Which of the following reaction is correct for a given electrochemical cell at `25^(@)C`?
`(PtBr_(2)(g)Br^()(g)||Cl^(-)(aq) Cl_(2) (g)pt`

Options

A. `2Br^(-) (aq) + C1_(2) ((g))rightarrow2Cl^(-) (aq) + Br_(2) (g)`
B. `(Br(g)) + 2C1^(-)(aq)rightarrow 2Br^(-) (aq) + Cl_(2) (g)`
C. `Br_(2)(g)+ C1_(2) (aq) rightarrow Br^(-) (aq) + 2C1 ^(-) (g)`
D. `2Br^(-)((aq)) + 2C1^(-) ((aq)) to Br_(2) ((g)) + C1_(2) ((g))`

Explanation

To determine the correct reaction for the given electrochemical cell at \(25^\circ C\), we need to analyze the half-reactions involved and their respective standard reduction potentials.

### Step 1: Identify the Half-Reactions
The electrochemical cell is represented as:
\[ \text{Pt} | \text{Br}_2(g) | \text{Br}^-(aq) || \text{Cl}^-(aq) | \text{Cl}_2(g) | \text{Pt} \]
From this, we can identify the half-reactions:
1. For bromine:
\[ \text{Br}^-(aq) \rightarrow \text{Br}_2(g) + 2e^- \]
2. For chlorine:
\[ \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \]

### Step 2: Determine Standard Reduction Potentials
Next, we need to look up the standard reduction potentials (\(E^\circ\)) for these half-reactions:
- For chlorine (\(Cl_2\)):
\[ E^\circ = +1.36 \, V \]
- For bromine (\(Br_2\)):
\[ E^\circ = +1.09 \, V \]

### Step 3: Identify the Cathode and Anode
In an electrochemical cell:
- The cathode is where reduction occurs (gain of electrons).
- The anode is where oxidation occurs (loss of electrons).

Since chlorine has a higher reduction potential than bromine, chlorine will be reduced at the cathode:
\[ \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \]
And bromine will be oxidized at the anode:
\[ 2\text{Br}^-(aq) \rightarrow \text{Br}_2(g) + 2e^- \]

### Step 4: Combine the Half-Reactions
Now, we can combine the half-reactions to find the overall reaction:
\[ 2\text{Br}^-(aq) + \text{Cl}_2(g) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(g) \]

### Final Reaction
The overall balanced reaction for the electrochemical cell is:
\[ 2\text{Br}^-(aq) + \text{Cl}_2(g) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(g) \]

### Conclusion
Thus, the correct reaction for the given electrochemical cell at \(25^\circ C\) is:
\[ \text{2Br}^-(aq) + \text{Cl}_2(g) \rightarrow \text{2Cl}^-(aq) + \text{Br}_2(g) \]

---