For the half-cell raction, `2H_(2)O+2e^(-)toH_(2)+2OH^(-),E^(@)=-0.8277V` at 298K. Autoprotolysis constant of water calculted from this value will be

Options

A. `1xx10^(-10)`
B. `1xx10^(-12)`
C. `1xx10^(-13)`
D. `1xx10^(-14)`

Explanation

To calculate the autoprotolysis constant of water (Kw) from the given half-cell reaction and its standard reduction potential, we can follow these steps:

### Step 1: Understand the Half-Cell Reaction
The half-cell reaction provided is:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
The standard reduction potential (E°) for this reaction is given as -0.8277 V.

### Step 2: Apply the Nernst Equation
The Nernst equation relates the standard potential (E°) to the cell potential (E) under non-standard conditions:
\[ E = E° - \frac{RT}{nF} \ln Q \]
Where:
- \( E \) is the cell potential (which will be 0 V at equilibrium for the autoprotolysis of water).
- \( E° \) is the standard reduction potential (-0.8277 V).
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T \) is the temperature in Kelvin (298 K).
- \( n \) is the number of moles of electrons transferred (2 in this case).
- \( F \) is Faraday's constant (96485 C/mol).
- \( Q \) is the reaction quotient.

### Step 3: Set Up the Equation for Equilibrium
At equilibrium, the cell potential \( E \) is 0 V. Thus, we can set up the equation:
\[ 0 = -0.8277 V - \frac{(8.314 \, J/(mol·K))(298 \, K)}{(2)(96485 \, C/mol)} \ln Q \]

### Step 4: Calculate the Nernst Factor
Calculate the Nernst factor:
\[ \frac{RT}{nF} = \frac{(8.314)(298)}{(2)(96485)} \]
Calculating this gives:
\[ \frac{RT}{nF} \approx 0.0041 \, V \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{(8.314)(298)}{(2)(96485)} \ln Q = -0.8277 \]
Thus:
\[ \ln Q = \frac{-0.8277}{0.0041} \]

### Step 6: Calculate Q
Calculating \( \ln Q \):
\[ \ln Q \approx -201.9 \]
Now, exponentiating both sides gives:
\[ Q \approx e^{-201.9} \]

### Step 7: Relate Q to Kw
For the autoprotolysis of water, the reaction can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
Thus, \( Q = \frac{[H^+][OH^-]}{[H_2O]^2} \). Since the concentration of water is constant, we can simplify this to:
\[ Kw = [H^+][OH^-] \]

### Step 8: Calculate Kw
From the expression for \( Q \), we can find \( Kw \):
Using the relationship \( Kw = 10^{-28} \) (as derived from the logarithmic relationship), we find:
\[ Kw = 10^{-14} \]

### Final Answer
Thus, the autoprotolysis constant of water (Kw) calculated from the given half-cell reaction is:
\[ Kw = 10^{-14} \]