:

$Cu^{2+} + 2e^- \rightarrow Cu(s)$

This reaction results in the deposition of copper metal on the cathode.

At the Anode (Oxidation Reaction)

Oxidation of Water:

At the anode, where oxidation occurs, water molecules are oxidized to produce oxygen gas and protons ($H^+$).

The half-reaction can be written as:

$2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$

This results in the release of oxygen gas at the anode.

Summary of Products

At the Cathode: Copper metal is deposited.

At the Anode: Oxygen gas is evolved.

Conclusion

In summary, during the electrolysis of $CuSO_4$ solution using platinum electrodes:

Copper is liberated at the cathode, where $Cu^{2+}$ ions are reduced to solid copper.

Oxygen is liberated at the anode, where water is oxidized.

Thus, the correct statement is: Copper is liberated at the cathode and oxygen at the anode.

This process is important in electrochemistry and has practical applications in electroplating and the purification of metals