`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. 1.70V
B. 0.91V
C. 1.37v
D. 0.548V

Answer

To find the standard reduction potential \( E_{MnO_4^{-}|MnO_2} \) for the half-reaction of \( MnO_4^{-} \) being reduced to \( MnO_2 \), we can use the given half-reactions and their standard potentials. Here’s the step-by-step solution:

### Step 1: Write down the half-reactions and their potentials
1. The first half-reaction is:
\[ MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \quad (E^0 = 1.51 \, V) \]
2. The second half-reaction is:
\[ MnO_2 + 4H^{+} + 2e^{-} \rightarrow Mn^{2+} + 2H_2O \quad (E^0 = 1.23 \, V) \]

### Step 2: Reverse the second half-reaction
To find the potential for the reduction of \( MnO_4^{-} \) to \( MnO_2 \), we need to reverse the second half-reaction since we want to go from \( MnO_2 \) to \( MnO_4^{-} \):
\[ Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^{+} + 2e^{-} \quad (E^0 = -1.23 \, V) \]

### Step 3: Balance the number of electrons
Now we need to balance the number of electrons in both half-reactions. The first half-reaction involves 5 electrons, while the reversed second half-reaction involves 2 electrons. To balance them, we can multiply the second half-reaction by 2.5:
\[ 2.5(Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^{+} + 2e^{-}) \quad (E^0 = -1.23 \, V) \]
This gives:
\[ 2.5Mn^{2+} + 5H_2O \rightarrow 2.5MnO_2 + 10H^{+} + 5e^{-} \quad (E^0 = -1.23 \, V) \]

### Step 4: Add the half-reactions
Now we can add the two half-reactions:
1. From \( MnO_4^{-} \):
\[ MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \]
2. From the modified \( MnO_2 \):
\[ 2.5Mn^{2+} + 5H_2O \rightarrow 2.5MnO_2 + 10H^{+} + 5e^{-} \]

Adding these gives:
\[ MnO_4^{-} + 2.5Mn^{2+} + 8H^{+} + 5H_2O \rightarrow 2.5MnO_2 + 10H^{+} + 5e^{-} + Mn^{2+} + 4H_2O \]

### Step 5: Simplify the equation
Now, we can simplify the equation:
- The \( Mn^{2+} \) cancels out.
- The \( H^{+} \) and \( H_2O \) terms can be simplified:
\[ MnO_4^{-} + 2.5Mn^{2+} + 8H^{+} + 5H_2O - 10H^{+} - 5H_2O \rightarrow 2.5MnO_2 \]
This simplifies to:
\[ MnO_4^{-} + 2.5Mn^{2+} - 2H^{+} \rightarrow 2.5MnO_2 \]

### Step 6: Calculate the overall potential
The overall standard cell potential \( E^0 \) is calculated using:
\[ E^0_{cell} = E^0_{MnO_4^{-}} - E^0_{MnO_2} \]
Substituting the values:
\[ E^0_{cell} = 1.51 \, V - (-1.23 \, V) = 1.51 + 1.23 = 2.74 \, V \]

### Final Answer
Thus, the standard reduction potential \( E_{MnO_4^{-}|MnO_2} \) is:
\[ E_{MnO_4^{-}|MnO_2} = 2.74 \, V \]