Find out `E^(@)` for `F^(-) rarr (1)/(2)F_(2)+e^(-)`, If `F_(2)+2e^(-)rarr 2F^(-),E^(@)=+2.7V`.

Options

A. 1.35 V
B. `-1.35V`
C. 2.7 V
D. `-2.7 V`

Answer

To find the standard electrode potential \( E^\circ \) for the reaction \( F^{-} \rightarrow \frac{1}{2} F_{2} + e^{-} \), given that \( F_{2} + 2e^{-} \rightarrow 2F^{-} \) has \( E^\circ = +2.7V \), we can follow these steps:

### Step 1: Write the given reaction and its standard potential
The given reaction is:
\[ F_{2} + 2e^{-} \rightarrow 2F^{-} \]
with a standard electrode potential \( E^\circ = +2.7V \).

### Step 2: Reverse the reaction
To find the potential for the reaction \( F^{-} \rightarrow \frac{1}{2} F_{2} + e^{-} \), we need to reverse the given reaction. When a reaction is reversed, the sign of the standard electrode potential changes:
\[ 2F^{-} \rightarrow F_{2} + 2e^{-} \]
This means:
\[ E^\circ = -2.7V \]

### Step 3: Adjust the stoichiometry
Now, we need to adjust the stoichiometry of the reversed reaction to match the desired reaction:
\[ F^{-} \rightarrow \frac{1}{2} F_{2} + e^{-} \]
To do this, we can divide the entire reaction by 2:
\[ \frac{1}{2} (2F^{-} \rightarrow F_{2} + 2e^{-}) \]
This gives us:
\[ F^{-} \rightarrow \frac{1}{2} F_{2} + e^{-} \]
The standard electrode potential remains the same because it is an intensive property:
\[ E^\circ = -2.7V \]

### Conclusion
Thus, the standard electrode potential \( E^\circ \) for the reaction \( F^{-} \rightarrow \frac{1}{2} F_{2} + e^{-} \) is:
\[ E^\circ = -2.7V \]