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

Options

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

Explanation

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

### Step 1: Write the Reduction Reaction
The reduction reaction for the hydrogen electrode can be written as:
\[ 2H^+ (aq) + 2e^- \rightarrow H_2 (g) \]

### Step 2: Understand the Concentration of \( H^+ \)
In pure water at 289 K, the concentration of \( H^+ \) ions is:
\[ [H^+] = 10^{-7} \, \text{mol/L} \]

### Step 3: Use the Nernst Equation
The Nernst equation for the hydrogen electrode is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) \]
where:
- \( E \) is the electrode potential,
- \( E^\circ \) is the standard electrode potential (0 V for the hydrogen electrode),
- \( n \) is the number of electrons transferred (2 for hydrogen),
- \( P_{H_2} \) is the pressure of hydrogen gas.

### Step 4: Set the Electrode Potential to Zero
To find the pressure of \( H_2 \) that makes the potential zero, we set \( E = 0 \):
\[ 0 = 0 - \frac{0.059}{2} \log \left( \frac{P_{H_2}}{(10^{-7})^2} \right) \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 0 = -\frac{0.059}{2} \log \left( \frac{P_{H_2}}{10^{-14}} \right) \]
This implies:
\[ \log \left( \frac{P_{H_2}}{10^{-14}} \right) = 0 \]

### Step 6: Solve for \( P_{H_2} \)
From the logarithmic equation, we find:
\[ \frac{P_{H_2}}{10^{-14}} = 1 \]
Thus:
\[ P_{H_2} = 10^{-14} \, \text{atm} \]

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

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