If `E_(ClO_(3)^(-)//ClO_(4)^(-))=-0.36 V & E_(ClO_(3)^(-)//ClO_(2)^(-))^(@)=0.33V` at 300 K.
The equilibrium concentration of perchlorate ion `(ClO_(4)^(-))` which was initially 1.0 M in `ClO_(3)^(-)` when the reaction starts to attain the equilibrium,
`2ClO_(3)^(-)hArr ClO_(2)^(-)+ClO_(4)^(-)`
Given : Anti `log(0.509)=3.329`

Options

A. `0.0236M`
B. `0.0190M`
C. `0.123M`
D. `0.191M`

Answer

To solve the problem, we need to determine the equilibrium concentration of the perchlorate ion (ClO₄⁻) when the reaction reaches equilibrium. The reaction is given as:

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

### Step 1: Write the expression for the equilibrium constant (K)

The equilibrium constant \( K \) for the reaction can be expressed as:

\[ K = \frac{[\text{ClO}_2^-][\text{ClO}_4^-]}{[\text{ClO}_3^-]^2} \]

### Step 2: Set up the initial concentrations and changes

Initially, we have:
- \([\text{ClO}_3^-] = 1.0 \, \text{M}\)
- \([\text{ClO}_2^-] = 0 \, \text{M}\)
- \([\text{ClO}_4^-] = 0 \, \text{M}\)

Let \( x \) be the change in concentration of \(\text{ClO}_4^-\) at equilibrium. Thus, at equilibrium:
- \([\text{ClO}_3^-] = 1.0 - 2x\)
- \([\text{ClO}_2^-] = x\)
- \([\text{ClO}_4^-] = x\)

### Step 3: Substitute the equilibrium concentrations into the K expression

Substituting the equilibrium concentrations into the expression for \( K \):

\[ K = \frac{x \cdot x}{(1.0 - 2x)^2} = \frac{x^2}{(1.0 - 2x)^2} \]

### Step 4: Calculate the standard cell potential (E°cell)

Using the given standard reduction potentials:
- \( E^\circ(\text{ClO}_3^- \rightarrow \text{ClO}_2^-) = 0.33 \, \text{V} \)
- \( E^\circ(\text{ClO}_3^- \rightarrow \text{ClO}_4^-) = -0.36 \, \text{V} \)

The overall cell potential \( E^\circ_{\text{cell}} \) is calculated as:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.33 - (-0.36) = 0.33 + 0.36 = 0.69 \, \text{V} \]

### Step 5: Use the Nernst equation to find K

The Nernst equation at standard conditions is given by:

\[ E = E^\circ - \frac{RT}{nF} \ln K \]

At equilibrium, \( E = 0 \), so:

\[ 0 = 0.69 - \frac{(8.314)(300)}{2(96500)} \ln K \]

Solving for \( \ln K \):

\[ \ln K = \frac{(0.69)(2)(96500)}{(8.314)(300)} \]

Calculating this gives:

\[ \ln K \approx 0.509 \implies K \approx e^{0.509} \approx 1.66 \]

### Step 6: Solve for x using the equilibrium constant

Substituting \( K \) back into the equilibrium expression:

\[ 1.66 = \frac{x^2}{(1.0 - 2x)^2} \]

Taking the square root of both sides gives:

\[ \sqrt{1.66} = \frac{x}{1.0 - 2x} \]

Let \( \sqrt{1.66} \approx 1.287 \):

\[ 1.287(1.0 - 2x) = x \]

Rearranging gives:

\[ 1.287 - 2.574x = x \implies 1.287 = 3.574x \implies x = \frac{1.287}{3.574} \approx 0.36 \]

### Step 7: Calculate the equilibrium concentration of ClO₄⁻

Since \( x \) represents the concentration of \(\text{ClO}_4^-\):

\[ [\text{ClO}_4^-] = x \approx 0.36 \, \text{M} \]

### Final Answer

The equilibrium concentration of the perchlorate ion \((\text{ClO}_4^-)\) is approximately **0.36 M**.

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