An excess of liquid mercury is added to an acidicfied solution of `1.0xx10^(-3) M Fe^(3+)` . It is found that `5%` of `Fe^(3+)` remains at equilibrium at `25^(@)C`. Calculate `E^(c-)._((Hg_(2)^(2+)|Hg))` assuming that the only reaction that occurs is
`2Hg+2Fe^(3+) rarr Hg_(2)^(2+)+2Fe^(2+)`
Given `: E^(c-)._((Fe^(3+)|Fe^(2+)))=0.77V`

Explanation

To solve the problem, we need to calculate the standard reduction potential \( E^{\circ} \) for the half-reaction \( \text{Hg}_2^{2+} + 2e^- \rightarrow 2\text{Hg} \) using the information provided about the equilibrium concentrations of the species involved in the reaction.

### Step-by-Step Solution:

1. **Identify Initial Concentrations**:
- The initial concentration of \( \text{Fe}^{3+} \) is given as \( 1.0 \times 10^{-3} \, \text{M} \).

2. **Determine Equilibrium Concentrations**:
- Since 5% of \( \text{Fe}^{3+} \) remains at equilibrium, we can calculate the concentration of \( \text{Fe}^{3+} \) at equilibrium:
\[ [\text{Fe}^{3+}]_{eq} = 0.05 \times (1.0 \times 10^{-3}) = 5.0 \times 10^{-5} \, \text{M} \]
- Therefore, the concentration of \( \text{Fe}^{2+} \) at equilibrium, which is produced from the reduction of \( \text{Fe}^{3+} \), is:
\[ [\text{Fe}^{2+}]_{eq} = 1.0 \times 10^{-3} - 5.0 \times 10^{-5} = 0.95 \times 10^{-3} \, \text{M} \]

3. **Calculate the Concentration of \( \text{Hg}_2^{2+} \)**:
- The stoichiometry of the reaction shows that 2 moles of \( \text{Fe}^{3+} \) react with 1 mole of \( \text{Hg}_2^{2+} \). Therefore, the concentration of \( \text{Hg}_2^{2+} \) produced at equilibrium is:
\[ [\text{Hg}_2^{2+}]_{eq} = \frac{1}{2} \times [\text{Fe}^{2+}]_{eq} = \frac{1}{2} \times (0.95 \times 10^{-3}) = 0.475 \times 10^{-3} \, \text{M} \]

4. **Set Up the Nernst Equation**:
- At equilibrium, the cell potential \( E \) is 0. The Nernst equation is given by:
\[ E = E^{\circ} - \frac{0.059}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
- For the reaction:
\[ 2\text{Hg} + 2\text{Fe}^{3+} \rightarrow \text{Hg}_2^{2+} + 2\text{Fe}^{2+} \]
- The number of electrons \( n = 2 \).

5. **Substituting Values into the Nernst Equation**:
- The products are \( [\text{Fe}^{2+}]^2 \) and \( [\text{Hg}_2^{2+}] \), and the reactants are \( [\text{Fe}^{3+}]^2 \):
\[ 0 = E^{\circ} - \frac{0.059}{2} \log \left( \frac{(0.95 \times 10^{-3})^2 \times (0.475 \times 10^{-3})}{(5.0 \times 10^{-5})^2} \right) \]

6. **Calculating the Logarithmic Term**:
- Calculate the logarithmic term:
\[ \frac{(0.95 \times 10^{-3})^2 \times (0.475 \times 10^{-3})}{(5.0 \times 10^{-5})^2} = \frac{(0.9025 \times 10^{-6}) \times (0.475 \times 10^{-3})}{(25 \times 10^{-10})} \]
\[ = \frac{0.4286875 \times 10^{-9}}{25 \times 10^{-10}} = 17.1475 \]
- Now take the logarithm:
\[ \log(17.1475) \approx 1.233 \]

7. **Final Calculation**:
- Substitute back into the Nernst equation:
\[ 0 = E^{\circ} - \frac{0.059}{2} \times 1.233 \]
\[ E^{\circ} = \frac{0.059}{2} \times 1.233 \approx 0.0363 \, \text{V} \]
- Adding the standard potential of \( \text{Fe}^{3+}/\text{Fe}^{2+} \):
\[ E^{\circ}_{\text{Hg}_2^{2+}/\text{Hg}} = 0.77 + 0.0363 \approx 0.792 \, \text{V} \]

### Final Answer:
\[ E^{\circ}_{\text{Hg}_2^{2+}/\text{Hg}} \approx 0.792 \, \text{V} \]