The standard reduction potential at `25^@C` of the reaction `2H_2O+2e^(-) leftrightarrow H_2 +2OH^(-)` is `-0.8277` volt. Calculate the equilibrium constant for the reaction `2H_2O=H_3O^+ +OH^(-)` at `25^@C`

Explanation

To calculate the equilibrium constant for the reaction \(2H_2O \rightleftharpoons H_3O^+ + OH^-\) at \(25^\circ C\), we can use the relationship between the standard reduction potential and the equilibrium constant. Here’s a step-by-step solution:

### Step 1: Identify the Standard Reduction Potential
The standard reduction potential for the reaction:
\[ 2H_2O + 2e^- \rightleftharpoons H_2 + 2OH^- \]
is given as \(E^\circ = -0.8277 \, \text{V}\).
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