`E^(@) = 0.93V` for the reactions:
`Fe(s) + 2M^(+)(aq) rightarrow Fe^(2+)(aq) + 2M(s)`
What is the standard potential for `M^(+) + e^(-) rightarrow M`, if `E_(Fe^(2+//Fe)^(@) = -0.44 V`?

Options

A. 0.26V
B. 0.49V
C. 0.67V
D. 1.34V

Answer

To find the standard potential for the reaction \( M^+ + e^- \rightarrow M \), we can use the information provided about the overall cell potential and the standard reduction potential for iron.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The standard cell potential \( E^\circ_{\text{cell}} = 0.93 \, \text{V} \).
- The standard reduction potential for iron \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \).

2. **Write the Half-Reactions**:
- The half-reaction for the reduction of \( M^+ \) to \( M \) is:
\[ M^+ + e^- \rightarrow M \quad (E^\circ_M) \]
- The half-reaction for the oxidation of iron is:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad (E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V}) \]

3. **Determine the Overall Cell Reaction**:
- The overall cell reaction can be written as:
\[ \text{Fe}(s) + 2M^+(aq) \rightarrow \text{Fe}^{2+}(aq) + 2M(s) \]

4. **Identify the Cathode and Anode**:
- In this cell, the reduction occurs at the cathode and oxidation occurs at the anode.
- Here, \( M^+ \) is reduced to \( M \) (cathode), and \( \text{Fe} \) is oxidized to \( \text{Fe}^{2+} \) (anode).

5. **Use the Nernst Equation**:
- The standard cell potential can be expressed as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the known values:
\[ 0.93 \, \text{V} = E^\circ_M - (-0.44 \, \text{V}) \]

6. **Solve for \( E^\circ_M \)**:
- Rearranging the equation gives:
\[ 0.93 \, \text{V} = E^\circ_M + 0.44 \, \text{V} \]
- Therefore:
\[ E^\circ_M = 0.93 \, \text{V} - 0.44 \, \text{V} = 0.49 \, \text{V} \]

7. **Conclusion**:
- The standard potential for the reaction \( M^+ + e^- \rightarrow M \) is \( E^\circ_M = 0.49 \, \text{V} \).

### Final Answer:
The standard potential for \( M^+ + e^- \rightarrow M \) is \( 0.49 \, \text{V} \).