`E^(0) = -08275V` for the reaction, `2H_2O + 2e^(-) to 2OH^(-) + H_(2)`.Calculate the ionic product for the reaction, `2H_2O

Answer

To solve the problem, we need to calculate the ionic product for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) using the given standard electrode potential \(E^{0} = -0.8275 \, V\) for the reaction \(2H_2O + 2e^- \rightarrow 2OH^- + H_2\).

### Step-by-Step Solution:

1. **Identify the Reaction**: The given reaction is:
\[ 2H_2O + 2e^- \rightarrow 2OH^- + H_2 \]
This is the half-reaction for the reduction of water to hydroxide ions and hydrogen gas.

2. **Determine the Number of Electrons (n)**: In the given reaction, 2 electrons are involved. Thus, \(n = 2\).

3. **Write the Expression for \(K_c\)**: The equilibrium constant \(K_c\) for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) can be expressed as:
\[ K_c = \frac{[H_3O^+][OH^-]}{[H_2O]} \]
Since the concentration of water is constant, we can relate \(K_c\) to the ionic product \(K_w\) (the product of the concentrations of \(H_3O^+\) and \(OH^-\)):
\[ K_c \cdot [H_2O] = K_w \]
Therefore, we can say:
\[ K_c = K_w \]

4. **Use the Nernst Equation**: The Nernst equation relates the standard electrode potential to the equilibrium constant:
\[ E^{0} = \frac{0.059}{n} \log K_c \]
Plugging in the values:
\[ -0.8275 = \frac{0.059}{2} \log K_c \]

5. **Rearranging the Equation**: To find \(\log K_c\), rearranging gives:
\[ \log K_c = \frac{-0.8275 \times 2}{0.059} \]

6. **Calculate \(\log K_c\)**:
\[ \log K_c = \frac{-1.655}{0.059} \approx -28.05 \]

7. **Find \(K_c\)**: Now, we can calculate \(K_c\) by taking the antilogarithm:
\[ K_c = 10^{-28.05} \approx 8.91 \times 10^{-29} \]

8. **Relate \(K_c\) to \(K_w\)**: Since \(K_c = K_w\), we can conclude that:
\[ K_w \approx 8.91 \times 10^{-29} \]

### Final Answer:
The ionic product \(K_w\) for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) is approximately:
\[ K_w \approx 8.91 \times 10^{-29} \]