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I don't get how does this reaction of hydrochloric acid and potassium permanganate proceeds:

$$\ce{HCl + KMnO4 -> KCl + MnCl2 + Cl2 + H2O}$$

What I know:

  1. $\ce{HCl}$ dissociates into $\ce{H+}$ and $\ce{Cl-}$.

  2. $\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^{2-}}$

What I don't get: How $\ce{MnCl2}$ and $\ce{Cl2}$ can appear there?

Could someone, please, guide me through this reaction, step by step?

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  • $\begingroup$ Take a look at the oxidation states. You will arrive at the conclusion that the reaction is of redox nature. I hope you can take it from there. $\endgroup$
    – tschoppi
    Sep 24, 2014 at 13:25
  • $\begingroup$ Correction: "$\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^-}$" $\endgroup$ Jul 12, 2018 at 2:43

3 Answers 3

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As others have pointed out, you don't have a "normal" reaction, you have a oxidation-reduction reaction (if you want to sound especially geeky, calls them RedOx (pronounced "ree-dox") reactions.

The electrons (which we don't normally balance in a equation) feature prominently because there are elements changing oxidation state.

First, split your reaction into its half reactions. Bring the "spectator ions" along for the ride and make sure they balance:

Oxidation: 2 $\ce{Cl-}$ gets oxidized to $\ce{Cl2}$ and gives up 2 electrons. $\ce{H+}$ is your spectator ion. $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$ Reduction: $\ce{Mn^7+}$ needs 5 electrons to reduce it to $\ce{Mn^2+}$. $\ce{K+}$ and $\ce{O^2-}$ are your spectator ions. $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$ Now, just add water. Oxidation reactions get $\ce{OH-}$; reduction reactions get $\ce{H+}$: $$\begin{align} \ce{2OH- + 2 H+ + 2 Cl- &-> Cl2 + 2e- + 2H2O}\\ \ce{8H+ + K+ + MnO4- + 5e- &-> Mn^2+ + 4 H2O + K+} \end{align}$$ Now we balance the electrons (remember least-common-multiple from math class? You thought you'd never use that, didn't you?)

LCM for 2 and 5 is 10, so multiply the Cl equation by 5 and the Mn reaction by 2.

$$\begin{align} \ce{10OH- + 10 H+ + 10 Cl- &-> 5Cl2 + 10e- + 10H2O}\\ \ce{16H+ + 2K+ + 2MnO4- + 10e- &-> 2Mn^2+ + 8 H2O + 2K+} \end{align}$$

Add everybody back together, make more water, and cancel out the electrons: $$\ce{10H2O + 10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 10H2O + 2Mn^2+ + 8 H2O + 2K+}$$ Take out the excess water: $$\ce{10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 2Mn^2+ + 8 H2O + 2K+}$$ Looks like we're short something to balance out the charges. We can add more chloride spectator ions to both sides of the equation. They're not part of the redox since they are $\ce{Cl-}$ on both sides of the equation: $$\ce{10 HCl + 6HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$ Collect terms one last time and we're done: $$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$

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  • $\begingroup$ @JulieThomasWhich equation can we use to calculate the amount of $\ce{HCl}$ reacted with $\pu{x mol}$ of $\ce{KMnO4}$? $$\ce{10HCl +6H+ +2KMnO4⟶5Cl2 +2Mn^{2+} +8H2O +2K+}\tag{1}$$ $$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}\tag{2}$$ $\endgroup$ Apr 12, 2019 at 17:38
  • $\begingroup$ Redox is also pronounced like it's written: "Red-Ox". $\endgroup$
    – Rafael
    Oct 6, 2021 at 14:54
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I'm assuming that you were expecting a double replacement. Clearly that doesn't happen here, and it's because of the permanganate ion. $\ce{KMnO4}$ is a powerful oxidizing agent. That means that it likes to steal electrons from other species, ultimately producing $\ce{Mn^2+}$ ions and water.

If we look at the species in solution with it, $\ce{K^{+}}$, $\ce{H^{+}}$, and $\ce{Cl^{-}}$, the only one that has electrons to get rid of is the chloride. If chloride loses its surplus electrons it will turn into chlorine, leaving behind manganese(II) ions and some oxygen ions that need a home. Those stick to hydrogen ions from the acid, making water. A portion of the chloride ions are not oxidized and remain in solution to balance the charges of the potassium ions.

I've simplified the mechanism, of course. For example, the permanganate ion doesn't just fly apart into $\ce{Mn^7+}$ and $\ce{O^2-}$ ions once the reaction starts, but the final effect is as if it did.

Oxidation-reduction reactions often have products that are difficult to predict without a significant amount of chemistry experience. Balancing these reactions often requires some new techniques (beyond inspection) such as balancing by oxidation number and balancing by half reactions.

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If you are interested in the mechanism it should look something like this ($\ce{HBr}$ was used instead of $\ce{HCl}$):

$$\begin{align} \ce{2H+ + Br- + MnO4- &<-> H2MnO4Br} \tag{$K$ fast}\\ \ce{H2MnO4Br + H+ + Br- &-> H3MnO4 + Br2} \tag{$k$ limiting}\\ \ce{H3MnO4 &-> \text{products}} \tag{fast} \end{align}$$

I suppose that the first two steps can be divided into more steps.

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