A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl or `pH = 10` and by passing bydrogen gas around the platinum wire at one atm pressure . The oxidation potential of electrode would be ?

Options

A. ` 0.59 V`
B. ` 0. 11 8 V`
C. ` 1.18 V`
D. ` 0.059 V`

Explanation

To find the oxidation potential of the hydrogen gas electrode in a solution with a pH of 10, we can follow these steps:

### Step 1: Understanding the Nernst Equation
The Nernst equation is used to calculate the potential of an electrochemical cell under non-standard conditions. The general form of the Nernst equation is:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where:
- \( E \) is the cell potential.
- \( E^\circ \) is the standard cell potential.
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( Q \) is the reaction quotient.

### Step 2: Determine the Standard Potential
For the hydrogen electrode, the standard potential \( E^\circ \) is 0.00 V.

### Step 3: Identify the Reaction
The half-reaction for the hydrogen electrode is:
\[ \text{H}_2 \rightarrow 2 \text{H}^+ + 2 e^- \]
This indicates that 2 moles of electrons are transferred, so \( n = 2 \).

### Step 4: Calculate the Concentration of \( \text{H}^+ \)
Given that the pH of the solution is 10, we can calculate the concentration of \( \text{H}^+ \):
\[ \text{pH} = -\log[\text{H}^+] \]
Thus,
\[ [\text{H}^+] = 10^{-10} \, \text{M} \]

### Step 5: Substitute Values into the Nernst Equation
Now, we can substitute the values into the Nernst equation. The reaction quotient \( Q \) for the reaction can be expressed as:
\[ Q = \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \]
Given that the pressure of hydrogen gas \( P_{\text{H}_2} \) is 1 atm:
\[ Q = \frac{(10^{-10})^2}{1} = 10^{-20} \]

Now substituting into the Nernst equation:
\[ E = 0.00 - \frac{0.0591}{2} \log(10^{-20}) \]

### Step 6: Calculate the Logarithm
Calculating the logarithm:
\[ \log(10^{-20}) = -20 \]

### Step 7: Final Calculation
Now substituting back into the equation:
\[ E = 0.00 - \frac{0.0591}{2} \times (-20) \]
\[ E = 0.00 + 0.0591 \times 10 \]
\[ E = 0.591 \, \text{V} \]

### Conclusion
The oxidation potential of the hydrogen gas electrode is:
\[ \boxed{0.591 \, \text{V}} \]