The potential of a hydrogen electrode in a solution with `pOH=4` at `25^(@)C` is

Options

A. `+0.59`
B. `-0.59`
C. `-0.295`
D. `0.295`

Explanation

To find the potential of a hydrogen electrode in a solution with a given pOH of 4 at 25°C, we can follow these steps:

### Step 1: Calculate pH from pOH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]

Given that:
\[ \text{pOH} = 4 \]

We can calculate pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4 = 10 \]

### Step 2: Calculate the concentration of H⁺ ions
The concentration of hydrogen ions \([H^+]\) can be calculated from the pH:
\[ \text{pH} = -\log[H^+] \]

Thus,
\[ [H^+] = 10^{-\text{pH}} = 10^{-10} \, \text{M} \]

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential (E) to the standard electrode potential (E°) and the concentrations of the reactants and products:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]

For the hydrogen electrode reaction:
\[ 2H^+ + 2e^- \leftrightarrow H_2 \]

Here, \( n = 2 \) (since 2 electrons are involved).

Given that the standard reduction potential \( E^\circ \) for the hydrogen electrode is 0 V, we can write:
\[ E = 0 - \frac{0.059}{2} \log \left( \frac{1}{[H^+]^2} \right) \]

### Step 4: Substitute the concentration of H⁺
Since \([H^+] = 10^{-10} \, \text{M}\):
\[ E = -\frac{0.059}{2} \log \left( \frac{1}{(10^{-10})^2} \right) \]

This simplifies to:
\[ E = -\frac{0.059}{2} \log \left( 10^{20} \right) \]

### Step 5: Calculate the logarithm
Using the property of logarithms:
\[ \log(10^{20}) = 20 \]

So we have:
\[ E = -\frac{0.059}{2} \times 20 \]

### Step 6: Calculate the final potential
Now, calculate:
\[ E = -\frac{0.059 \times 20}{2} = -0.59 \, \text{V} \]

Thus, the potential of the hydrogen electrode in the solution with pOH = 4 at 25°C is:
\[ \boxed{-0.59 \, \text{V}} \]

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