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

Explanation

To solve the problem, we need to find the pH of the solution at which the measured electromotive force (e.m.f.) of the cell combining the nickel electrode and the hydrogen electrode is zero. Here are the steps to derive the solution:

### Step 1: Understand the Standard Potentials
The standard oxidation potential of the `Ni/Ni^(2+)` electrode is given as `0.236 V`. The standard reduction potential of the hydrogen electrode is `0 V`.

### Step 2: Write the Half-Reactions
1. **Oxidation at Anode (Nickel):**
\[ Ni \rightarrow Ni^{2+} + 2e^- \]
(This is the oxidation reaction, and its potential is `0.236 V`.)

2. **Reduction at Cathode (Hydrogen):**
\[ 2H^+ + 2e^- \rightarrow H_2 \]
(This is the reduction reaction, and its potential is `0 V`.)

### Step 3: Write the Overall Cell Reaction
The overall cell reaction can be written as:
\[ Ni + 2H^+ \rightarrow Ni^{2+} + H_2 \]

### Step 4: Calculate the Standard Cell Potential (E°cell)
The standard cell potential (E°cell) is calculated as:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
Substituting the values:
\[ E°_{cell} = 0 - 0.236 = -0.236 \, V \]

### Step 5: Use the Nernst Equation
The Nernst equation for the cell can be expressed as:
\[ E = E°_{cell} - \frac{0.059}{n} \log \left( \frac{[Ni^{2+}]}{[H^+]^2} \cdot P_{H_2} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( [Ni^{2+}] = 1 \, M \)
- \( P_{H_2} = 1 \, atm \)

### Step 6: Set E = 0 for the Condition Given
For the e.m.f. to be zero:
\[ 0 = -0.236 - \frac{0.059}{2} \log \left( \frac{1}{[H^+]^2} \cdot 1 \right) \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ 0.236 = \frac{0.059}{2} \log \left( \frac{1}{[H^+]^2} \right) \]
\[ 0.236 = \frac{0.059}{2} (-2 \log [H^+]) \]
\[ 0.236 = -0.059 \log [H^+] \]

### Step 8: Solve for [H+]
Now, solving for \([H^+]\):
\[ \log [H^+] = -\frac{0.236}{0.059} \]
Calculating the right side:
\[ \log [H^+] \approx -4 \]
Thus:
\[ [H^+] = 10^{-4} \, M \]

### Step 9: Calculate pH
The pH is defined as:
\[ pH = -\log [H^+] \]
Substituting the value of \([H^+]\):
\[ pH = -\log (10^{-4}) = 4 \]

### Final Answer
Thus, the pH of the solution at which the measured e.m.f. will be zero is:
\[ \text{pH} = 4 \]