Emf of a cell corresponding to the reaction:-
`Zn(s) +2H^(+) (aq) rarr Zn^(2+)(aq) [0.1M] +H_(2)(g) [1.0atm]` is `0.49V` at `25^(@)C, E_(Zn^(2+)//Zn)^(@) = - 0.76 V`.
The `pH` of solution in cathode chamber is.

Explanation

To solve the problem, we need to find the pH of the solution in the cathode chamber corresponding to the given cell reaction. We will use the Nernst equation for this purpose.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell reaction is given as:
\[ \text{Zn(s)} + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) [0.1M] + \text{H}_2(g) [1.0 atm] \]
Here, zinc (Zn) is oxidized to zinc ions (\(\text{Zn}^{2+}\)), and hydrogen ions (\(\text{H}^+\)) are reduced to hydrogen gas (\(\text{H}_2\)).

2. **Determine the Standard Electrode Potentials**:
The standard reduction potential for zinc is given as:
\[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \]
The standard reduction potential for the hydrogen electrode (cathode) is:
\[ E^\circ_{\text{H}^+/H_2} = 0 \, \text{V} \]

3. **Calculate the Standard Cell Potential**:
The standard cell potential (\(E^\circ_{\text{cell}}\)) can be calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0 - (-0.76) = 0.76 \, \text{V} \]

4. **Apply the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left(\frac{[\text{products}]}{[\text{reactants}]}\right) \]
Here, \(n\) (the number of electrons transferred) is 2, as 2 electrons are involved in the reaction.

5. **Substitute Known Values**:
We know:
- \(E_{\text{cell}} = 0.49 \, \text{V}\)
- \(E^\circ_{\text{cell}} = 0.76 \, \text{V}\)
- Concentration of \(\text{Zn}^{2+} = 0.1 \, \text{M}\)
- Pressure of \(\text{H}_2 = 1.0 \, \text{atm}\)

The Nernst equation becomes:
\[ 0.49 = 0.76 - \frac{0.059}{2} \log \left(\frac{0.1 \times 1}{(1)^2}\right) \]

6. **Simplify the Equation**:
\[ 0.49 = 0.76 - 0.0295 \log(0.1) \]
Since \(\log(0.1) = -1\):
\[ 0.49 = 0.76 + 0.0295 \]
\[ 0.49 = 0.76 + 0.0295 \]
\[ 0.49 = 0.7895 \]
Rearranging gives us:
\[ 0.49 - 0.76 = -0.0295 \log(0.1) \]
\[ -0.27 = -0.0295 \log(0.1) \]

7. **Calculate the Concentration of \(\text{H}^+\)**:
\[ -0.27 = -0.0295 \cdot (-1) \]
\[ \log[\text{H}^+] = \frac{-0.27}{-0.0295} \approx 9.15 \]
Therefore:
\[ [\text{H}^+] = 10^{-9.15} \approx 7.08 \times 10^{-10} \, \text{M} \]

8. **Calculate the pH**:
\[ \text{pH} = -\log[\text{H}^+] \approx -\log(7.08 \times 10^{-10}) \approx 9.15 \]

### Final Answer:
The pH of the solution in the cathode chamber is approximately **9.15**.