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^(o)._((Fe^(3+)|Fe^(2+)))=0.77V`

Explanation

To solve the problem, we need to find the standard reduction potential \( E^{\circ} \) for the half-reaction \( \text{Hg}_2^{2+} + 2\text{e}^- \rightarrow 2\text{Hg} \) given the reaction and the equilibrium conditions.

### Step-by-Step Solution:

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

2. **Determine Equilibrium Concentration of \( \text{Fe}^{3+} \)**:
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