Given :
`Hg_(2)^(2+) rightarrow 2Hg`, `E^(@) = 0.789 V` and `Hg^(2+) + 2e^(-) rightarrow Hg`, `E^(@) = 0.854V`
Calculate the equilibrium consant for `Hg_(2)^(2+) rightarrow Hg + Hg^(2+)`.

Options

A. `3.13xx10^(-3)`
B. `3.13xx10^(-4)`
C. `6.26xx10^(-3)`
D. `6.26xx10^(-4)`

Explanation

To calculate the equilibrium constant for the reaction \( \text{Hg}_2^{2+} \rightleftharpoons \text{Hg} + \text{Hg}^{2+} \), we will follow these steps:

### Step 1: Write the half-reactions and their standard reduction potentials.

1. The first half-reaction is:
\[ \text{Hg}_2^{2+} + 2e^- \rightarrow 2\text{Hg} \quad E^\circ = 0.789 \, \text{V} \]

2. The second half-reaction is:
\[ \text{Hg}^{2+} + 2e^- \rightarrow \text{Hg} \quad E^\circ = 0.854 \, \text{V} \]

### Step 2: Reverse the second half-reaction to represent oxidation.

To find the overall reaction, we need to reverse the second half-reaction because we want to obtain \(\text{Hg}^{2+}\) on the product side:

\[ \text{Hg} \rightarrow \text{Hg}^{2+} + 2e^- \quad E^\circ = -0.854 \, \text{V} \]

### Step 3: Add the half-reactions.

Now, we can add the two half-reactions together:

1. From the first half-reaction:
\[ \text{Hg}_2^{2+} + 2e^- \rightarrow 2\text{Hg} \]

2. From the reversed second half-reaction:
\[ \text{Hg} \rightarrow \text{Hg}^{2+} + 2e^- \]

Adding these gives:
\[ \text{Hg}_2^{2+} \rightarrow \text{Hg} + \text{Hg}^{2+} \]

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

The standard cell potential for the overall reaction is calculated by adding the potentials of the half-reactions:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} = 0.789 \, \text{V} + (-0.854 \, \text{V}) = -0.065 \, \text{V} \]

### Step 5: Relate \(E^\circ\) to the equilibrium constant \(K\).

The relationship between the standard cell potential and the equilibrium constant is given by the equation:
\[ \Delta G^\circ = -nFE^\circ \]
and
\[ \Delta G^\circ = -RT \ln K \]

Setting these equal gives:
\[ -nFE^\circ = -RT \ln K \]

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

Rearranging the equation gives:
\[ \ln K = \frac{nFE^\circ}{RT} \]

Where:
- \(n = 2\) (number of electrons transferred)
- \(F = 96485 \, \text{C/mol}\) (Faraday's constant)
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 298 \, \text{K}\)

Substituting the values:
\[ \ln K = \frac{2 \times 96485 \times (-0.065)}{8.314 \times 298} \]

Calculating this gives:
\[ \ln K \approx -6.3 \]

Thus:
\[ K \approx e^{-6.3} \approx 6.3 \times 10^{-3} \]

### Final Answer:
The equilibrium constant \(K\) for the reaction \( \text{Hg}_2^{2+} \rightleftharpoons \text{Hg} + \text{Hg}^{2+} \) is approximately \( 6.3 \times 10^{-3} \).

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