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Recently, I am playing with redox reaction and I come over a compound acidified potassium manganate (VII) ($\ce{KMnO4/H^+}$). I get confused that why they put $\ce{H^+}$ beside $\ce{KMnO4}$. Doesn't the hydrogen ions flow freely in the solution or something else.

And so, when we are writing a chemical reaction with occur $\ce{KMnO4}$, we sometimes exclude the $\ce{H^+}$. Why that so?

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  • $\begingroup$ Permanganic acid ($\ce{HMnO4}$) is a strong acid, so it's not easy to form via protonation of $\ce{MnO4^{-}}$. When adding permanganate ions to very very acidic mixtures, for example pure liquid $\ce{H2SO4}$, you can consider that the reaction $\ce{MnO4^{-} + H^+ <=> HMnO4}$ happens to an appreciable extent, producing permanganic acid. In the specific case I mentioned, however, $\ce{HMnO4}$ is a short lived entity, as it is quickly dehydrated by the sulfuric acid to form the anhydride $\ce{Mn2O7}$, a very reactive species. $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 18:11
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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:

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

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