This link explains the process behind balancing redox equations.

Now you have to add things to the half-equation in order to make it balance completely.

All you are allowed to add are:

  • electrons
  • water
  • hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead)

Adding electrons makes sense, because this whole process models electron transfer. Any added electrons will be accounted for elsewhere in the reaction.

Adding water makes sense as well, because it's an entire molecule -- it is essentially part of the reaction at this point.

I don't understand why $\ce{H^+}$ ions can be added simply for balancing purposes.

In the second example on the same site, the final reaction is this:

$$\ce{2MnO4- + 6H+ +5H2O2 -> 2Mn^{2+} + 8H2O + 5O2}$$

Where does the $\ce{6H^+}$ come from, in the real world? It seems like it just appears out of nowhere to make the math work out.

What is actually happening, on a physical basis, to make this type of balancing check out?

  • $\begingroup$ Hello and welcome to Chemistry.SE. If you have any questions regarding the site, you can visit the help center. Best of luck with your problem! $\endgroup$
    – airhuff
    Commented Mar 15, 2017 at 2:54

1 Answer 1


You should really only add $\ce{H+}$ to balance if a reaction is explicitly said to take place in acidic solution. For example, in your link it says.

The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid.

So the sulfuric acid allows you to add these $\ce{H+}$. Alternatively, if it is in basic solution, $\ce{OH-}$ atoms can be added instead.

  • 2
    $\begingroup$ To clarify, you can add these $H^+$ and $OH^-$ in acids/bases because they are necessarily present in acids/bases, similar to adding just a whole $H_2O$ molecule? $\endgroup$
    – Aaa
    Commented Mar 15, 2017 at 3:51
  • 2
    $\begingroup$ Yes, if you are in acidic solution, you know there are available $\ce{H+}$ just as you know that in an aqueous solution you can always find water molecules to help you balance the reaction. $\endgroup$
    – Tyberius
    Commented Mar 15, 2017 at 4:28

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