Given that the standard reduction potentials `E ^(0) of Fe^(+2)|Feis 0.26V and Fe ^(+3) |Fe is 0.76V` respectively. The `E ^(@)of Fe ^(+2)|Fe^(+3)` is:

Options

A. `+1.76V`
B. `-1.76V`
C. `+0.182V`
D. `-0.182V`

Explanation

To find the standard reduction potential \( E^\circ \) for the half-reaction \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \), we can use the given standard reduction potentials for the half-reactions involving iron.

### Step-by-Step Solution:

1. **Identify the Given Half-Reactions and Their Potentials:**
- For the half-reaction \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \), the standard reduction potential \( E^\circ_A = 0.26 \, \text{V} \).
- For the half-reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \), the standard reduction potential \( E^\circ_B = 0.76 \, \text{V} \).

2. **Write the Reactions:**
- Reaction A: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) (with \( E^\circ_A = 0.26 \, \text{V} \))
- Reaction B: \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \) (with \( E^\circ_B = 0.76 \, \text{V} \))

3. **Reverse Reaction B:**
- When we reverse Reaction B, we get:
\[ \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \]
- The potential for the reversed reaction becomes negative:
\[ E^\circ_{B, \text{rev}} = -0.76 \, \text{V} \]

4. **Combine the Reactions:**
- Now we can add Reaction A and the reversed Reaction B:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \]
\[ \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \]
- Adding these two reactions gives:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

5. **Calculate the Standard Reduction Potential for the Combined Reaction:**
- The overall cell potential \( E^\circ_C \) can be calculated using the formula:
\[ E^\circ_C = E^\circ_A + E^\circ_{B, \text{rev}} \]
- Substituting the values:
\[ E^\circ_C = 0.26 \, \text{V} + (-0.76 \, \text{V}) = 0.26 \, \text{V} - 0.76 \, \text{V} = -0.50 \, \text{V} \]

6. **Final Result:**
- The standard reduction potential \( E^\circ \) for the half-reaction \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \) is:
\[ E^\circ = -0.50 \, \text{V} \]