A hydrogen electrode is prepared by using a sample of HCl solution with `pH=4` and `H_(2)` gas at 1 atm pressure. What is the electrode potential of the electrode?

Options

A. 0.814 V
B. 0.236 V
C. `-0.06V`
D. `-0.618V`

Explanation

To find the electrode potential of a hydrogen electrode prepared with a given HCl solution and hydrogen gas, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Understand the given data
- pH of the HCl solution = 4
- Pressure of H2 gas = 1 atm

### Step 2: Calculate the concentration of H⁺ ions
The pH is related to the concentration of hydrogen ions (H⁺) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that pH = 4, we can find [H⁺]:
\[ 4 = -\log[H^+] \]
To find [H⁺], we take the antilogarithm:
\[ [H^+] = 10^{-4} \, \text{M} \]

### Step 3: Write the half-cell reaction
The half-cell reaction for the hydrogen electrode is:
\[ \text{H}_2(g) \rightleftharpoons 2H^+(aq) + 2e^- \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log Q \]
Where:
- \( E^\circ \) = standard electrode potential (for hydrogen electrode, \( E^\circ = 0 \) V)
- \( n \) = number of electrons transferred in the half-reaction (for H₂, \( n = 2 \))
- \( Q \) = reaction quotient

### Step 5: Calculate the reaction quotient (Q)
For the reaction:
\[ Q = \frac{[H^+]^2}{P_{H_2}} \]
Where:
- \( [H^+] = 10^{-4} \, \text{M} \)
- \( P_{H_2} = 1 \, \text{atm} \)

Substituting the values:
\[ Q = \frac{(10^{-4})^2}{1} = 10^{-8} \]

### Step 6: Substitute values into the Nernst equation
Now substituting \( E^\circ = 0 \), \( n = 2 \), and \( Q = 10^{-8} \) into the Nernst equation:
\[ E = 0 - \frac{0.059}{2} \log(10^{-8}) \]

### Step 7: Calculate the logarithm
We know that:
\[ \log(10^{-8}) = -8 \]
Thus:
\[ E = -\frac{0.059}{2} \times (-8) \]
\[ E = \frac{0.059 \times 8}{2} \]
\[ E = \frac{0.472}{2} = 0.236 \, \text{V} \]

### Final Answer
The electrode potential of the hydrogen electrode is:
\[ E = 0.236 \, \text{V} \]