`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 determine the standard electrode potential (E°) for the half-reaction involving the reduction of fluorine gas (F₂) to fluoride ions (F⁻) given the standard electrode potential for the reaction:

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

is \( E° = 2.8 \, \text{V} \).

We also have the half-reaction:

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

and we need to find the standard electrode potential for this half-reaction.

### Step-by-Step Solution:

1. **Understanding the Given Reaction**:
The first reaction provides us with the standard electrode potential for the reduction of fluorine gas to fluoride ions. The equation is:
\[ F_2 + 2e^- \rightleftharpoons 2F^- \]
with \( E° = 2.8 \, \text{V} \).

2. **Relating the Two Reactions**:
The second reaction is a scaled-down version of the first:
\[ \frac{1}{2} F_2 + e^- \rightleftharpoons F^- \]
This reaction is essentially half of the first reaction.

3. **Using the Property of Standard Electrode Potential**:
The standard electrode potential (E°) is an intensive property, meaning it does not depend on the amount of substance involved. Therefore, when we scale down the reaction by a factor of 2 (from 2 electrons to 1 electron), the E° value remains the same.

4. **Conclusion**:
Since the E° value is an intensive property and does not change with the amount of substance, we can conclude that:
\[ E° \text{ for } \frac{1}{2} F_2 + e^- \rightleftharpoons F^- = 2.8 \, \text{V} \]

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