The value of `E_("cell")` of hydrogen electrode at `pH=0`, 298 K
and 1 atm, is

Options

A. 0.59V
B. zero volt
C. `-0.59V`
D. `-0.059V`

Explanation

To find the value of \( E_{\text{cell}} \) of the hydrogen electrode at \( \text{pH} = 0 \), 298 K, and 1 atm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between pH and H\(^+\) concentration**:
\[ \text{pH} = -\log[\text{H}^+] \]
Given that \( \text{pH} = 0 \), we can find the concentration of H\(^+\) ions.

2. **Calculate the concentration of H\(^+\)**:
\[ 0 = -\log[\text{H}^+] \]
This implies:
\[ [\text{H}^+] = 10^{0} = 1 \, \text{M} \]

3. **Use the Nernst equation**:
The Nernst equation for the hydrogen electrode is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{1}{[\text{H}^+]} \right) \]
Here, \( E^\circ_{\text{cell}} = 0 \, \text{V} \) for the standard hydrogen electrode, and \( n = 1 \) (since one electron is involved in the half-reaction).

4. **Substituting the values into the Nernst equation**:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{1} \log \left( \frac{1}{1} \right) \]
Since \( [\text{H}^+] = 1 \, \text{M} \), we have:
\[ E_{\text{cell}} = 0 - 0.0591 \log(1) \]

5. **Calculate the logarithm**:
The logarithm of 1 is 0:
\[ \log(1) = 0 \]
Therefore:
\[ E_{\text{cell}} = 0 - 0.0591 \times 0 = 0 \]

6. **Final answer**:
\[ E_{\text{cell}} = 0 \, \text{V} \]

### Conclusion:
The value of \( E_{\text{cell}} \) of the hydrogen electrode at \( \text{pH} = 0 \), 298 K, and 1 atm is \( 0 \, \text{V} \).