If `No_(3)^(-) rarr NO_(2)` (acid medium) , `E^(@) = 0.790 V`
and `NO_(3)^(-) rarr NH_(2)OH` (acid medium) , `E^(@) = 0.731 V`
At what `pH` of the above two half reaction will have some `E` values? Assume the concentrations of all other species be unity.

Answer

To solve the problem, we need to find the pH at which the two half-reactions have the same electrode potential (E). We will use the Nernst equation for both half-reactions and set them equal to each other.

### Step-by-Step Solution:

1. **Write the half-reactions and their standard potentials**:
- For the first half-reaction:
\[ \text{NO}_3^{-} + 2\text{H}^+ + e^{-} \rightarrow \text{NO}_2 + \text{H}_2\text{O} \quad E^\circ = 0.790 \, \text{V} \]
- For the second half-reaction:
\[ \text{NO}_3^{-} + 7\text{H}^+ + 6e^{-} \rightarrow \text{NH}_2\text{OH} + 2\text{H}_2\text{O} \quad E^\circ = 0.731 \, \text{V} \]

2. **Apply the Nernst equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
- For the first half-reaction:
\[ E_1 = 0.790 - \frac{0.059}{1} \log \frac{1}{[\text{H}^+]^2} \]
\[ E_1 = 0.790 - 0.059 \log \frac{1}{[\text{H}^+]^2} = 0.790 + 0.118 \log [\text{H}^+] \]

- For the second half-reaction:
\[ E_2 = 0.731 - \frac{0.059}{6} \log \frac{1}{[\text{H}^+]^7} \]
\[ E_2 = 0.731 - \frac{0.059}{6} \cdot (-7 \log [\text{H}^+]) = 0.731 + \frac{0.059 \cdot 7}{6} \log [\text{H}^+] \]
\[ E_2 = 0.731 + 0.0685 \log [\text{H}^+] \]

3. **Set the two equations equal**:
\[ 0.790 + 0.118 \log [\text{H}^+] = 0.731 + 0.0685 \log [\text{H}^+] \]

4. **Rearranging the equation**:
\[ 0.790 - 0.731 = 0.0685 \log [\text{H}^+] - 0.118 \log [\text{H}^+] \]
\[ 0.059 = (0.0685 - 0.118) \log [\text{H}^+] \]
\[ 0.059 = -0.0495 \log [\text{H}^+] \]

5. **Solving for \(\log [\text{H}^+]\)**:
\[ \log [\text{H}^+] = \frac{0.059}{-0.0495} \approx -1.19 \]

6. **Finding \([\text{H}^+]\)**:
\[ [\text{H}^+] = 10^{-1.19} \approx 0.064 \]

7. **Calculating pH**:
\[ \text{pH} = -\log [\text{H}^+] \approx -\log(0.064) \approx 1.19 \]

### Final Answer:
The pH at which the two half-reactions have the same E values is approximately **1.19**.