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

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

given that

\[ F_{2} + 2 e^{-} \rightleftharpoons 2 F^{-}, \quad E^\circ = +2.7 \, \text{V} \]

we can follow these steps:

### Step 1: Write the given half-reaction and its standard electrode potential.
The half-reaction provided is:

\[ F_{2} + 2 e^{-} \rightleftharpoons 2 F^{-} \]

with a standard electrode potential \( E^\circ = +2.7 \, \text{V} \).

### Step 2: Reverse the reaction to find the potential for the desired reaction.
To find the potential for the reaction we are interested in, we need to reverse the given half-reaction. When reversing a reaction, the sign of the standard electrode potential changes. Thus, we have:

\[ 2 F^{-} \rightleftharpoons F_{2} + 2 e^{-} \]

The new standard electrode potential becomes:

\[ E^\circ = -2.7 \, \text{V} \]

### Step 3: Adjust the stoichiometry of the reaction.
The desired reaction is:

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

To adjust the stoichiometry, we can divide the entire reaction by 2:

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

### Step 4: Calculate the new standard electrode potential.
When we divide the reaction by 2, the standard electrode potential is adjusted according to the Nernst equation. The new standard electrode potential becomes:

\[ E^\circ = \frac{-2.7 \, \text{V}}{2} = -1.35 \, \text{V} \]

### Conclusion
Thus, the standard electrode potential \( E^\circ \) for the reaction

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

is

\[ E^\circ = -1.35 \, \text{V} \]