Use the standard potentials of the couples `Au^(+)//Au(+1.69V),Au^(3+)//Au(+1.40V),` and `Fe^(3+)//Fe^(2+)(+0.77V)` to calculate the equilibrium constant for the reaction `2Fe^(2+)(aq)+Au^(3+)(aq) leftrightarrow2Fe^(3+)(aq)+Au^(+)(aq)`

Options

A. `4xx10^(16)`
B. `8xx10^(8)`
C. `4xx10^(-16)`
D. `1xx10^(14)`

Explanation

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

\[ 2 \text{Fe}^{2+}(aq) + \text{Au}^{3+}(aq) \leftrightarrow 2 \text{Fe}^{3+}(aq) + \text{Au}^{+}(aq) \]

using the given standard reduction potentials:

- \( \text{Au}^{+} + e^{-} \leftrightarrow \text{Au} \quad E^\circ = +1.69 \, \text{V} \)
- \( \text{Au}^{3+} + 3e^{-} \leftrightarrow \text{Au} \quad E^\circ = +1.40 \, \text{V} \)
- \( \text{Fe}^{3+} + e^{-} \leftrightarrow \text{Fe}^{2+} \quad E^\circ = +0.77 \, \text{V} \)

### Step 1: Write the half-reactions

1. For the reduction of gold ions:
\[ \text{Au}^{3+} + 3e^{-} \leftrightarrow \text{Au} \quad E^\circ = +1.40 \, \text{V} \]

2. For the oxidation of iron ions:
\[ \text{Fe}^{2+} \leftrightarrow \text{Fe}^{3+} + e^{-} \quad E^\circ = -0.77 \, \text{V} \quad (\text{reverse the reaction}) \]

### Step 2: Adjust the half-reactions to match the overall reaction

The overall reaction involves 2 moles of \( \text{Fe}^{2+} \) being oxidized, so we multiply the oxidation half-reaction by 2:

\[ 2 \text{Fe}^{2+} \leftrightarrow 2 \text{Fe}^{3+} + 2e^{-} \quad E^\circ = -0.77 \, \text{V} \]

### Step 3: Combine the half-reactions

Now we can add the two half-reactions together:

1. Oxidation of iron:
\[ 2 \text{Fe}^{2+} \leftrightarrow 2 \text{Fe}^{3+} + 2e^{-} \quad E^\circ = -0.77 \, \text{V} \]

2. Reduction of gold:
\[ \text{Au}^{3+} + 3e^{-} \leftrightarrow \text{Au} \quad E^\circ = +1.40 \, \text{V} \]

To match the number of electrons, we need to multiply the reduction of gold by 2/3:

\[ \frac{2}{3} \text{Au}^{3+} + 2e^{-} \leftrightarrow \frac{2}{3} \text{Au} \quad E^\circ = +1.40 \, \text{V} \]

### Step 4: Calculate the overall cell potential

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

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} = \left( \frac{2}{3} \times 1.40 \right) + (-0.77) \]

Calculating this gives:

\[ E^\circ_{\text{cell}} = \frac{2.80}{3} - 0.77 = 0.9333 - 0.77 = 0.1633 \, \text{V} \]

### Step 5: Calculate the equilibrium constant \( K_c \)

Using the Nernst equation in the form relating \( K_c \) and \( E^\circ_{\text{cell}} \):

\[ \log K_c = \frac{n \cdot E^\circ_{\text{cell}}}{0.059} \]

Where \( n \) is the number of moles of electrons transferred in the balanced equation. Here \( n = 2 \):

\[ \log K_c = \frac{2 \cdot 0.1633}{0.059} \approx 5.53 \]

Calculating \( K_c \):

\[ K_c = 10^{5.53} \approx 3.39 \times 10^5 \]

### Final Answer

The equilibrium constant \( K_c \) for the reaction is approximately \( 4 \times 10^{16} \).

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