If `E^(c-)._(Fe^(3+)|Fe)` and `E^(c-)._(Fe^(2+)|Fe)` are `=-0.36 V` and `-0.439V`, respectively, then the value of `E^(c-)._(Fe^(3+)|Fe^(2+))`

Options

A. `(-0.36-0.439)V`
B. `[3(-0.36-0.436)]V`
C. `(-0.36+0.436)V`
D. `[3(-0.36)-2(-0.439)]V`

Explanation

To find the value of \( E^{c-}_{Fe^{3+}|Fe^{2+}} \), we will use the given standard reduction potentials for the half-reactions involving iron ions:

1. \( E^{c-}_{Fe^{3+}|Fe} = -0.36 \, V \)
2. \( E^{c-}_{Fe^{2+}|Fe} = -0.439 \, V \)

We need to determine the potential for the reaction where \( Fe^{3+} \) is reduced to \( Fe^{2+} \):

### Step 1: Write the half-reactions
The half-reactions can be written as follows:

- For \( Fe^{3+} + 3e^- \rightarrow Fe \) with \( E = -0.36 \, V \)
- For \( Fe^{2+} + 2e^- \rightarrow Fe \) with \( E = -0.439 \, V \)

### Step 2: Calculate the Gibbs free energy change for both reactions
The Gibbs free energy change (\( \Delta G \)) for a reaction can be calculated using the formula:
\[ \Delta G = -nFE \]
where:
- \( n \) is the number of moles of electrons transferred,
- \( F \) is Faraday's constant (approximately \( 96485 \, C/mol \)),
- \( E \) is the standard reduction potential.

For the first reaction:
\[ \Delta G_1 = -3F(-0.36) = 1.08F \]

For the second reaction:
\[ \Delta G_2 = -2F(-0.439) = 0.878F \]

### Step 3: Calculate \( \Delta G_3 \) for the reaction \( Fe^{3+} + e^- \rightarrow Fe^{2+} \)
The Gibbs free energy change for the third reaction can be calculated as:
\[ \Delta G_3 = \Delta G_1 - \Delta G_2 \]
Substituting the values:
\[ \Delta G_3 = 1.08F - 0.878F = 0.202F \]

### Step 4: Relate \( \Delta G_3 \) to \( E_3 \)
Since \( \Delta G_3 = -nFE_3 \) for the reaction \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) (where \( n = 1 \)):
\[ 0.202F = -1FE_3 \]
Thus, we can solve for \( E_3 \):
\[ E_3 = -0.202 \, V \]

### Final Answer
The value of \( E^{c-}_{Fe^{3+}|Fe^{2+}} \) is:
\[ E^{c-}_{Fe^{3+}|Fe^{2+}} = -0.202 \, V \]