The standard oxidation potential of `Ni//Ni^(+2)` electrode is `0.236` V. If this electrode 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)] = 1`M

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to determine the pH at which the measured EMF of the combined nickel and hydrogen electrode system is zero at 25°C. We will use the Nernst equation and the given standard oxidation potential for the nickel electrode.

### Step-by-step Solution:

1. **Identify the Standard Potentials**:
- The standard oxidation potential for the nickel electrode (Ni/Ni²⁺) is given as \( E^\circ_{Ni/Ni^{2+}} = 0.236 \, V \).
- The standard reduction potential for the hydrogen electrode (H⁺/H₂) is \( E^\circ_{H^+/H_2} = 0.00 \, V \).

2. **Set Up the Overall Cell Reaction**:
- The half-reaction for nickel oxidation is:
\[ Ni \rightarrow Ni^{2+} + 2e^- \]
- The half-reaction for hydrogen reduction is:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
- The overall cell reaction is:
\[ Ni + 2H^+ \rightarrow Ni^{2+} + H_2 \]

3. **Calculate the Standard Cell Potential**:
- The standard cell potential \( E^\circ_{cell} \) can be calculated as:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{H^+/H_2} - E^\circ_{Ni/Ni^{2+}} = 0.00 - 0.236 = -0.236 \, V \]

4. **Use the Nernst Equation**:
- The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
- Here, \( n = 2 \) (number of electrons transferred), and \( Q \) is the reaction quotient:
\[ Q = \frac{[Ni^{2+}]}{[H^+]^2 \cdot P_{H_2}} \]
- Given that \([Ni^{2+}] = 1 \, M\) and \(P_{H_2} = 1 \, atm\), we can simplify \( Q \) to:
\[ Q = \frac{1}{[H^+]^2} \]

5. **Set the EMF to Zero**:
- We want to find the pH when the measured EMF is zero:
\[ 0 = -0.236 - \frac{0.0591}{2} \log \left( \frac{1}{[H^+]^2} \right) \]
- Rearranging gives:
\[ 0.236 = \frac{0.0591}{2} \log \left( \frac{1}{[H^+]^2} \right) \]

6. **Calculate the Logarithm**:
- Simplifying further:
\[ 0.236 = 0.02955 \log \left( \frac{1}{[H^+]^2} \right) \]
- This implies:
\[ \log \left( \frac{1}{[H^+]^2} \right) = \frac{0.236}{0.02955} \approx 8 \]
- Thus,
\[ \log [H^+]^2 = -8 \implies [H^+]^2 = 10^{-8} \implies [H^+] = 10^{-4} \, M \]

7. **Determine the pH**:
- The pH is calculated as:
\[ pH = -\log [H^+] = -\log (10^{-4}) = 4 \]

### Final Answer:
The pH of the solution at which the measured EMF will be zero is **4**.