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 hydrogen gas at 1 atm, we can follow these steps:

### Step 1: Understand the Standard Reduction Potential
The standard reduction potential (E°) for the hydrogen electrode reaction is defined as:
\[ \text{2H}^+ + 2e^- \leftrightarrow \text{H}_2 \]
At standard conditions (1 atm, pH 0), the standard reduction potential \( E^\circ \) is 0 V.

### Step 2: Use the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( E \) is the half-cell potential we want to find.
- \( E^\circ \) is the standard reduction potential (0 V for hydrogen).
- \( n \) is the number of electrons transferred (2 for the hydrogen electrode).
- The concentration of \( \text{H}^+ \) ions can be derived from the pH.

### Step 3: Calculate the Concentration of \( \text{H}^+ \)
Given that pH = 10:
\[ [\text{H}^+] = 10^{-10} \, \text{M} \]

### Step 4: Substitute Values into the Nernst Equation
Using the Nernst equation:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{P_{\text{H}_2}}{[\text{H}^+]^2} \right) \]
Where:
- \( P_{\text{H}_2} = 1 \, \text{atm} \)
- \( [\text{H}^+] = 10^{-10} \, \text{M} \)

Substituting these values:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{1}{(10^{-10})^2} \right) \]
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{1}{10^{-20}} \right) \]
\[ E = 0 - \frac{0.0591}{2} \log (10^{20}) \]
\[ E = 0 - \frac{0.0591}{2} \times 20 \]

### Step 5: Calculate the Final Value
\[ E = 0 - \frac{0.0591 \times 20}{2} \]
\[ E = 0 - 0.591 \]
\[ E = -0.591 \, \text{V} \]

### Final Answer
The half-cell reduction potential of the hydrogen electrode at pH 10 and hydrogen gas at 1 atm is:
\[ \boxed{-0.591 \, \text{V}} \]

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