At what pH of HCl solution, will hydrogen gas electrode show electrode potential of -0.118 V ? `H_(2)` gas is bubbled at 298 K and 1 atm pressure.

Explanation

To determine the pH of the HCl solution at which the hydrogen gas electrode shows an electrode potential of -0.118 V, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Electrode potential (E_cell) = -0.118 V
- Temperature (T) = 298 K
- Pressure of H2 gas = 1 atm

2. **Use the Nernst Equation:**
The Nernst equation for the hydrogen electrode can be expressed as:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \left( \frac{1}{[H^+]}\right) \]
where:
- \(E^{\circ}_{cell}\) = standard electrode potential (0 V for the hydrogen electrode)
- \(n\) = number of electrons transferred (for H2, n = 2, but since we are considering the half-reaction, n = 1)
- \([H^+]\) = concentration of hydrogen ions

3. **Substituting Known Values:**
Since \(E^{\circ}_{cell} = 0\) for the standard hydrogen electrode, we can simplify the equation:
\[ E_{cell} = 0 - \frac{0.0591}{1} \log \left( \frac{1}{[H^+]}\right) \]
This can be rewritten as:
\[ -0.118 = -0.0591 \log \left( \frac{1}{[H^+]}\right) \]

4. **Rearranging the Equation:**
To isolate the logarithm, we can multiply both sides by -1:
\[ 0.118 = 0.0591 \log \left( [H^+] \right) \]

5. **Solving for \([H^+]\):**
Divide both sides by 0.0591:
\[ \log \left( [H^+] \right) = \frac{0.118}{0.0591} \]
Calculate the right side:
\[ \log \left( [H^+] \right) \approx 2 \]

6. **Finding \([H^+]\):**
To find \([H^+]\), we take the antilogarithm:
\[ [H^+] = 10^{-2} = 0.01 \, \text{M} \]

7. **Calculating pH:**
The pH is calculated using the formula:
\[ \text{pH} = -\log [H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(0.01) = 2 \]

### Final Answer:
The pH of the HCl solution at which the hydrogen gas electrode shows an electrode potential of -0.118 V is **2**.