What is the standard reduction potential `(E^(@))` for `Fe^(3+) to Fe`?
Given that :
`Fe^(2+) + 2e^(-) to Fe, E_(Fe^(2+)//Fe)^(@)`=-0.47V
`Fe^(3+) + e^(-) to Fe^(2+), E_(Fe^(3+)//Fe^(2+))^(@) = +0.77V`

Options

A. –0.057 V
B. + 0.30 V
C. – 0.30 V
D. + 0.057 V

Answer

To find the standard reduction potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \), we will use the provided standard reduction potentials for the following half-reactions:

1. \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) with \( E^\circ = -0.47 \, \text{V} \)
2. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with \( E^\circ = +0.77 \, \text{V} \)

### Step 1: Write the half-reactions
We have the following half-reactions:
- Reaction 1 (for \( \text{Fe}^{2+} \)):
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ_1 = -0.47 \, \text{V}) \]
- Reaction 2 (for \( \text{Fe}^{3+} \)):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (E^\circ_2 = +0.77 \, \text{V}) \]

### Step 2: Combine the half-reactions
To find the overall reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \), we need to combine the two half-reactions. We will multiply the first reaction by 1 and the second reaction by 2 to balance the electrons:

1. \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) (multiplied by 1)
2. \( 2 \cdot (\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) \):
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \]

Now, we can add these reactions:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (E^\circ_2 = +0.77 \, \text{V}) \]
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ_1 = -0.47 \, \text{V}) \]

### Step 3: Write the overall reaction
Adding these gives:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \]

### Step 4: Calculate the overall standard reduction potential
The overall standard reduction potential \( E^\circ \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_2 + E^\circ_1 \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (+0.77 \, \text{V}) + (-0.47 \, \text{V}) = +0.30 \, \text{V} \]

### Step 5: Adjust for the number of electrons
However, we need to adjust for the total number of electrons transferred in the overall reaction. Since we are dealing with three electrons in total, we divide the total potential by 3:
\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}} = \frac{E^\circ_{\text{cell}}}{3} = \frac{+0.30 \, \text{V}}{3} = +0.10 \, \text{V} \]

### Final Answer
Thus, the standard reduction potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \) is:
\[ E^\circ = +0.10 \, \text{V} \]