When $CuSO_4$ solution is electrolysed using platinum electrodes, which of the following is true? |
Copper is liberated at cathode and sulphur at anode Sulphur is liberated at cathode and copper at anode Oxygen is liberated at cathode and copper at anode Copper is liberated at cathode and oxygen at anode |
Copper is liberated at cathode and oxygen at anode |
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 |