`E^(@)` for `F_(2)+2e^(-) hArr 2F^(-)` is `2.8V,E^(@)` for `(1)/(2)F+e^(-)=F^(-)` is `-`

Options

A. `2.8V`
B. `1.4V`
C. `-2.8V`
D. `-1.4V`

Explanation

To solve the problem, we need to find the standard electrode potential (E°) for the reaction:

\[ F_2 + 2e^- \rightleftharpoons 2F^- \]

We are given that the standard electrode potential for this reaction is 2.8 V. We also have another half-reaction:

\[ \frac{1}{2}F_2 + e^- \rightleftharpoons F^- \]

with an unknown standard electrode potential.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The standard electrode potential (E°) for the reaction \( F_2 + 2e^- \rightleftharpoons 2F^- \) is 2.8 V.
- We need to find the standard electrode potential for the half-reaction \( \frac{1}{2}F_2 + e^- \rightleftharpoons F^- \).

2. **Recognize the Relationship Between the Reactions**:
- The second reaction is essentially half of the first reaction. When you halve a balanced redox reaction, the standard electrode potential remains the same.

3. **Apply the Concept of Intensive Properties**:
- Standard electrode potential (E°) is an intensive property, which means it does not change with the scale of the reaction. Therefore, if we halve the stoichiometry of the reaction, the E° remains unchanged.

4. **Conclude the Value of E° for the Half-Reaction**:
- Since the half-reaction \( \frac{1}{2}F_2 + e^- \rightleftharpoons F^- \) is half of the original reaction, the standard electrode potential for this half-reaction is also 2.8 V.

5. **Final Answer**:
- The standard electrode potential for the reaction \( \frac{1}{2}F_2 + e^- \rightleftharpoons F^- \) is **2.8 V**.