Match the column :
`{:("Column-I","Column-II"),((A)"Molten" PbCl_2 "using inert electrode", (p)"Metal of salt will reduced"),((B)"Sodium chloride solution using inert electrode ",(q)H_2O+2e to H_2(g)+2OH^(-)),((C )"Silver nitrate solution with silver electrode",(r)"Solution become basic after electrolysis"),((D)"Sodium nitrate solution using inert electrode",(s)"Solution become neutral after electrolysis "),(,(t)"Solution become acidic after electrolysis"):}`

Explanation

To solve the problem of matching the items in Column-I with those in Column-II, we will analyze each item in Column-I and determine the corresponding statements in Column-II based on the electrolysis reactions.

### Step-by-Step Solution:

1. **Analyzing Column-I:**
- **(A) Molten PbCl₂ using inert electrode**
- At the cathode: \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb (s)} \) (Lead is reduced)
- At the anode: \( 2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^- \) (Chloride ions are oxidized)
- The solution remains neutral after electrolysis as Pb is deposited and Cl₂ is released.
- **Matches with:** (p) "Metal of salt will reduce" and (s) "Solution becomes neutral after electrolysis."

2. **(B) Sodium chloride solution using inert electrode**
- At the cathode: \( \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^- \) (Water is reduced)
- At the anode: \( 2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^- \) (Chloride ions are oxidized)
- The solution becomes basic due to the formation of hydroxide ions.
- **Matches with:** (q) \( \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^- \) and (r) "Solution becomes basic after electrolysis."

3. **(C) Silver nitrate solution with silver electrode**
- At the cathode: \( \text{Ag}^+ + e^- \rightarrow \text{Ag (s)} \) (Silver ions are reduced)
- At the anode: \( \text{Ag (s)} \rightarrow \text{Ag}^+ + e^- \) (Silver is oxidized)
- The solution remains neutral as there are no additional ions produced.
- **Matches with:** (p) "Metal of salt will reduce" and (s) "Solution becomes neutral after electrolysis."

4. **(D) Sodium nitrate solution using inert electrode**
- At the cathode: \( \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^- \) (Water is reduced)
- At the anode: \( 2\text{H}_2\text{O} \rightarrow \text{O}_2(g) + 4\text{H}^+ + 4e^- \) (Water is oxidized)
- The solution becomes neutral as the number of hydroxide ions produced equals the number of hydrogen ions produced.
- **Matches with:** (q) \( \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^- \) and (s) "Solution becomes neutral after electrolysis."

### Final Matches:
- (A) → (p), (s)
- (B) → (q), (r)
- (C) → (p), (s)
- (D) → (q), (s)

### Summary of Matches:
- A → p, s
- B → q, r
- C → p, s
- D → q, s