If given a half reaction, how do you determine if an element can exist in a acidic conditions?
Doing so will result in the two half equations

$\ce{2MnO_4^2- -&gt; 2MnO_4^- + 2e^-}$
and

$\ce{MnO_4^2- + 4H^+ +2e^- -&gt; MnO_2 + 2H2O}$

which you then combine into the total equation

$\ce{3MnO_4^2- + 4H+ &lt;=&gt; 2MnO4- + MnO2 + 2 H2O}$.

The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells.