The half cell reduction potential of a hydrogen electrode at pH= 10 and `H_2` gas at 1 atmi will be

Options

A. 0.59V
B. `-0.59V`
C. `0.0059V`
D. `-0.0059V`

Answer

To find the half-cell reduction potential of a hydrogen electrode at pH = 10 and H₂ gas at 1 atm, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the half-cell reaction
The half-cell reaction for the hydrogen electrode can be written as:
\[ \text{2H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 2: Identify the standard reduction potential
The standard reduction potential (E°) for the hydrogen electrode is defined as:
\[ E° = 0 \, \text{V} \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E = E° - \frac{0.0591}{n} \log Q \]
where:
- \( E \) is the cell potential,
- \( E° \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred,
- \( Q \) is the reaction quotient.

### Step 4: Determine the values for the Nernst equation
In our case:
- \( n = 2 \) (since 2 electrons are involved in the reaction),
- The reaction quotient \( Q \) can be expressed as:
\[ Q = \frac{P_{\text{H}_2}}{[H^+]^2} \]
Given:
- \( P_{\text{H}_2} = 1 \, \text{atm} \),
- The concentration of \( H^+ \) can be calculated from the pH:
\[ \text{pH} = -\log[H^+] \implies [H^+] = 10^{-10} \, \text{M} \]

### Step 5: Calculate the reaction quotient \( Q \)
Substituting the values into the expression for \( Q \):
\[ Q = \frac{1}{(10^{-10})^2} = \frac{1}{10^{-20}} = 10^{20} \]

### Step 6: Substitute values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{2} \log(10^{20}) \]
\[ E = 0 - \frac{0.0591}{2} \times 20 \]
\[ E = 0 - 0.0591 \times 10 \]
\[ E = -0.591 \, \text{V} \]

### Final Answer
Thus, the half-cell reduction potential of the hydrogen electrode at pH = 10 and H₂ gas at 1 atm is:
\[ E = -0.59 \, \text{V} \]

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