Given : `E_(cu^(+2)//cu)^0=0.34 V`
`E_(Cl_2//Cl^(-))^0`=1.36 V
`E_(Br_2//Br^(-))^0=1.08 V`
`E_(l_2//l^(-))^0=0.54 V`
`{:("Column-I","Column-II"),((A)Cu^(+2)+2Cl to Cu+Cl_2, (p)"Can produce electricity in the galvanic cell"),((B)Cl_2+CutoCu^(+2)+2Cl^(-),(q)"Can be made to occur in electrolysis cell"),((C )2I^(-)+"starch solution + chlorine water",(r)"Appearance of brown colour"),((D)2Br^(-)+C Cl_4+"Chlorine water",(s)"Appearance of violet colour"):}`

Explanation

To solve the problem, we need to match the reactions from Column I with the corresponding descriptions in Column II based on the standard electrode potentials provided. Let's analyze each reaction step by step.

### Step 1: Analyze Reaction A
**Reaction:** \( \text{Cu}^{2+} + 2\text{Cl}^- \rightarrow \text{Cu} + \text{Cl}_2 \)

- **Standard Electrode Potentials:**
- \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \)
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \)

- **Analysis:**
- Since \( E^\circ_{\text{Cl}_2/\text{Cl}^-} > E^\circ_{\text{Cu}^{2+}/\text{Cu}} \), Cl2 will be reduced and Cu will be oxidized.
- This reaction cannot produce electricity in a galvanic cell but can be made to occur in an electrolysis cell.

**Match:** A → Q

### Step 2: Analyze Reaction B
**Reaction:** \( \text{Cl}_2 + \text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{Cl}^- \)

- **Standard Electrode Potentials:**
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \)
- \( E^\circ_{\text{Cu}/\text{Cu}^{2+}} = -0.34 \, \text{V} \) (reverse of Cu reduction)

- **Analysis:**
- Here, Cl2 is reduced and Cu is oxidized, which means this reaction can produce electricity in a galvanic cell.

**Match:** B → P

### Step 3: Analyze Reaction C
**Reaction:** \( 2\text{I}^- + \text{starch solution} + \text{chlorine water} \)

- **Standard Electrode Potentials:**
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \)
- \( E^\circ_{\text{I}_2/\text{I}^-} = 0.54 \, \text{V} \)

- **Analysis:**
- Cl2 will be reduced to Cl^- and I^- will be oxidized to I2, resulting in the appearance of a violet color due to I2 formation.

**Match:** C → S

### Step 4: Analyze Reaction D
**Reaction:** \( 2\text{Br}^- + \text{CCl}_4 + \text{chlorine water} \)

- **Standard Electrode Potentials:**
- \( E^\circ_{\text{Br}_2/\text{Br}^-} = 1.08 \, \text{V} \)
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \)

- **Analysis:**
- Cl2 will be reduced to Cl^- and Br^- will be oxidized to Br2, which has a reddish-brown color.

**Match:** D → R

### Final Matches:
- A → Q (Can be made to occur in an electrolysis cell)
- B → P (Can produce electricity in the galvanic cell)
- C → S (Appearance of violet color)
- D → R (Appearance of brown color)