`1/2 F_(2)+e^(-) rarr F^(-)" "E^(@)=+3.02V`
Electrode potential for given reaction
`F_(2)+2e^(-) rarr 2F^(-)`

Options

A. 3.02 V
B. 6.04 V
C. 1.5 V
D. `-3.02` V

Answer

To find the electrode potential for the reaction \( F_2 + 2e^- \rightarrow 2F^- \) given that the electrode potential for the reaction \( \frac{1}{2} F_2 + e^- \rightarrow F^- \) is \( E^\circ = +3.02 \, V \), we can follow these steps:

### Step 1: Understand the Relationship Between the Reactions
The first reaction is:
\[ \frac{1}{2} F_2 + e^- \rightarrow F^- \]
The second reaction is:
\[ F_2 + 2e^- \rightarrow 2F^- \]
The second reaction is simply the first reaction multiplied by 2.

### Step 2: Apply the Concept of Intensive Properties
Electrode potential (\( E^\circ \)) is an intensive property, which means it does not change when the coefficients in the balanced equation are multiplied. Therefore, the electrode potential for the second reaction will remain the same as that of the first reaction.

### Step 3: Write the Final Answer
Since the electrode potential for the first reaction is \( +3.02 \, V \), the electrode potential for the second reaction will also be:
\[ E^\circ = +3.02 \, V \]

### Final Answer
The electrode potential for the reaction \( F_2 + 2e^- \rightarrow 2F^- \) is \( +3.02 \, V \).

---