Match List I with List II: List I List II A. aq. NaCl solution with P…

The correct answer is option 4. A-III, B-I, C-IV, D-II.

List I	List II
A. aq. \(NaCl\) solution with \(Pt\) electrodes	III. Hydrogen gas evolved at cathode
B. Dilute solution of \(H_2SO_4\) with \(Pt\) electrodes	I. Oxygen gas evolved at anode
C. aq. \(CuCl_2\) with \(Pt\) electrode	IV. Metal deposited at cathode
D. aq. \(CuCl_2\) with \(Cu\) electrodes	II. Metal loss at anode

Let us dive into the electrochemical processes involved in each case to better understand the matching of List I with List II.

A. Aqueous \( \text{NaCl} \) Solution with \( \text{Pt} \) Electrodes

Electrolysis Process:

Dissociation in Water:

Sodium chloride (\( \text{NaCl} \)) dissociates into sodium ions (\( \text{Na}^+ \)) and chloride ions (\( \text{Cl}^- \)) in water:

\(\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-\)

At the Cathode (Negative Electrode):

The cathode attracts cations, so \( \text{Na}^+ \) ions and \( \text{H}_2\text{O} \) molecules will move toward the cathode.

However, sodium ions (\( \text{Na}^+ \)) are not reduced because water is more easily reduced. The reduction of water occurs:

\(2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-\)

Result: Hydrogen gas (\( \text{H}_2 \)) is evolved at the cathode, and hydroxide ions (\( \text{OH}^- \)) remain in solution.

At the Anode (Positive Electrode):

The anode attracts anions, so \( \text{Cl}^- \) ions will move toward the anode.

Chloride ions are oxidized to form chlorine gas (\( \text{Cl}_2 \)):

\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)

Result: Chlorine gas (\( \text{Cl}_2 \)) is evolved at the anode.

Conclusion:

Hydrogen gas is evolved at the cathode.

Correct Match:III (Hydrogen gas evolved at cathode)

B. Dilute Solution of \( \text{H}_2\text{SO}_4 \) with \( \text{Pt} \) Electrodes

Electrolysis Process:

Dissociation in Water:

Sulfuric acid (\( \text{H}_2\text{SO}_4 \)) dissociates into hydrogen ions (\( \text{H}^+ \)) and sulfate ions (\( \text{SO}_4^{2-} \)):

\(\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}\)

At the Cathode (Negative Electrode):

The cathode attracts \( \text{H}^+ \) ions. These hydrogen ions are reduced to form hydrogen gas:

\(2\text{H}^+ + 2e^- \rightarrow \text{H}_2\)

Result: Hydrogen gas (\( \text{H}_2 \)) is evolved at the cathode.

At the Anode (Positive Electrode):

The anode attracts \( \text{SO}_4^{2-} \) ions, but sulfate ions are not easily oxidized. Instead, water molecules (\( \text{H}_2\text{O} \)) are oxidized:

\(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-\)

Result: Oxygen gas (\( \text{O}_2 \)) is evolved at the anode.

Conclusion:

Oxygen gas is evolved at the anode. Correct Match: I (Oxygen gas evolved at anode)

C. Aqueous \( \text{CuCl}_2 \) with \( \text{Pt} \) Electrodes

Electrolysis Process:

Dissociation in Water:

Copper(II) chloride (\( \text{CuCl}_2 \)) dissociates into copper ions (\( \text{Cu}^{2+} \)) and chloride ions (\( \text{Cl}^- \)):

\(\text{CuCl}_2 \rightarrow \text{Cu}^{2+} + 2\text{Cl}^-\)

At the Cathode (Negative Electrode):

The cathode attracts \( \text{Cu}^{2+} \) ions. These copper ions are reduced to form solid copper:

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

Result: Copper metal is deposited at the cathode.

At the Anode (Positive Electrode):

The anode attracts \( \text{Cl}^- \) ions. These chloride ions are oxidized to form chlorine gas:

\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)

Result: Chlorine gas (\( \text{Cl}_2 \)) is evolved at the anode.

Conclusion:

Metal (copper) is deposited at the cathode. Correct Match: IV (Metal deposited at cathode)

D. Aqueous \( \text{CuCl}_2 \) with \( \text{Cu} \) Electrodes

Electrolysis Process:

Dissociation in Water:

Copper(II) chloride (\( \text{CuCl}_2 \)) dissociates into copper ions (\( \text{Cu}^{2+} \)) and chloride ions (\( \text{Cl}^- \)):

\(\text{CuCl}_2 \rightarrow \text{Cu}^{2+} + 2\text{Cl}^-\)

At the Cathode (Negative Electrode):

The cathode attracts \( \text{Cu}^{2+} \) ions. These copper ions are reduced to form solid copper:

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

Result: Copper metal is deposited at the cathode.

At the Anode (Positive Electrode):

Here, the anode is made of copper metal. Copper atoms from the anode are oxidized to form \( \text{Cu}^{2+} \) ions, which go into the solution:

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

Result: The copper anode loses metal as it dissolves into the solution.

Conclusion:

Metal loss occurs at the anode as copper dissolves into the solution. Correct Match: II (Metal loss at anode).

Thus, the correct answer is Option 4.