`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 \( K \) 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 Nernst equation in its logarithmic form:

\[ E^\circ_{cell} = \frac{0.0591}{n} \log K \]

where:
- \( E^\circ_{cell} \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( K \) is the equilibrium constant.

### Step 1: Identify the number of electrons transferred
In the reaction, we can see that one electron is involved in the half-reaction. Therefore, \( n = 1 \).

### Step 2: Rearrange the Nernst equation to solve for \( K \)
We can rearrange the equation to isolate \( \log K \):

\[ \log K = \frac{n \cdot E^\circ_{cell}}{0.0591} \]

### Step 3: Substitute the values into the equation
Substituting \( E^\circ_{cell} = -0.8277 \) V and \( n = 1 \):

\[ \log K = \frac{1 \cdot (-0.8277)}{0.0591} \]

### Step 4: Calculate \( \log K \)
Calculating the right side:

\[ \log K = \frac{-0.8277}{0.0591} \approx -14.0 \]

### Step 5: Convert from logarithmic form to exponential form
To find \( K \), we convert from the logarithmic form:

\[ K = 10^{\log K} = 10^{-14} \]

### Conclusion
Thus, the equilibrium constant \( K \) for the reaction is:

\[ K = 10^{-14} \]