The standard reduction potential for the half cell `:`
`NO_(3)^(c-)(aq)+2H^(c-) +e^(-) rarr NO_(2)(g)+H_(2)O` is `0.78V`.
Calculate the reduction potential in `8M H^(o+)`.

Explanation

To calculate the reduction potential in 8M H⁺, we will use the Nernst equation. The Nernst equation relates the standard reduction potential to the concentrations of the reactants and products in a redox reaction.

### Step-by-Step Solution:

1. **Identify the given values:**
- Standard reduction potential (E°) = 0.78 V
- Concentration of H⁺ ions = 8 M
- Number of electrons transferred (n) = 1 (from the half-reaction)

2. **Write the Nernst equation:**
The Nernst equation is given by:
\[ E = E° + \frac{0.059}{n} \log \left(\frac{[\text{products}]}{[\text{reactants}]}\right) \]
For our half-reaction:
\[ \text{NO}_3^{-} + 2 \text{H}^+ + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \]
The reactants are H⁺ ions and the products are NO₂ and H₂O. However, since H₂O is a liquid, it does not appear in the expression.

3. **Set up the Nernst equation for our reaction:**
Since we only have H⁺ ions as reactants, we can simplify the equation:
\[ E = E° + \frac{0.059}{1} \log \left(\frac{1}{[\text{H}^+]^2}\right) \]
Here, we take the concentration of H⁺ ions raised to the power of its stoichiometric coefficient (which is 2).

4. **Substitute the values into the Nernst equation:**
\[ E = 0.78 + 0.059 \log \left(\frac{1}{8^2}\right) \]
\[ E = 0.78 + 0.059 \log \left(\frac{1}{64}\right) \]
\[ E = 0.78 + 0.059 \log (64^{-1}) \]
\[ E = 0.78 + 0.059 \cdot (-2 \log(8)) \]

5. **Calculate the logarithm:**
\[ \log(8) = 0.903 \]
Therefore:
\[ E = 0.78 + 0.059 \cdot (-2 \cdot 0.903) \]
\[ E = 0.78 - 0.1062 \]

6. **Final calculation:**
\[ E = 0.78 - 0.1062 = 0.6738 \text{ V} \]

### Final Answer:
The reduction potential in 8M H⁺ is approximately **0.6738 V**.