For the reactions
`{:(MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)4H_(2)O","E^(o) = + 1.51 V),(MnO_(2)+4H^(+)+ 2e^(-) rarr Mn^(2+)+2H_(2)O"," E^(o)= + 1.23 V ):}`
then for the reaction :
`MnO_(4)^(-) + 4H^(+) + 3e^(-) rarr MnO_(2)+2H_(2)O","E^(o)`

Options

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

Answer

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

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

We are given two half-reactions with their standard electrode potentials:

1. \( \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) with \( E^{\circ} = +1.51 \, \text{V} \)
2. \( \text{MnO}_2 + 4\text{H}^{+} + 2\text{e}^{-} \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} \) with \( E^{\circ} = +1.23 \, \text{V} \)

### Step 1: Reverse the second half-reaction
To use the second half-reaction in our desired reaction, we need to reverse it:

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

When we reverse a reaction, the sign of the standard electrode potential changes:

\[ E^{\circ} = -1.23 \, \text{V} \]

### Step 2: Adjust the number of electrons
Now we have:

1. \( \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (1)
2. \( \text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^{+} + 2\text{e}^{-} \) (2 reversed)

To combine these reactions, we need to multiply the second reaction by 1.5 (to equalize the number of electrons):

\[ 1.5 \times (\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^{+} + 2\text{e}^{-}) \]

This gives us:

\[ 1.5\text{Mn}^{2+} + 3\text{H}_2\text{O} \rightarrow 1.5\text{MnO}_2 + 6\text{H}^{+} + 3\text{e}^{-} \]

### Step 3: Combine the half-reactions
Now we can add the two half-reactions together:

1. \( \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
2. \( 1.5\text{Mn}^{2+} + 3\text{H}_2\text{O} \rightarrow 1.5\text{MnO}_2 + 6\text{H}^{+} + 3\text{e}^{-} \)

When we add these, we cancel out the common species:

- The \( \text{Mn}^{2+} \) cancels out.
- The \( 4\text{H}_2\text{O} \) from the first reaction and \( 3\text{H}_2\text{O} \) from the second gives \( 1\text{H}_2\text{O} \).
- The \( 8\text{H}^{+} \) and \( 6\text{H}^{+} \) gives \( 2\text{H}^{+} \).

The resulting reaction is:

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

### Step 4: Calculate the standard potential for the overall reaction
The overall standard potential \( E^{\circ}_{\text{cell}} \) is calculated using the formula:

\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}} \]

Where:

- \( E^{\circ}_{\text{reduction}} = +1.51 \, \text{V} \) (from the first reaction)
- \( E^{\circ}_{\text{oxidation}} = -1.23 \, \text{V} \) (from the reversed second reaction)

Thus,

\[ E^{\circ}_{\text{cell}} = 1.51 - 1.23 = 0.28 \, \text{V} \]

### Final Answer
The standard electrode potential \( E^{\circ} \) for the reaction \( \text{MnO}_4^{-} + 4\text{H}^{+} + 3\text{e}^{-} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \) is \( +0.28 \, \text{V} \).

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