The standard reduction potential at `25^(@)C` of the reaction
`2H_(2)O+2e^(-)hArrH_(2)+2overset(Θ)(O)H` is `-0.8277V`. Calculate the equilibrium constant for the reaction.
`2H_(2)OhArrH_(3)O^(o+)+overset(Θ)(O)H` at `25^(@)C` .

Explanation

To calculate the equilibrium constant for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) at \(25^\circ C\), we will use the standard reduction potential provided and the Nernst equation.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- The standard reduction potential (\(E^\circ\)) for the reaction \(2H_2O + 2e^- \rightleftharpoons H_2 + 2OH^-\) is \(-0.8277 \, V\).

2. **Understand the Reaction:**
- The reaction we are interested in is \(2H_2O \rightleftharpoons H_3O^+ + OH^-\).
- This can be derived from the reduction and oxidation half-reactions involving water.

3. **Determine the Half-Reactions:**
- Oxidation: \(H_2O \rightarrow H_3O^+ + e^-\)
- Reduction: \(H_2O + e^- \rightarrow OH^-\)

4. **Calculate the Standard Cell Potential (\(E^\circ_{cell}\)):**
- For the overall reaction, the standard cell potential at equilibrium is \(0 \, V\) (as \(E_{cell} = 0\) at equilibrium).
- The net standard potential can be calculated as:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Here, we can assume \(E^\circ_{cathode} = 0\) (for the reduction of \(H_2O\)) and \(E^\circ_{anode} = -0.8277 \, V\):
\[ E^\circ_{cell} = 0 - (-0.8277) = 0.8277 \, V \]

5. **Use the Nernst Equation:**
- The Nernst equation relates the standard electrode potential to the equilibrium constant \(K\):
\[ E^\circ_{cell} = \frac{0.0591}{n} \log K \]
- Here, \(n\) is the number of moles of electrons transferred. In this case, \(n = 1\).

6. **Rearranging the Nernst Equation:**
- Rearranging gives:
\[ \log K = \frac{n \cdot E^\circ_{cell}}{0.0591} \]
- Substituting the values:
\[ \log K = \frac{1 \cdot (-0.8277)}{0.0591} \]

7. **Calculate \(\log K\):**
- Performing the calculation:
\[ \log K = -14.01 \quad (\text{approximately}) \]

8. **Find the Equilibrium Constant \(K\):**
- Therefore, \(K\) can be calculated as:
\[ K = 10^{-14.01} \approx 10^{-14} \]

### Final Answer:
The equilibrium constant \(K\) for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) at \(25^\circ C\) is approximately \(10^{-14}\).