`E_"cell"^@` for the reaction , `2H_2O to H_3O^(+) + OH^(-)` at `25^@C` is -0.8277 V. The equilibrium constant for the reaction is

Options

A. `10^(-14)`
B. `10^(-23)`
C. `10^(-7)`
D. `10^(-21)`

Explanation

To find the equilibrium constant for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) given that the standard cell potential \(E^\circ_{cell}\) is \(-0.8277 \, V\), we can use the relationship between the standard cell potential and the equilibrium constant, which is derived from the Nernst equation.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
- The reaction is:
\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]
- Given:
\[ E^\circ_{cell} = -0.8277 \, V \]

2. **Determine the Number of Electrons Transferred (n)**:
- In this reaction, the change in charge is from neutral water to one hydronium ion and one hydroxide ion. Therefore, the number of electrons transferred \(n\) is 1.

3. **Use the Nernst Equation**:
- The Nernst equation in relation to the equilibrium constant \(K\) is given by:
\[ E^\circ_{cell} = \frac{0.059}{n} \log K \]
- Rearranging this gives:
\[ \log K = \frac{n \cdot E^\circ_{cell}}{0.059} \]

4. **Substituting the Values**:
- Substitute \(n = 1\) and \(E^\circ_{cell} = -0.8277 \, V\):
\[ \log K = \frac{1 \cdot (-0.8277)}{0.059} \]

5. **Calculate the Value**:
- Performing the calculation:
\[ \log K = -14.03 \quad (\text{approximately}) \]

6. **Convert Logarithmic Value to Equilibrium Constant**:
- To find \(K\), we convert from log form:
\[ K = 10^{-14.03} \approx 10^{-14} \]

### Final Answer:
The equilibrium constant \(K\) for the reaction is approximately:
\[ K \approx 10^{-14} \]