The hydrogen ion in acidified potassium manganate (VII)
I think you have known that the permanganate(VII) ion is a strong oxidizing agent because manganese is in the highest +7 oxidation state.

But in different pH conditions, $\ce{MnO_4^-}$ Will be reduced to different product:


In a acidic solution, it is reduced to the colourless manganese(II) ion, $\ce{Mn^{2+}}$.
$\ce{8H^+ + MnO_4^{−} + 5e^{−} → Mn^{2+} + 4H_2O}$
In a strongly basic solution, it is reduced to the green manganate ion, $\ce{MnO_4^{2−}}$.
$\ce{MnO_4^{−} + e^{−} → MnO_4^{2−}}$
In a neutral solution, it is reduced to the brown manganese dioxide $\ce{MnO_2}$.
$\ce{2H_2O + MnO_4^{−} + 3e^{−} → MnO_2 + 4OH^{−}}$


But the more understandable way to say, I think, is that $\ce{MnO_4^-}$ can be reduced to different oxidation states; however, it is an idea about chemical balance. If there is $\ce{H^+}$, the first reaction dominates (since we put in more reactant). While if there is $\ce{OH^-}$, neither reaction 1 nor 3 happens (1: lack of reactant; 3: too many product), so the third dominates.

Hope this helps.

Addition: in those different reactions, there are also different tendency to take place and make $\ce{MnO_4^-}$ reduced. In fact, the first reaction (happening with acidic solution) have a much larger tendency to take place. Thus, if $\ce{H^+}$ presents, $\ce{MnO_4^-}$ will have a larger tendency to oxidate other compounds. Thus some reactions require $\ce{MnO_4^-}$ with $\ce{H^+}$ instead of only $\ce{MnO_4^-}$.