What is the difference between balancing a redox reaction in acidic solution vs basic solution?
You basically just have to add hydroxide ions to either sides of your equilibrium.

$$\ce{2MnO4 + 5H2O2 + 6H+ + \color{blue}{6OH-} -&gt; 2Mn^2+ +5O2 +8H2O + \color{blue}{6OH-}}$$

$\ce{H+}$ and $\ce{OH-}$ ions can be collectively re-written as $\ce{H2O}$, so we have:

$$\ce{2MnO4 + 5H2O2 + \color{blue}{6H2O} -&gt; 2Mn^2+ +5O2 +8H2O + 6OH-}$$

Finally, we can cancel out the excess water molecules which leaves us with:

$$\ce{2MnO4 + 5H2O2  -&gt; 2Mn^2+ +5O2 +2H2O + 6OH-}$$