The standard potential of the reaction
`H_(2)O + e^(-) rightarrow (1/2)H_(2) + OH^(-)` at 298 K by using `k_(w) (H_(2)O) = 10^(-14)`, is:

Options

A. `-0.828V`
B. `0.828V`
C. `0V`
D. `-0.5V`

Explanation

To find the standard potential of the reaction

\[ H_2O + e^- \rightarrow \frac{1}{2} H_2 + OH^- \]

at 298 K using \( K_w (H_2O) = 10^{-14} \), we can follow these steps:

### Step 1: Identify the Half-Reactions
We can break down the overall reaction into two half-reactions:

1. The reduction half-reaction:
\[ H^+ + e^- \rightarrow \frac{1}{2} H_2 \]
This half-reaction has a standard electrode potential \( E^\circ = 0 \, \text{V} \) (standard hydrogen electrode).

2. The oxidation half-reaction:
\[ H_2O \rightarrow H^+ + OH^- \]
We need to find the standard potential for this reaction.

### Step 2: Relate Gibbs Free Energy to Electrode Potential
The relationship between Gibbs free energy change (\( \Delta G^\circ \)) and the standard electrode potential (\( E^\circ \)) is given by:

\[ \Delta G^\circ = -nFE^\circ \]

where:
- \( n \) = number of moles of electrons transferred (1 mole in this case),
- \( F \) = Faraday's constant (\( 96500 \, \text{C/mol} \)),
- \( E^\circ \) = standard electrode potential.

### Step 3: Calculate \( \Delta G^\circ \) for the Oxidation Half-Reaction
For the oxidation half-reaction, we can use the relationship with the ion product of water \( K_w \):

\[ \Delta G^\circ = -2.303RT \log K_w \]

Substituting the values:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( K_w = 10^{-14} \)

### Step 4: Substitute the Values
Now we can calculate \( \Delta G^\circ \):

\[ \Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log(10^{-14}) \]

Calculating \( \log(10^{-14}) = -14 \):

\[ \Delta G^\circ = -2.303 \times 8.314 \times 298 \times (-14) \]

### Step 5: Calculate \( \Delta G^\circ \)
Calculating the values:

\[ \Delta G^\circ = 2.303 \times 8.314 \times 298 \times 14 \]

Calculating the product:

\[ \Delta G^\circ \approx 2.303 \times 8.314 \times 298 \times 14 \approx 79881.87 \, \text{J/mol} \]

### Step 6: Find the Standard Potential \( E^\circ \)
Now, we can find the standard potential for the oxidation reaction:

\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]

Substituting the values:

\[ E^\circ = -\frac{79881.87}{96500} \]

Calculating:

\[ E^\circ \approx -0.828 \, \text{V} \]

### Conclusion
Thus, the standard potential of the reaction

\[ H_2O + e^- \rightarrow \frac{1}{2} H_2 + OH^- \]

at 298 K is

\[ \boxed{-0.828 \, \text{V}} \]