For the reactions
`MnO_4^_+8H^++5e^(-) rarrMn^(2+)+4H_2O,E^0=1.51V`
`MnO_2^_+4H^++2e^(-) rarrMn^(2+)+2H_2O,E^0=1.23V` then the reaction `MnO_4^_+4H^++3e^(-) rarrMnO_(2)+2H_2O,E^0` is -

Options

A. `1.70V`
B. `5.09V`
C. `0.28V`
D. `0.84V`

Answer

To solve the problem, we need to find the standard electrode potential \( E^0 \) for the reaction:

\[ \text{MnO}_4^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \]

We will use the two given half-reactions and manipulate them to derive the desired reaction.

### Step 1: Write down the given half-reactions

1. **First half-reaction:**
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}, \quad E^0 = 1.51 \text{ V} \]

2. **Second half-reaction:**
\[ \text{MnO}_2 + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}, \quad E^0 = 1.23 \text{ V} \]

### Step 2: Reverse the second half-reaction

To combine these reactions, we need to reverse the second half-reaction because we want to produce \( \text{MnO}_2 \) instead of consuming it:

\[ \text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + 2\text{e}^-, \quad E^0 = -1.23 \text{ V} \]

### Step 3: Adjust the number of electrons

Now, we need to adjust the number of electrons in the first and reversed second half-reaction so that they can be combined. We can multiply the reversed second half-reaction by 1.5 (or 3/2) to make the number of electrons equal to 3:

\[ \frac{3}{2} \left( \text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + 2\text{e}^- \right) \]

This gives us:

\[ \frac{3}{2} \text{Mn}^{2+} + 3\text{H}_2\text{O} \rightarrow \frac{3}{2} \text{MnO}_2 + 6\text{H}^+ + 3\text{e}^-, \quad E^0 = -1.23 \text{ V} \times \frac{3}{2} = -1.845 \text{ V} \]

### Step 4: Combine the reactions

Now we can combine the first half-reaction and the adjusted second half-reaction:

1. **First half-reaction:**
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

2. **Adjusted reversed second half-reaction:**
\[ \frac{3}{2} \text{Mn}^{2+} + 3\text{H}_2\text{O} \rightarrow \frac{3}{2} \text{MnO}_2 + 6\text{H}^+ + 3\text{e}^- \]

Now, we add these two reactions:

\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- + \frac{3}{2} \text{Mn}^{2+} + 3\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + \frac{3}{2} \text{MnO}_2 + 6\text{H}^+ + 3\text{e}^- \]

### Step 5: Simplify the equation

After canceling out the common species, we get:

\[ \text{MnO}_4^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \]

### Step 6: Calculate \( E^0 \)

Using the equation for the standard potentials:

\[ E^0_{\text{cell}} = E^0_1 + E^0_2 \]

Where \( E^0_1 = 1.51 \text{ V} \) and \( E^0_2 = -1.845 \text{ V} \):

\[ E^0_{\text{cell}} = 1.51 - 1.845 = 1.699 \text{ V} \]

### Final Answer

Thus, the standard electrode potential \( E^0 \) for the reaction \( \text{MnO}_4^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \) is approximately:

\[ E^0 \approx 1.70 \text{ V} \]