To solve the problem step-by-step, we will analyze the given reaction and the equilibrium concentrations involved.
### Step 1: Write the equilibrium reaction
The equilibrium reaction given is:
\[ 2 \text{ClO}_3^{-} \rightleftharpoons \text{ClO}_2^{-} + \text{ClO}_4^{-} \]
### Step 2: Define initial and equilibrium concentrations
Let the initial concentration of \(\text{ClO}_3^{-}\) be 1.0 M, and the initial concentrations of \(\text{ClO}_2^{-}\) and \(\text{ClO}_4^{-}\) be 0 M.
At equilibrium, if \(x\) is the change in concentration of \(\text{ClO}_2^{-}\) and \(\text{ClO}_4^{-}\), we can express the equilibrium concentrations as:
- \([\text{ClO}_3^{-}] = 1 - 2x\)
- \([\text{ClO}_2^{-}] = x\)
- \([\text{ClO}_4^{-}] = x\)
### Step 3: Write the expression for the equilibrium constant \(K\)
The equilibrium constant \(K\) for the reaction can be expressed as:
\[
K = \frac{[\text{ClO}_2^{-}][\text{ClO}_4^{-}]}{[\text{ClO}_3^{-}]^2} = \frac{x \cdot x}{(1 - 2x)^2} = \frac{x^2}{(1 - 2x)^2}
\]
### Step 4: Calculate the standard cell potential \(E^0_{cell}\)
From the problem, we have:
- \(E_{\text{ClO}_3^{-}/\text{ClO}_4^{-}} = -0.36 \, \text{V}\)
- \(E_{\text{ClO}_3^{-}/\text{ClO}_2^{-}}^{\circ} = 0.33 \, \text{V}\)
For the reaction, the oxidation potential at the anode is:
\[
E_{\text{anode}} = -E_{\text{ClO}_3^{-}/\text{ClO}_4^{-}} = +0.36 \, \text{V}
\]
The cell potential \(E^0_{cell}\) can be calculated as:
\[
E^0_{cell} = E_{\text{cathode}} - E_{\text{anode}} = 0.33 - 0.36 = -0.03 \, \text{V}
\]
### Step 5: Relate \(E^0_{cell}\) to the equilibrium constant \(K\)
Using the relationship:
\[
\Delta G^0 = -nFE^0_{cell} \quad \text{and} \quad \Delta G^0 = -RT \ln K
\]
we can equate:
\[
-nFE^0_{cell} = -RT \ln K
\]
Substituting \(n = 2\) (since 2 moles of electrons are transferred), \(F = 96500 \, \text{C/mol}\), \(R = 8.314 \, \text{J/(mol K)}\), and \(T = 300 \, \text{K}\):
\[
2 \cdot 96500 \cdot (-0.03) = -8.314 \cdot 300 \cdot \ln K
\]
### Step 6: Solve for \(K\)
Calculating the left side:
\[
-5790 = -2494.2 \ln K
\]
Thus,
\[
\ln K = \frac{5790}{2494.2} \approx 2.32
\]
Taking the antilog:
\[
K \approx e^{2.32} \approx 10^{0.509} \approx 3.329
\]
### Step 7: Substitute \(K\) into the equilibrium expression
Now we have:
\[
K = \frac{x^2}{(1 - 2x)^2} \approx 3.329
\]
Cross-multiplying gives:
\[
x^2 = 3.329(1 - 2x)^2
\]
### Step 8: Solve for \(x\)
Expanding and rearranging:
\[
x^2 = 3.329(1 - 4x + 4x^2)
\]
\[
x^2 = 3.329 - 13.316x + 13.316x^2
\]
\[
0 = 12.316x^2 - 13.316x + 3.329
\]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
x = \frac{13.316 \pm \sqrt{(-13.316)^2 - 4 \cdot 12.316 \cdot 3.329}}{2 \cdot 12.316}
\]
Calculating the discriminant and solving for \(x\) yields:
\[
x \approx 0.193 \, \text{M}
\]
### Step 9: Find the equilibrium concentration of \(\text{ClO}_4^{-}\)
Since at equilibrium \([\text{ClO}_4^{-}] = x\):
\[
[\text{ClO}_4^{-}] \approx 0.193 \, \text{M}
\]
### Final Answer
The equilibrium concentration of perchlorate ion \(\text{ClO}_4^{-}\) is approximately **0.193 M**.
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