`E_(Mn^(2+)//MnO_(4)^(-))^(0)=-1.51V`
`E_(MnO_(2)//Mn^(+2))^(0)=+1.23V`
`E_(MnO_(4)^(-)//MnO_(2)=?` (All in acidic medium)

Explanation

To find the standard electrode potential \( E_{MnO_4^{-}/MnO_2}^0 \), we will use the given standard electrode potentials and the reactions involved. Here's a step-by-step solution:

### Step 1: Write the half-reactions
We are given the following half-reactions:

1. For the conversion of \( Mn^{2+} \) to \( MnO_4^{-} \):
\[
Mn^{2+} + 4H_2O \rightarrow MnO_4^{-} + 8H^+ + 5e^- \quad (E^0 = -1.51 \, V)
...