The reduction potential of hydrogen electrode containing a solution of pH=4 is

Options

A. `0.236 V`
B. `4.059 V`
C. `-0.236V`
D. `3.941V`

Explanation

To find the reduction potential of a hydrogen electrode in a solution with a pH of 4, we can follow these steps:

### Step 1: Understand the Reaction
The half-reaction for the reduction of hydrogen ions (H⁺) is:
\[ \text{2H}^+ + 2e^- \rightarrow \text{H}_2 \]
This indicates that two moles of hydrogen ions gain two electrons to form one mole of hydrogen gas.

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

### Step 3: Use the Nernst Equation
The Nernst equation relates the electrode potential (E) to the standard electrode potential (E°) and the concentration of the reactants and products:
\[ E = E° - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \( n \) is the number of moles of electrons exchanged (for the hydrogen reaction, \( n = 2 \)).
- The concentration of hydrogen gas (\( H_2 \)) is taken as 1 (since it is a gas).

### Step 4: Calculate the Concentration of H⁺ Ions
Given that pH = 4, we can find the concentration of hydrogen ions:
\[ \text{pH} = -\log [\text{H}^+] \]
Thus,
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-4} \, \text{M} \]

### Step 5: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{1}{10^{-4}} \right) \]
This simplifies to:
\[ E = -\frac{0.0591}{2} \log (10^4) \]

### Step 6: Simplify the Logarithm
Since \( \log (10^4) = 4 \):
\[ E = -\frac{0.0591}{2} \times 4 \]

### Step 7: Calculate the Final Value
Now, calculate the value:
\[ E = -0.0591 \times 2 = -0.1182 \, \text{V} \]

### Step 8: Conclusion
Thus, the reduction potential of the hydrogen electrode at pH = 4 is:
\[ E = -0.236 \, \text{V} \]

### Final Answer
The reduction potential of the hydrogen electrode containing a solution of pH = 4 is \(-0.236 \, \text{V}\).

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