The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)

Options

A. `0.51` volt
B. 0 volt
C. `-0.591` volt
D. `0.059` volt

Explanation

To calculate the reduction potential of a hydrogen electrode at pH 10 and 298 K, we will use the Nernst equation. Here’s the step-by-step solution:

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

### Step 2: Determine the Concentration of Hydrogen Ions
Given that pH = 10, we can calculate the concentration of hydrogen ions \([\text{H}^+]\) using the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-10} \, \text{M} \]

### Step 3: Write the Nernst Equation
The Nernst equation for the hydrogen electrode can be expressed as:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
Where:
- \(E^\circ\) is the standard reduction potential (0 V for the hydrogen electrode).
- \(n\) is the number of electrons transferred (1 for this reaction).

### Step 4: Substitute Values into the Nernst Equation
For our reaction:
- The product is \(\frac{1}{2} \text{H}_2\) at a pressure of 1 atm, which can be expressed as \((1)^{1/2} = 1\).
- The reactant is \([\text{H}^+] = 10^{-10} \, \text{M}\).

Substituting these values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{1} \log \left( \frac{1^{1/2}}{10^{-10}} \right) \]

### Step 5: Simplify the Logarithm
This simplifies to:
\[ E = -0.0591 \log \left( \frac{1}{10^{-10}} \right) \]
\[ E = -0.0591 \log(10^{10}) \]
\[ E = -0.0591 \times 10 \]

### Step 6: Calculate the Final Value
Calculating this gives:
\[ E = -0.591 \, \text{V} \]

### Conclusion
Thus, the reduction potential of the hydrogen electrode at pH 10 and 298 K is:
\[ \boxed{-0.591 \, \text{V}} \]

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