What is the potential of an electrode which originally contained `0.1 M NO_(3)^(-)` and `0.4 M H^(+)` and which has been treated by `80%` of the cadmium necessary to reduce all the `NO_(3)^(-)` to `NO(g)` at 1 bar ?
Given : `NO_(3)^(-)+4H^(+)+3e^(-)rarr NO+2H_(2)O`,
`E^(@)=0.96 , log 2 = 0.3`

Options

A. `0.84 V`
B. `1.08 V`
C. `1.23 V`
D. `1.36 V`

Explanation

To solve the problem, we need to determine the potential of the electrode after it has been treated with 80% of the cadmium necessary to reduce all the nitrate ions to nitric oxide (NO).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reduction reaction is given as:
\[ \text{NO}_3^{-} + 4\text{H}^{+} + 3\text{e}^{-} \rightarrow \text{NO} + 2\text{H}_2\text{O} \]

2. **Initial Concentrations**:
- Concentration of \(\text{NO}_3^{-}\) = 0.1 M
- Concentration of \(\text{H}^{+}\) = 0.4 M

3. **Calculate the Amount of Cadmium Used**:
Since 80% of the cadmium necessary to reduce all \(\text{NO}_3^{-}\) to \(\text{NO}\) is used, we first need to determine how many moles of \(\text{NO}_3^{-}\) are present:
- Moles of \(\text{NO}_3^{-}\) = 0.1 M
- Moles of \(\text{H}^{+}\) = 0.4 M

The stoichiometry of the reaction shows that 1 mole of \(\text{NO}_3^{-}\) requires 3 moles of electrons. Therefore, to reduce 0.1 M of \(\text{NO}_3^{-}\), we need:
\[ \text{Electrons required} = 0.1 \, \text{M} \times 3 = 0.3 \, \text{M} \]

Since 80% of the cadmium is used:
\[ \text{Electrons used} = 0.8 \times 0.3 \, \text{M} = 0.24 \, \text{M} \]

4. **Calculate Remaining Concentrations**:
After using 80% of the cadmium:
- Remaining \(\text{NO}_3^{-}\):
\[ \text{NO}_3^{-} = 0.1 - 0.08 = 0.02 \, \text{M} \]
- Remaining \(\text{H}^{+}\):
\[ \text{H}^{+} = 0.4 - 4 \times 0.08 = 0.4 - 0.32 = 0.08 \, \text{M} \]

5. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^{\circ} - \frac{0.0591}{n} \log Q \]
where \(E^{\circ} = 0.96 \, \text{V}\), \(n = 3\) (number of electrons transferred), and \(Q\) is the reaction quotient.

6. **Calculate the Reaction Quotient \(Q\)**:
The reaction quotient \(Q\) can be calculated as:
\[ Q = \frac{P_{\text{NO}}}{[\text{NO}_3^{-}][\text{H}^{+}]^4} \]
Given that the pressure of \(\text{NO}\) is 1 bar:
\[ Q = \frac{1}{0.02 \times (0.08)^4} \]

Calculate \((0.08)^4\):
\[ (0.08)^4 = 0.0004096 \]
Therefore,
\[ Q = \frac{1}{0.02 \times 0.0004096} = \frac{1}{0.000008192} \approx 122,000 \]

7. **Substitute into the Nernst Equation**:
\[ E = 0.96 - \frac{0.0591}{3} \log(122,000) \]
Calculate \(\log(122,000) \approx 5.086\):
\[ E = 0.96 - \frac{0.0591}{3} \times 5.086 \]
\[ E = 0.96 - 0.0197 \times 5.086 \approx 0.96 - 0.100 \]
\[ E \approx 0.84 \, \text{V} \]

### Final Answer:
The potential of the electrode is approximately **0.84 V**.