The pressure of `H_(2)` required to make the potential of `H_(2)-`electrode zero in pure water at 289K is :

Options

A. `10^(-12) atm `
B. `10^(-10) atm `
C. `10^(-4) atm `
D. `10^(-14)atm`

Explanation

To solve the problem of determining the pressure of \( H_2 \) required to make the potential of the hydrogen electrode zero in pure water at 289 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Standard Hydrogen Electrode (SHE)**:
The SHE is defined by the half-reaction:
\[ 2H^+ + 2e^- \leftrightarrow H_2(g) \]
At standard conditions, the potential \( E^\circ \) of this electrode is 0 V.

2. **Use the Nernst Equation**:
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where \( n \) is the number of moles of electrons transferred and \( Q \) is the reaction quotient.

3. **Identify Parameters**:
For the SHE:
- \( E^\circ = 0 \) V
- \( n = 2 \) (since 2 electrons are involved)
- The reaction quotient \( Q \) can be expressed as:
\[ Q = \frac{[H_2]}{[H^+]^2} \]
Here, \( [H_2] \) is the pressure of hydrogen gas and \( [H^+] \) is the concentration of hydrogen ions.

4. **Determine \( [H^+] \) in Pure Water**:
In pure water at 289 K, the concentration of \( H^+ \) ions is given by:
\[ [H^+] = 10^{-7} \, \text{mol/L} \]
This is derived from the ion product of water \( K_w = 10^{-14} \, \text{mol}^2/\text{L}^2 \).

5. **Set Up the Nernst Equation**:
Since we want the potential \( E \) to be zero:
\[ 0 = 0 - \frac{0.0591}{2} \log \left( \frac{P_{H_2}}{(10^{-7})^2} \right) \]
Simplifying gives:
\[ 0 = -\frac{0.0591}{2} \log \left( \frac{P_{H_2}}{10^{-14}} \right) \]

6. **Solve for \( P_{H_2} \)**:
Rearranging the equation:
\[ \log \left( \frac{P_{H_2}}{10^{-14}} \right) = 0 \]
This implies:
\[ \frac{P_{H_2}}{10^{-14}} = 1 \]
Therefore:
\[ P_{H_2} = 10^{-14} \, \text{atm} \]

### Final Answer:
The pressure of \( H_2 \) required to make the potential of the hydrogen electrode zero in pure water at 289 K is:
\[ \boxed{10^{-14} \, \text{atm}} \]