The standard redox potentials for the reactions,
`MN^(2+) + 2e^(-)to Mn and Mn^(3+) + e^(-) are `-1.18V and 1.51V respectively. What is the redox potential for the reaction `Mn^(3+)+ 3e^(-) to Mn`?

Options

A. `(0.33V)`
B. `(1.69)`
C. `(-0.28)`
D. `(-0.85)`

Explanation

To find the redox potential for the reaction \( \text{Mn}^{3+} + 3e^{-} \rightarrow \text{Mn} \), we will use the given standard redox potentials for the 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 1: Write the half-reactions and their potentials

The half-reactions are as follows:
- For \( \text{Mn}^{2+} + 2e^{-} \rightarrow \text{Mn} \):
\[ E^\circ_1 = -1.18 \, \text{V} \]

- For \( \text{Mn}^{3+} + e^{-} \rightarrow \text{Mn}^{2+} \):
\[ E^\circ_2 = 1.51 \, \text{V} \]

### Step 2: Determine the overall reaction

We want to find the potential for the reaction:
\[ \text{Mn}^{3+} + 3e^{-} \rightarrow \text{Mn} \]

This can be viewed as the sum of the two half-reactions above. We can express this reaction as:
\[ \text{Mn}^{3+} + e^{-} \rightarrow \text{Mn}^{2+} \quad (E^\circ = 1.51 \, \text{V}) \]
\[ \text{Mn}^{2+} + 2e^{-} \rightarrow \text{Mn} \quad (E^\circ = -1.18 \, \text{V}) \]

### Step 3: Combine the half-reactions

To combine these half-reactions, we need to ensure the electrons cancel out. The first half-reaction involves one electron, and the second involves two electrons. Therefore, we can multiply the first half-reaction by 2 to balance the electrons:

\[ 2(\text{Mn}^{3+} + e^{-} \rightarrow \text{Mn}^{2+}) \quad (E^\circ = 1.51 \, \text{V}) \]
This gives:
\[ 2\text{Mn}^{3+} + 2e^{-} \rightarrow 2\text{Mn}^{2+} \quad (E^\circ = 1.51 \, \text{V}) \]

Now, we can add this to the second half-reaction:
\[ 2\text{Mn}^{2+} + 2e^{-} \rightarrow 2\text{Mn} \quad (E^\circ = -1.18 \, \text{V}) \]

### Step 4: Calculate the overall potential

Now we can find the overall potential for the reaction:
\[ \text{Mn}^{3+} + 3e^{-} \rightarrow \text{Mn} \]

Using the formula for the overall cell potential:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

Where:
- \( E^\circ_{\text{reduction}} \) is the potential for the reduction of \( \text{Mn}^{3+} \) to \( \text{Mn}^{2+} \) (1.51 V)
- \( E^\circ_{\text{oxidation}} \) is the potential for the oxidation of \( \text{Mn}^{2+} \) to \( \text{Mn} \) (-(-1.18 V) = 1.18 V)

Thus:
\[ E^\circ_{\text{cell}} = 1.51 \, \text{V} - (-1.18 \, \text{V}) = 1.51 \, \text{V} + 1.18 \, \text{V} = 2.69 \, \text{V} \]

### Step 5: Adjust for the number of electrons

Since the overall reaction involves 3 electrons, we divide the total potential by 3:
\[ E^\circ_{\text{reaction}} = \frac{2.69 \, \text{V}}{3} = 0.8967 \, \text{V} \]

### Conclusion

Therefore, the redox potential for the reaction \( \text{Mn}^{3+} + 3e^{-} \rightarrow \text{Mn} \) is approximately:
\[ E^\circ \approx 0.90 \, \text{V} \]