`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 \( E^\circ \) for the half-reaction \( \text{MnO}_4^- \rightarrow \text{MnO}_2 \) in acidic medium, we can use the given standard electrode potentials for the reactions involving manganese species.

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Potentials:**
- \( E^\circ (\text{Mn}^{2+} \rightarrow \text{MnO}_4^-) = -1.51 \, \text{V} \)
- \( E^\circ (\text{MnO}_2 \rightarrow \text{Mn}^{2+}) = +1.23 \, \text{V} \)

...