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:

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

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$\ce{KMnO4}$ 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?

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.

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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:$$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$