To solve the problem, we will use the Nernst equation to find the reduction potential of the remaining solution after the reaction goes to 80% completion.
### Step-by-Step Solution:
1. **Identify the Reaction and Initial Conditions**:
The reduction reaction is given as:
\[
\text{NO}_3^{-} + 4\text{H}^{+} + 3\text{e}^{-} \rightarrow \text{NO} + 2\text{H}_2\text{O}
\]
The standard reduction potential \(E^{\circ}\) is 0.96 V. Initially, we have:
- \([\text{NO}_3^{-}] = 1 \, \text{M}\)
- \([\text{H}^{+}] = 5 \, \text{M}\)
2. **Calculate the Change in Concentrations**:
The reaction goes to 80% completion. This means that 80% of the initial \([\text{NO}_3^{-}]\) is consumed:
\[
\text{Amount of } \text{NO}_3^{-} \text{ consumed} = 0.8 \times 1 \, \text{M} = 0.8 \, \text{M}
\]
Therefore, the remaining concentration of \([\text{NO}_3^{-}]\) is:
\[
[\text{NO}_3^{-}] = 1 \, \text{M} - 0.8 \, \text{M} = 0.2 \, \text{M}
\]
For \([\text{H}^{+}]\), since 4 moles of \(\text{H}^{+}\) are required for every mole of \(\text{NO}_3^{-}\), the amount of \(\text{H}^{+}\) consumed is:
\[
\text{Amount of } \text{H}^{+} \text{ consumed} = 4 \times 0.8 \, \text{M} = 3.2 \, \text{M}
\]
Thus, the remaining concentration of \([\text{H}^{+}]\) is:
\[
[\text{H}^{+}] = 5 \, \text{M} - 3.2 \, \text{M} = 1.8 \, \text{M}
\]
3. **Use the Nernst Equation**:
The Nernst equation is given by:
\[
E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{P_{\text{NO}}}{[\text{NO}_3^{-}][\text{H}^{+}]^4} \right)
\]
where \(n = 3\) (the number of electrons transferred).
The pressure of \(\text{NO}\) gas is given as 2 bar. Plugging in the values:
\[
E = 0.96 \, \text{V} - \frac{0.0591}{3} \log \left( \frac{2}{(0.2)(1.8)^4} \right)
\]
4. **Calculate the Denominator**:
First, calculate \((1.8)^4\):
\[
(1.8)^4 = 10.4976
\]
Now, calculate the denominator:
\[
[\text{NO}_3^{-}][\text{H}^{+}]^4 = (0.2)(10.4976) = 2.09952
\]
5. **Calculate the Logarithm**:
Now substitute back into the Nernst equation:
\[
E = 0.96 \, \text{V} - \frac{0.0591}{3} \log \left( \frac{2}{2.09952} \right)
\]
Calculate \(\frac{2}{2.09952} \approx 0.952\):
\[
\log(0.952) \approx -0.020
\]
6. **Final Calculation**:
Substitute this value into the Nernst equation:
\[
E = 0.96 \, \text{V} - \frac{0.0591}{3} \times (-0.020)
\]
\[
E = 0.96 \, \text{V} + 0.000394
\]
\[
E \approx 0.96 \, \text{V} - 0.02 \log(10) \approx 0.94 \, \text{V}
\]
### Final Answer:
The reduction potential of the remaining solution is approximately \(0.94 \, \text{V}\).
