The standard oxidation potential of Ni/Ni2+ electrode is 0.3 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution with the measured e.m.f. be zero at 25°C? (Assume [Ni2+] = 1M)

Options

A. 5.08
B. 4
C. 4.5
D. 5.25

Answer

To solve the problem, we will use the Nernst equation and the information provided about the standard oxidation potential of the Ni/Ni²⁺ electrode and the hydrogen electrode.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Standard oxidation potential of Ni/Ni²⁺ (E°_ox) = 0.3 V
- The concentration of Ni²⁺ ([Ni²⁺]) = 1 M
- The standard reduction potential of the hydrogen electrode (E°_red) = 0 V (by definition).

2. **Write the Half-Reactions:**
- Oxidation half-reaction: Ni → Ni²⁺ + 2e⁻
- Reduction half-reaction: 2H⁺ + 2e⁻ → H₂

3. **Calculate the Standard Cell Potential (E°_cell):**
\[ E°_{cell} = E°_{ox} + E°_{red} = 0.3 \, \text{V} + 0 \, \text{V} = 0.3 \, \text{V} \]

4. **Apply the Nernst Equation:**
The Nernst equation is given by:
\[ E_{cell} = E°_{cell} - \frac{RT}{nF} \ln Q \]
At 25°C (298 K), we can use the simplified version:
\[ E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q \]
Here, \( n = 2 \) (number of moles of electrons transferred).

5. **Determine the Reaction Quotient (Q):**
For the given reaction:
\[ Q = \frac{[H^+]^2}{[Ni^{2+}]} \]
Given that [Ni²⁺] = 1 M, we can simplify Q to:
\[ Q = [H^+]^2 \]

6. **Set E_cell to Zero:**
To find the pH at which the e.m.f. is zero:
\[ 0 = E°_{cell} - \frac{0.0592}{2} \log([H^+]^2) \]
Substituting the values:
\[ 0 = 0.3 - \frac{0.0592}{2} \log([H^+]^2) \]

7. **Rearranging the Equation:**
\[ \frac{0.0592}{2} \log([H^+]^2) = 0.3 \]
\[ \log([H^+]^2) = \frac{0.3 \times 2}{0.0592} \]
\[ \log([H^+]^2) = \frac{0.6}{0.0592} \approx 10.12 \]

8. **Calculate [H⁺]:**
\[ [H^+]^2 = 10^{10.12} \]
\[ [H^+] \approx 10^{5.06} \]

9. **Determine the pH:**
\[ pH = -\log([H^+]) \approx -\log(10^{5.06}) = 5.06 \]

### Final Answer:
The pH of the solution at which the measured e.m.f. will be zero is approximately **5.06**.