`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O`, `E^(@) = 1.51V`
`MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O` `E^(@) = 1.23V`
`E_(MnO_(4)^(-)|MnO_(2)`

Options

A. 1.70 V
B. 0.91 V
C. 1.37 V
D. 0.548 V

Answer

To find the standard electrode potential \( E^0 \) for the half-reaction \( \text{MnO}_4^{-} + 2 \text{MnO}_2 \rightarrow 3 \text{Mn}^{2+} + 4 \text{H}_2\text{O} \), we can use the given reactions and their standard electrode potentials.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Potentials:**
- Reaction 1:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \quad (E^0 = 1.51 \, \text{V}) \]
- Reaction 2:
\[ \text{MnO}_2 + 4 \text{H}^{+} + 2 e^{-} \rightarrow \text{Mn}^{2+} + 2 \text{H}_2\text{O} \quad (E^0 = 1.23 \, \text{V}) \]

2. **Calculate \( \Delta G \) for Each Reaction:**
- For Reaction 1:
\[ \Delta G_1 = -nFE^0 = -5F \cdot 1.51 \]
- For Reaction 2:
\[ \Delta G_2 = -nFE^0 = -2F \cdot 1.23 \]

3. **Determine the Overall Reaction:**
- We want to find the potential for the reaction:
\[ \text{MnO}_4^{-} + 2 \text{MnO}_2 \rightarrow 3 \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
- The overall \( \Delta G \) for this reaction can be expressed as:
\[ \Delta G_3 = \Delta G_1 - \Delta G_2 \]

4. **Substituting the Values:**
- Substitute \( \Delta G_1 \) and \( \Delta G_2 \):
\[ \Delta G_3 = (-5F \cdot 1.51) - (-2F \cdot 1.23) \]
- Simplifying gives:
\[ \Delta G_3 = -5F \cdot 1.51 + 2F \cdot 1.23 = -F(5 \cdot 1.51 - 2 \cdot 1.23) \]

5. **Calculate the Result:**
- Calculate the expression:
\[ 5 \cdot 1.51 = 7.55 \]
\[ 2 \cdot 1.23 = 2.46 \]
\[ 7.55 - 2.46 = 5.09 \]
- Thus,
\[ \Delta G_3 = -F \cdot 5.09 \]

6. **Relate \( \Delta G_3 \) to \( E^0 \):**
- Since \( \Delta G_3 = -nFE^0 \) and \( n \) for the overall reaction is 3 (since 5 - 2 = 3):
\[ -3F \cdot E^0 = -F \cdot 5.09 \]
- Cancel \( F \):
\[ 3E^0 = 5.09 \]
- Therefore,
\[ E^0 = \frac{5.09}{3} \approx 1.697 \, \text{V} \]

7. **Final Answer:**
- The standard electrode potential \( E^0 \) for the reaction \( \text{MnO}_4^{-} + 2 \text{MnO}_2 \rightarrow 3 \text{Mn}^{2+} + 4 \text{H}_2\text{O} \) is approximately \( 1.70 \, \text{V} \).