If `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V` and `0.439 V` respectively, then value of `E_(Fe^(3+)//Fe^(2+))^(@)` is

Options

A. `-0.057` V
B. `+0.30 V`
C. `-0.30 `V
D. `+0.057` V

Answer

To find the value of \( E^{\circ}_{Fe^{3+} // Fe^{2+}} \), we can use the given standard electrode potentials and apply the concept of the Latimer diagram. Here’s the step-by-step solution:

### Step 1: Identify the Given Values
We are given:
- \( E^{\circ}_{Fe^{3+} // Fe} = -0.36 \, V \)
- \( E^{\circ}_{Fe^{2+} // Fe} = 0.439 \, V \)

### Step 2: Write the Half-Reactions
The half-reactions involved are:
1. \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) (Reduction of \( Fe^{3+} \) to \( Fe^{2+} \))
2. \( Fe^{2+} + 2e^- \rightarrow Fe \) (Reduction of \( Fe^{2+} \) to \( Fe \))

### Step 3: Set Up the Equation Using the Latimer Diagram
Using the Latimer diagram, we can express the relationship between the standard potentials:
\[ n_1 E^{\circ}_{Fe^{3+} // Fe^{2+}} + n_2 E^{\circ}_{Fe^{2+} // Fe} = n E^{\circ}_{Fe^{3+} // Fe} \]
Where:
- \( n_1 = 1 \) (for the first half-reaction)
- \( n_2 = 2 \) (for the second half-reaction)
- \( n = 3 \) (total electrons transferred from \( Fe^{3+} \) to \( Fe \))

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ 1 \cdot E^{\circ}_{Fe^{3+} // Fe^{2+}} + 2 \cdot 0.439 = 3 \cdot (-0.36) \]

### Step 5: Simplify the Equation
Now, simplify the equation:
\[ E^{\circ}_{Fe^{3+} // Fe^{2+}} + 0.878 = -1.08 \]

### Step 6: Solve for \( E^{\circ}_{Fe^{3+} // Fe^{2+}} \)
Rearranging gives:
\[ E^{\circ}_{Fe^{3+} // Fe^{2+}} = -1.08 - 0.878 \]
\[ E^{\circ}_{Fe^{3+} // Fe^{2+}} = -1.958 \, V \]

### Step 7: Conclusion
Thus, the value of \( E^{\circ}_{Fe^{3+} // Fe^{2+}} \) is:
\[ \boxed{-1.958 \, V} \]