To find the standard redox potential for the reaction \( \text{Mn}^{3+} + 3e^- \rightarrow \text{Mn} \), we can use the given standard redox potentials for the half-reactions:
1. \( \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \) with \( E^\circ = -1.18 \, \text{V} \)
2. \( \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \) with \( E^\circ = 1.51 \, \text{V} \)
### Step-by-Step Solution:
**Step 1: Write the half-reaction for the reduction of \( \text{Mn}^{3+} \) to \( \text{Mn}^{2+} \)**
The half-reaction is:
\[
\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}
\]
The standard potential for this reaction is given as \( E^\circ = 1.51 \, \text{V} \).
**Step 2: Write the half-reaction for the reduction of \( \text{Mn}^{2+} \) to \( \text{Mn} \)**
The half-reaction is:
\[
\text{Mn}^{2+} + 2e^- \rightarrow \text{Mn}
\]
The standard potential for this reaction is given as \( E^\circ = -1.18 \, \text{V} \).
**Step 3: Combine the half-reactions to find the overall reaction**
To find the overall reaction \( \text{Mn}^{3+} + 3e^- \rightarrow \text{Mn} \), we need to add the two half-reactions. We need to ensure that the electrons cancel out.
1. The first half-reaction needs to be multiplied by 2 to match the number of electrons:
\[
2(\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}) \implies 2\text{Mn}^{3+} + 2e^- \rightarrow 2\text{Mn}^{2+}
\]
with \( E^\circ = 1.51 \, \text{V} \).
2. The second half-reaction remains as is:
\[
\text{Mn}^{2+} + 2e^- \rightarrow \text{Mn}
\]
with \( E^\circ = -1.18 \, \text{V} \).
Now we can add these two half-reactions:
\[
2\text{Mn}^{3+} + 2e^- + \text{Mn}^{2+} + 2e^- \rightarrow 2\text{Mn}^{2+} + \text{Mn}
\]
**Step 4: Calculate the overall standard redox potential**
The overall standard redox potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[
E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}
\]
Here, we have:
- \( E^\circ_{\text{reduction}} = 1.51 \, \text{V} \) (for \( \text{Mn}^{3+} \rightarrow \text{Mn}^{2+} \))
- \( E^\circ_{\text{oxidation}} = -1.18 \, \text{V} \) (for \( \text{Mn}^{2+} \rightarrow \text{Mn} \))
So,
\[
E^\circ_{\text{cell}} = 1.51 \, \text{V} + 1.18 \, \text{V} = 2.69 \, \text{V}
\]
**Step 5: Adjust for the number of electrons**
Since we are looking for the potential for the reaction \( \text{Mn}^{3+} + 3e^- \rightarrow \text{Mn} \), we need to consider that the total number of electrons transferred is 3. Therefore, we can use the Nernst equation or simply divide the total potential by 3 to find the potential for the overall reaction:
\[
E^\circ = \frac{2.69 \, \text{V}}{3} = 0.8967 \, \text{V}
\]
However, since we are looking for the potential for the reaction \( \text{Mn}^{3+} + 3e^- \rightarrow \text{Mn} \), we will keep the potential as is.
### Final Answer:
The standard redox potential for the reaction \( \text{Mn}^{3+} + 3e^- \rightarrow \text{Mn} \) is approximately \( 0.8967 \, \text{V} \).
