Redox: $\ce{Fe + 2HCl -> FeCl2 + H2}$

Oxidation: $\ce{Fe -> Fe^{2+} + 2e-}$

Reduction: $\ce{2H+ + 2e- -> H2}$

The $\ce{HCl}$ and $\ce{H2}$ are formed by covalent bonds. The $\ce{FeCl2}$ forms an ionic bond. The iron actually loses 2 electrons to the two chlorines, so the oxidation equation makes sense to me. But the hydrogen is always in a covalent bond with the chlorine or another hydrogen atom, it is always sharing electrons, never losing or gaining, so the reduction equation doesn't make sense to me. When the $\ce{H}$ is bonded with the $\ce{Cl}$, it "has" two electrons because it is sharing a pair with the $\ce{Cl}$. After the reaction, when the $\ce{H}$ is bonded with another $\ce{H}$, it still "has" two electrons because it is sharing a pair with another $\ce{H}$, so no electron was gained as the reduction equation indicates.

After a little a thinking I came up with an answer, which raised even more questions:

Since the covalent bond between the $\ce{H}$ and the $\ce{Cl}$ is a polar bond and it behaves more like an ionic compound because of that, when $\ce{HCl}$ is dissolved in water, the $\ce{H}$ will lose an electron to the $\ce{Cl}$, making $\ce{H+}$ and $\ce{Cl-}$ ions, so there is no covalent bond when $\ce{HCl}$ is in water. The $\ce{H+}$ ions can gain one electron and then make a covalent bond with each other, making $\ce{H2}$ molecules.

But what about the $\ce{Cl-}$ ions in the solution? they have already formed an octet in their outer shell by getting an electron off the hydrogen when they got separated, so they are satisfied. Then why $\ce{Fe}$ atoms oxidize? who is the oxidizing agent? The $\ce{H2}$ molecules are already comfortable with their covalent bond, and the $\ce{Cl-}$ ions are also satisfied with their octet.

I'm very confused.


1 Answer 1


To answer your initial question:

The oxidizing agent is the reagent that gets reduced. Conversely: The reducing agent is the reagent that gets oxidized.

Now, addressing your trouble with the concept: You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so: $$\ce{HCl ->[][\ce{H2O}] H+ + Cl-}$$

Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution: $$\ce{2H+ + 2e- -> H2 ^}$$ This is basically our electron sink, because that's where they go "to die", i.e. to vanish from our solution.

But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?

The answer we already know: Iron.

Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.1) $$\ce{Fe^0 -> Fe+ + e- -> Fe^{2+} + 2e-}$$

The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better! $$\ce{Fe^{2+} + 2Cl- <=> FeCl2}$$

Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.


  1. Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -> Fe^{2+} + 2e-}$ and $\ce{Cl- -> Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.
  • 1
    $\begingroup$ Thank you!! It is much clearer now. I was forgetting about the fact that the Chlorine atom had already gained an electron from the Hydrogen when the Hydrocloric acid was dissolved in water, so I thought they were getting the extra electron from the Iron (to form the ionic bond in FeCl2), thus leaving the Hydrogen ions with no source to get the electrons from. $\endgroup$
    – Lucas M.
    Commented Mar 20, 2014 at 15:18

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