pH of the solution in the anode compartment of the following cell at `25^@C` is x when `E_("cell")-E_("cell")^(@)=0.0591V`.
`underset("1 atm")(Pt,H_(2)|pH=X|| Ni^(2+)(M)|Ni`
Find the value of x.

Explanation

To solve the problem, we need to find the pH (denoted as x) of the solution in the anode compartment of the given electrochemical cell at 25°C when the cell potential difference is 0.0591 V.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- At the cathode, the reduction reaction is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]
The standard reduction potential (\(E^\circ\)) for this reaction is -0.25 V.
- At the anode, the oxidation reaction is:
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]
The standard reduction potential (\(E^\circ\)) for hydrogen is 0 V.

2. **Calculate the Cell Potential**:
- The overall cell potential \(E_{\text{cell}}\) can be expressed using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
- Here, \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.25) - (0) = -0.25 \, \text{V}\).
- Given that \(E_{\text{cell}} - E^\circ_{\text{cell}} = 0.0591 \, \text{V}\), we can write:
\[ 0.0591 = -\frac{0.0591}{2} \log Q \]

3. **Determine the Reaction Quotient (Q)**:
- The reaction quotient \(Q\) for the anode reaction can be expressed as:
\[ Q = \frac{[\text{H}^+]^2}{[\text{Ni}^{2+}]} \]
- Given that the concentration of \(\text{Ni}^{2+}\) is \(M\) (1 M in this case), we can substitute:
\[ Q = \frac{[\text{H}^+]^2}{1} = [\text{H}^+]^2 \]

4. **Substituting into the Nernst Equation**:
- Rearranging the equation:
\[ 0.0591 = -\frac{0.0591}{2} \log [\text{H}^+]^2 \]
- Simplifying gives:
\[ 2 = -\log [\text{H}^+]^2 \]
- This can be rewritten as:
\[ 2 = -2 \log [\text{H}^+] \]

5. **Solving for pH**:
- Dividing both sides by 2:
\[ 1 = -\log [\text{H}^+] \]
- Thus, we find:
\[ \text{pH} = 1 \]

6. **Conclusion**:
- Therefore, the value of \(x\) (the pH of the solution in the anode compartment) is:
\[ x = 1 \]