The standard oxidation potential of `Ni//Ni^(2+)` electrode is `0.236V`. If this is combined with a hydrogen electrode in acid solution. At what pH of the solution will the measured emf be zero at `25^(@)C` ?
Assume `[Ni^(2+)]=1M` and `p_(H_(2))=1` atm

Explanation

To solve the problem, we need to find the pH at which the measured emf of the combined `Ni//Ni^(2+)` electrode and the hydrogen electrode is zero. We will use the Nernst equation for this purpose.

### Step-by-Step Solution:

1. **Understanding the Electrode Potentials:**
The standard oxidation potential of the `Ni//Ni^(2+)` electrode is given as `0.236 V`. The standard reduction potential for the hydrogen electrode (SHE) is `0 V`.

2. **Setting up the Nernst Equation:**
The Nernst equation for the half-cell reaction is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \(E\) is the cell potential,
- \(E^\circ\) is the standard electrode potential,
- \(R\) is the universal gas constant (8.314 J/(mol·K)),
- \(T\) is the temperature in Kelvin (25°C = 298 K),
- \(n\) is the number of moles of electrons transferred (for Ni, \(n = 2\)),
- \(F\) is Faraday's constant (96485 C/mol),
- \(Q\) is the reaction quotient.

3. **Calculating the Reaction Quotient (Q):**
The reaction for the `Ni//Ni^(2+)` half-cell can be represented as:
\[ Ni^{2+} + 2e^- \leftrightarrow Ni \]
The reaction quotient \(Q\) is given by:
\[ Q = \frac{1}{[Ni^{2+}]} \]
Given that \([Ni^{2+}] = 1 M\), we have:
\[ Q = 1 \]

4. **Substituting Values into the Nernst Equation:**
At the point where the measured emf is zero, we set \(E = 0\):
\[ 0 = 0.236 - \frac{(8.314)(298)}{(2)(96485)} \ln(1) \]
Since \(\ln(1) = 0\), the equation simplifies to:
\[ 0 = 0.236 \]
This indicates that we need to consider the hydrogen electrode's potential in relation to the pH.

5. **Considering the Hydrogen Electrode:**
The potential of the hydrogen electrode in terms of pH is given by:
\[ E_{H_2} = 0 - 0.0591 \times \text{pH} \]
Setting the total cell potential to zero:
\[ 0 = 0.236 - 0.0591 \times \text{pH} \]

6. **Solving for pH:**
Rearranging the equation gives:
\[ 0.0591 \times \text{pH} = 0.236 \]
\[ \text{pH} = \frac{0.236}{0.0591} \approx 4.00 \]

### Final Answer:
The pH of the solution at which the measured emf will be zero is approximately **4.00**.