The standard electrode potentials, `E^(@)` of `Fe^(3+)//Fe^(2+)` and `Fe^(2+)//Fe` at 300 K are +0.77 V and `-0.44V`, respectively, The `E^(@)` of `Fe^(3+)//Fe)` at the same temperature is:

Options

A. 1.21 V
B. 0.33 V
C. `-0.036 V`
D. 0.036 V

Answer

To find the standard electrode potential \( E^\circ \) for the half-reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \), we can use the given standard electrode potentials for the other half-reactions:

1. **Given Data:**
- \( E^\circ (\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) = +0.77 \, \text{V} \)
- \( E^\circ (\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}) = -0.44 \, \text{V} \)

2. **Step 1: Write the half-reactions.**
- For \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) (Reduction)
- For \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) (Reduction)

3. **Step 2: Reverse the second half-reaction to find the oxidation potential.**
- The oxidation reaction is \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \)
- The standard electrode potential for this oxidation reaction will be the negative of the reduction potential:
\[ E^\circ (\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-) = -(-0.44) = +0.44 \, \text{V} \]

4. **Step 3: Combine the half-reactions to find \( E^\circ (\text{Fe}^{3+} \rightarrow \text{Fe}) \).**
- The overall reaction combining both half-reactions is:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (1) \]
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad (2) \]
- To combine these, we need to multiply the first half-reaction by 2 to balance the electrons:
\[ 2(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) \quad \Rightarrow \quad 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \]
- Now we can add the two half-reactions:
\[ 2\text{Fe}^{3+} + 2e^- + \text{Fe} \rightarrow 2\text{Fe}^{2+} + 2e^- + \text{Fe} \]
- Cancel out the \( 2e^- \) and \( \text{Fe}^{2+} \):
\[ 2\text{Fe}^{3+} + \text{Fe} \rightarrow 2\text{Fe}^{2+} \]

5. **Step 4: Calculate the overall standard electrode potential using the Nernst equation.**
- The overall standard potential \( E^\circ \) for the reaction can be calculated using:
\[ E^\circ = E^\circ (\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) + E^\circ (\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-) \]
- Substitute the values:
\[ E^\circ = 0.77 \, \text{V} + 0.44 \, \text{V} = 1.21 \, \text{V} \]

6. **Step 5: Adjust for the overall reaction.**
- Since we are looking for the potential from \( \text{Fe}^{3+} \) to \( \text{Fe} \):
\[ E^\circ (\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}) = E^\circ (\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) + E^\circ (\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}) \]
- Thus, we need to adjust the final potential:
\[ E^\circ = 0.77 + (-0.44) = 0.33 \, \text{V} \]

7. **Final Answer:**
- The standard electrode potential \( E^\circ (\text{Fe}^{3+} \rightarrow \text{Fe}) \) is approximately \( +0.33 \, \text{V} \).