If `E_(CIO_(3)^(-)//ClO_(4)^(-))^(@) = -0.36V` and `E_(ClO_(3)^(-)//ClO_(2)^(-))^(@)= 0.33V` at 300 K . Find the equilibrium constant for the following reaction .
`2ClO_(3)^(-)to ClO_(2)^(-)_ClO_(4)^(-)`

Options

A. `(0.04)`
B. `(0.001)`
C. `(1.0)`
D. `(4.0)`

Explanation

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ 2 \text{ClO}_3^{-} \rightarrow \text{ClO}_2^{-} + \text{ClO}_4^{-} \]

Given:
- \( E^\circ_{\text{ClO}_3^{-} // \text{ClO}_4^{-}} = -0.36 \, \text{V} \)
- \( E^\circ_{\text{ClO}_3^{-} // \text{ClO}_2^{-}} = 0.33 \, \text{V} \)

### Step 1: Write the half-reactions

1. **Oxidation half-reaction** (for \( \text{ClO}_3^{-} \) to \( \text{ClO}_4^{-} \)):
\[ \text{ClO}_3^{-} \rightarrow \text{ClO}_4^{-} + 2e^{-} \quad (E^\circ = -0.36 \, \text{V}) \]

2. **Reduction half-reaction** (for \( \text{ClO}_3^{-} \) to \( \text{ClO}_2^{-} \)):
\[ \text{ClO}_3^{-} + e^{-} \rightarrow \text{ClO}_2^{-} \quad (E^\circ = 0.33 \, \text{V}) \]

### Step 2: Adjust the half-reactions to have the same number of electrons

To combine the two half-reactions, we need to multiply the reduction half-reaction by 2 so that the number of electrons cancels out:

\[ 2 \text{ClO}_3^{-} + 2e^{-} \rightarrow 2 \text{ClO}_2^{-} \quad (E^\circ = 0.33 \, \text{V}) \]

Now, we have:

1. Oxidation:
\[ \text{ClO}_3^{-} \rightarrow \text{ClO}_4^{-} + 2e^{-} \quad (E^\circ = -0.36 \, \text{V}) \]

2. Adjusted Reduction:
\[ 2 \text{ClO}_3^{-} + 2e^{-} \rightarrow 2 \text{ClO}_2^{-} \quad (E^\circ = 0.33 \, \text{V}) \]

### Step 3: Add the half-reactions

Now, we can add the two half-reactions:

\[ 2 \text{ClO}_3^{-} \rightarrow \text{ClO}_4^{-} + \text{ClO}_2^{-} \]

### Step 4: Calculate the overall standard cell potential \( E^\circ \)

The overall standard cell potential \( E^\circ \) is calculated by adding the standard potentials of the half-reactions:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} = 0.33 \, \text{V} + (-0.36 \, \text{V}) = 0.33 - 0.36 = -0.03 \, \text{V} \]

### Step 5: Use the Nernst equation to find the equilibrium constant \( K \)

At equilibrium, the cell potential \( E \) is 0. Therefore, we can use the Nernst equation:

\[ E = E^\circ - \frac{0.0591}{n} \log K \]

Where \( n \) is the number of electrons transferred. In this case, \( n = 2 \):

\[ 0 = -0.03 - \frac{0.0591}{2} \log K \]

Rearranging gives:

\[ 0.03 = \frac{0.0591}{2} \log K \]

### Step 6: Solve for \( \log K \)

\[ \log K = \frac{0.03 \times 2}{0.0591} = \frac{0.06}{0.0591} \approx 1.014 \]

### Step 7: Calculate \( K \)

Taking the antilog:

\[ K \approx 10^{1.014} \approx 10.26 \]

### Final Answer

The equilibrium constant \( K \) for the reaction \( 2 \text{ClO}_3^{-} \rightarrow \text{ClO}_2^{-} + \text{ClO}_4^{-} \) is approximately **10.26**.

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