Calculate the equilibrium constant for the following cell reaction.
`(1)/(2)Cl_(2) (g) +Br^(-) to (1)/(2) Br_(2) (l) +Cl^(-) , E^(Theta) =0.30V`

Explanation

To calculate the equilibrium constant (K) for the given cell reaction:

\[ \frac{1}{2} \text{Cl}_2 (g) + \text{Br}^- \rightarrow \frac{1}{2} \text{Br}_2 (l) + \text{Cl}^- \]

with a standard cell potential \( E^\Theta = 0.30 \, \text{V} \), we can follow these steps:

### Step 1: Identify the Reaction and the Standard Cell Potential
We have the cell reaction and the standard cell potential given as \( E^\Theta = 0.30 \, \text{V} \).

### Step 2: Use the Nernst Equation
The Nernst equation relates the cell potential to the equilibrium constant:

\[ E_{\text{cell}} = E^\Theta_{\text{cell}} - \frac{0.0591}{n} \log K \]

At equilibrium, \( E_{\text{cell}} = 0 \). Thus, we can rewrite the equation as:

\[ 0 = E^\Theta_{\text{cell}} - \frac{0.0591}{n} \log K \]

### Step 3: Rearrange the Equation
Rearranging gives us:

\[ \frac{0.0591}{n} \log K = E^\Theta_{\text{cell}} \]

### Step 4: Solve for \( \log K \)
Now we can solve for \( \log K \):

\[ \log K = \frac{n \cdot E^\Theta_{\text{cell}}}{0.0591} \]

### Step 5: Determine the Number of Electrons Transferred (n)
In the given reaction, we can see that one mole of electrons is transferred (as the oxidation state of chlorine changes from 0 in \( \text{Cl}_2 \) to -1 in \( \text{Cl}^- \) and bromine changes from -1 to 0). Therefore, \( n = 1 \).

### Step 6: Substitute Values into the Equation
Substituting the values into the equation:

\[ \log K = \frac{1 \cdot 0.30}{0.0591} \]

### Step 7: Calculate \( \log K \)
Calculating this gives:

\[ \log K = \frac{0.30}{0.0591} \approx 5.07 \]

### Step 8: Calculate \( K \)
Now, to find \( K \), we take the antilog:

\[ K = 10^{5.07} \approx 1.17 \times 10^5 \]

### Conclusion
Thus, the equilibrium constant \( K \) for the reaction is approximately:

\[ K \approx 1.17 \times 10^5 \]

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