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 Nernst Equation
The Nernst equation relates the reduction potential of an electrochemical cell to the standard reduction potential and the concentrations of the reactants and products involved in the half-reaction.

The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[ \text{products} ]}{[ \text{reactants} ]} \right) \]

### Step 2: Identify the Reaction
For the hydrogen electrode, the half-reaction is:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
Here, \( n = 2 \) (the number of electrons transferred).

### Step 3: Calculate the Concentration of \( H^+ \)
Given that pH = 5, we can calculate the concentration of hydrogen ions (\( H^+ \)):
\[ \text{pH} = -\log[H^+] \]
\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

### Step 4: Substitute Values into the Nernst Equation
Since the standard reduction potential \( E^\circ \) for the hydrogen electrode is 0 V, we can substitute the values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{1}{(10^{-5})^2} \right) \]

### Step 5: Simplify the Logarithmic Expression
Calculating the logarithm:
\[ \log \left( \frac{1}{(10^{-5})^2} \right) = \log(10^{10}) = 10 \]

### Step 6: Calculate the Reduction Potential
Now substituting back into the equation:
\[ E = -\frac{0.0591}{2} \times 10 \]
\[ E = -0.2955 \, \text{V} \]

### Final Answer
The half-cell reduction potential of a hydrogen electrode at pH = 5 is:
\[ E = -0.2955 \, \text{V} \]

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