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 step by step, we need to calculate the standard reduction potential \( E^{\circ}_{(Hg_2^{2+}|Hg)} \) based on the given information.

### Step 1: Determine the initial and equilibrium concentrations of \( Fe^{3+} \)

Given:
- Initial concentration of \( Fe^{3+} = 1.0 \times 10^{-3} \, M \)
- At equilibrium, \( 5\% \) of \( Fe^{3+} \) remains.

Calculating the concentration of \( Fe^{3+} \) at equilibrium:
\[ \text{Remaining } Fe^{3+} = 5\% \text{ of } 1.0 \times 10^{-3} = 0.05 \times 1.0 \times 10^{-3} = 5.0 \times 10^{-5} \, M \]

### Step 2: Calculate the change in concentration of \( Fe^{3+} \)

The change in concentration of \( Fe^{3+} \) is:
\[ \Delta [Fe^{3+}] = [Fe^{3+}]_{initial} - [Fe^{3+}]_{equilibrium} = 1.0 \times 10^{-3} - 5.0 \times 10^{-5} = 0.95 \times 10^{-3} \, M \]

### Step 3: Determine the concentration of \( Fe^{2+} \)

From the reaction:
\[ 2Hg + 2Fe^{3+} \rightarrow Hg_2^{2+} + 2Fe^{2+} \]
The stoichiometry indicates that for every 2 moles of \( Fe^{3+} \) reduced, 2 moles of \( Fe^{2+} \) are produced.

Thus, the concentration of \( Fe^{2+} \) formed is equal to the change in concentration of \( Fe^{3+} \):
\[ [Fe^{2+}] = 0.95 \times 10^{-3} \, M \]

### Step 4: Calculate the concentration of \( Hg_2^{2+} \)

Since 2 moles of \( Fe^{3+} \) produce 1 mole of \( Hg_2^{2+} \):
\[ [Hg_2^{2+}] = \frac{0.95 \times 10^{-3}}{2} = 0.475 \times 10^{-3} \, M \]

### Step 5: Use the Nernst equation to find \( E_{cell} \)

The Nernst equation for the cell reaction is:
\[ E_{cell} = E^{\circ} - \frac{RT}{nF} \ln Q \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96485 \, C/mol \)

The reaction quotient \( Q \) is given by:
\[ Q = \frac{[Hg_2^{2+}]}{[Fe^{3+}]^2} \]
Substituting the concentrations:
\[ Q = \frac{0.475 \times 10^{-3}}{(5.0 \times 10^{-5})^2} = \frac{0.475 \times 10^{-3}}{25 \times 10^{-10}} = 1.9 \times 10^{6} \]

Now substituting into the Nernst equation:
\[ E_{cell} = E^{\circ}_{(Fe^{3+}|Fe^{2+})} - \frac{0.0257}{2} \ln(1.9 \times 10^{6}) \]
Where \( E^{\circ}_{(Fe^{3+}|Fe^{2+})} = 0.77 \, V \).

Calculating:
\[ E_{cell} = 0.77 - 0.01285 \ln(1.9 \times 10^{6}) \]
Calculating \( \ln(1.9 \times 10^{6}) \approx 14.4 \):
\[ E_{cell} = 0.77 - 0.01285 \times 14.4 \approx 0.77 - 0.18504 \approx 0.58496 \, V \]

### Step 6: Relate \( E_{cell} \) to \( E^{\circ}_{(Hg_2^{2+}|Hg)} \)

Using the relationship:
\[ E_{cell} = E^{\circ}_{(Fe^{3+}|Fe^{2+})} - E^{\circ}_{(Hg_2^{2+}|Hg)} \]
Substituting the known values:
\[ 0.58496 = 0.77 - E^{\circ}_{(Hg_2^{2+}|Hg)} \]
Rearranging gives:
\[ E^{\circ}_{(Hg_2^{2+}|Hg)} = 0.77 - 0.58496 \approx 0.18504 \, V \]

### Final Answer
\[ E^{\circ}_{(Hg_2^{2+}|Hg)} \approx 0.185 \, V \]