The standard oxidation potential for the half-cell
`NO_(2)^(-)(g)+H_(2)OtoNO_(3)^(-)(aq)+2H^(+)(aq)+2e` is `-0.78V`.
Calculate the reduction in 9 molar `H^(+)` assuming all other species at unit concentration. What will be the reduction potential in neutral medium?

Explanation

To solve the problem, we need to calculate the reduction potential for the half-cell reaction given the standard oxidation potential and the concentration of H⁺ ions. We will also find the reduction potential in neutral medium (pH 7).

### Step-by-Step Solution:

1. **Identify the Standard Oxidation Potential**:
The standard oxidation potential (E°_ox) for the half-cell reaction is given as:
\[ E°_{ox} = -0.78 \, V \]

2. **Use the Nernst Equation for Oxidation**:
The Nernst equation for the oxidation reaction is:
\[ E_{ox} = E°_{ox} - \frac{0.0591}{n} \log \left( \frac{[H^+]^2}{1} \right) \]
Here, \( n = 2 \) (the number of electrons transferred).

3. **Substitute the Values**:
Given that the concentration of \( H^+ \) is 9 M, we can substitute the values into the equation:
\[ E_{ox} = -0.78 - \frac{0.0591}{2} \log (9^2) \]
Calculate \( 9^2 = 81 \):
\[ E_{ox} = -0.78 - \frac{0.0591}{2} \log (81) \]

4. **Calculate the Logarithm**:
Using the logarithm:
\[ \log (81) = 4 \log (3) \approx 4 \times 0.477 = 1.908 \]
Substitute this back into the equation:
\[ E_{ox} = -0.78 - \frac{0.0591}{2} \times 1.908 \]

5. **Calculate the Final Oxidation Potential**:
\[ E_{ox} = -0.78 - 0.02955 \times 1.908 \approx -0.78 - 0.0585 \approx -0.8385 \, V \]

6. **Calculate the Reduction Potential**:
The reduction potential (E_reduction) is the negative of the oxidation potential:
\[ E_{red} = -E_{ox} = 0.8385 \, V \]

7. **Calculate the Reduction Potential in Neutral Medium**:
In neutral medium (pH = 7), the concentration of \( H^+ \) is \( 10^{-7} \, M \). We will use the Nernst equation again:
\[ E_{ox} = E°_{ox} - \frac{0.0591}{2} \log \left( \frac{(10^{-7})^2}{1} \right) \]
\[ E_{ox} = -0.78 - \frac{0.0591}{2} \log (10^{-14}) \]
Since \( \log (10^{-14}) = -14 \):
\[ E_{ox} = -0.78 - \frac{0.0591}{2} \times (-14) \]
\[ E_{ox} = -0.78 + 0.0591 \times 7 = -0.78 + 0.4137 \approx -0.3663 \, V \]

8. **Calculate the Reduction Potential in Neutral Medium**:
\[ E_{red} = -E_{ox} = 0.3663 \, V \]

### Final Answers:
- Reduction potential in 9 M \( H^+ \): **0.8385 V**
- Reduction potential in neutral medium: **0.3663 V**