The electrode potential of hydrogen electrode at the pH=12 will be

Options

A. 0
B. ` +ve`
C. ` -ve`
D. unpredictable.

Explanation

To calculate the electrode potential of the hydrogen electrode at pH = 12, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Nernst Equation:**
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{1}{[H^+]} \]
For the hydrogen electrode, \(E^{\circ}_{\text{cell}} = 0\) V.

2. **Determine the Concentration of Hydrogen Ions:**
We know that pH is related to the concentration of hydrogen ions \([H^+]\) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that pH = 12, we can find \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} = 10^{-12} \, \text{M} \]

3. **Substitute Values into the Nernst Equation:**
Now we substitute the values into the Nernst equation. Since \(n = 1\) (one electron is involved in the half-reaction):
\[ E_{\text{cell}} = 0 - \frac{0.0591}{1} \log \frac{1}{10^{-12}} \]

4. **Simplify the Logarithmic Term:**
The logarithm can be simplified:
\[ \log \frac{1}{10^{-12}} = \log 10^{12} = 12 \]

5. **Calculate the Electrode Potential:**
Now we can substitute this back into the equation:
\[ E_{\text{cell}} = 0 - 0.0591 \times 12 \]
\[ E_{\text{cell}} = -0.7092 \, \text{V} \]

6. **Conclusion:**
The electrode potential of the hydrogen electrode at pH = 12 is approximately \(-0.709 \, \text{V}\).

### Final Answer:
The electrode potential of the hydrogen electrode at pH = 12 is \(-0.709 \, \text{V}\).

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