The half cell reduction potential of a hydrogen electrode at `pH = 5 ` will be :

Options

A. `0.29V`
B. `-0.29V`
C. `0.059V`
D. `-0.059V`

Answer

To find the half-cell reduction potential of a hydrogen electrode at pH = 5, we can follow these steps:

### Step 1: Understand the relationship between pH and [H⁺]
The pH is defined as:
\[ \text{pH} = -\log[\text{H}^+] \]
Given that pH = 5, we can find the concentration of hydrogen ions \([\text{H}^+]\):
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

### Step 2: Write the half-cell reaction for the hydrogen electrode
The half-cell reaction for the hydrogen electrode can be represented as:
\[ \text{H}^+ + e^- \rightarrow \frac{1}{2} \text{H}_2 \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log\left(\frac{[\text{oxidized form}]}{[\text{reduced form}]}\right) \]
For the hydrogen electrode, the standard reduction potential \(E^\circ\) is 0 V, and \(n = 1\) (since one electron is involved). The oxidized form is \([\text{H}^+]\) and the reduced form is \([\text{H}_2]\), which we take as 1 (since it is in its standard state).

### Step 4: Substitute values into the Nernst equation
Substituting the values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{1} \log\left(\frac{1}{[\text{H}^+]}\right) \]
Since \([\text{H}^+] = 10^{-5}\):
\[ E = -0.0591 \log\left(\frac{1}{10^{-5}}\right) \]
\[ E = -0.0591 \log(10^5) \]
\[ E = -0.0591 \times 5 \]
\[ E = -0.2955 \, \text{V} \]

### Step 5: Final answer
Thus, the half-cell reduction potential of the hydrogen electrode at pH = 5 is:
\[ E \approx -0.30 \, \text{V} \]