`{:(Given Fe^(+2)+2e^(-)toFe(s),,,,E^(@)=-0.447V),(Fe^(+2)toFe^(+3)+e^(-),,,,E^(@)=-0.771V):}`
Find `E^(@)` for the reaction `Fe^(+3)+3e^(-)toFe(s)`

Options

A. 0.34V
B. `-0.041V`
C. `+0.39V`
D. `-0.47V`

Explanation

To find the standard electrode potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}(s) \), we can use the given half-reactions and their standard electrode potentials. Here’s a step-by-step solution:

### Step 1: Write the given half-reactions and their potentials
1. The first half-reaction is:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.447 \, \text{V} \quad \text{(Reaction 1)} \]

2. The second half-reaction is:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad E^\circ = -0.771 \, \text{V} \quad \text{(Reaction 2)} \]

### Step 2: Reverse the second half-reaction
To use the second half-reaction in the desired form, we need to reverse it:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = +0.771 \, \text{V} \]

### Step 3: Combine the half-reactions
Now we will combine the two half-reactions. We need to multiply the reversed second reaction by 2 so that we can cancel out \( \text{Fe}^{2+} \):
1. From the reversed second reaction:
\[ 2(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) \quad E^\circ = +0.771 \, \text{V} \]
becomes:
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \quad E^\circ = +0.771 \, \text{V} \]

2. Now, we add this to the first reaction:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.447 \, \text{V} \]

### Step 4: Write the overall reaction
Adding these two reactions gives:
\[ 2\text{Fe}^{3+} + 2e^- + \text{Fe}^{2+} + 2e^- \rightarrow 2\text{Fe}^{2+} + \text{Fe}(s) \]
This simplifies to:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}(s) \]

### Step 5: Calculate the overall standard electrode potential
Now we can calculate the overall standard electrode potential \( E^\circ \):
\[ E^\circ_{\text{overall}} = E^\circ_{\text{reduced}} + E^\circ_{\text{oxidized}} = +0.771 \, \text{V} - 0.447 \, \text{V} \]
Calculating this gives:
\[ E^\circ_{\text{overall}} = 0.771 - 0.447 = 0.324 \, \text{V} \]

### Step 6: Final answer
The standard electrode potential for the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}(s) \) is approximately:
\[ E^\circ \approx 0.34 \, \text{V} \]