I asked this question on Math Stackexchange, but thought I'd get a more chemical view:


I understand that for redox reactions in acidic conditions, water molecules and hydrogen ions can be added to a reaction in order to make balancing it easier. However, my teacher specifically told us that we should only add in water molecules and hydrogen ions if and only if at least one of these conditions are met (the accuracy of this may be debatable, but he specifically told my class that marks will be taken off on assessments if we do not follow this instruction):

  • There are zero hydrogen atoms/ions on one side of the reaction
  • There are zero oxygen atoms/ions on one side of the reaction
  • The reaction is impossible to balance by manipulating coefficients

The third point is what is giving me trouble, and why I asked for a mathematical explanation. How can we just look at a reaction and determine whether it is possible to balance or not?

  • 3
    $\begingroup$ Balancing just means that every atom on the left side appears also on the right side and net charge on the left side equals net charge on the right side. It does not mean that the balanced chemical equation is realistic or even plausible. The third condition may just mean that it is assumed that an acidic aqueous solution has water and hydrogen ions freely available for balancing purposes. $\endgroup$
    – Ed V
    Dec 8, 2021 at 0:06
  • $\begingroup$ I stress that the teacher was just giving a set of "to do list". The meaning is in fact that all chemical reactions conserve atoms and charge and, thus, are inherently balanced. Further, all the linked threads have nothing to do with chemistry, in one case blatantly, in the other case is just simple mathematics that, perhaps disguised by the use of chemical symbols, was left open in Maths SE. $\endgroup$
    – Alchimista
    Dec 8, 2021 at 9:03

3 Answers 3


The reaction is impossible to balance by manipulating coefficients

I feel you over-interpreted the statement. Instead of a pure mathematical reasoning, think in terms of mass balance. New atoms cannot be created or destroyed during a chemical reaction.

Let us see the classical $\ce{Fe^{2+}}$ and $\ce{MnO4-}$ reaction.

The case of $\ce{Fe^{2+}->Fe^{3+} + e-}$ is trivial.

For $\ce{MnO4^{-}-> Mn^{2+}}$

Now what choices we can have? We have to rely on chemical knowledge and experimentally observed facts (i) the reaction is carried out in acidic medium and (ii) Mn(II) is formed.

Since oxygen atoms cannot be destroyed, and we know by experimentation that the reaction has to be carried out in acidic medium (in basic medium this reaction does not proceed), so $$\ce{MnO4^{-} + H+-> Mn^{2+} + hydrogen containing molecule}$$

What choices do you have expect to add water on the right hand side? Oxygen has to go somewhere. This is what your instructor meant.

I know a lot of mathematically inclined chemists have developed matrix algenra methods, but a matrix can only be set up when you know beforehand what products and reactants are there. Mathematics cannot tell us which molecule will be formed.

Go to Google Scholar and search balancing chemical equations + matrix.


Before any chemical equation can be balanced the reactants and products must be defined; otherwise, it is just a math problem and might not even be possible. A simple reaction: stannous ion is used to reduce ferric ion to ferrous ion in some iron analyses: Fe+3 + Sn+2 = Fe+2 + Sn+4. This equation is easy to balance and does not require any water even though Sn+4 hydrates and water is most likely involved in the reaction. 2Fe+3 + Sn+2 = 2Fe+2 + Sn+4. Lets write a wrong equation without oxidation and reduction: Fe+3 + Sn+4 = Fe+2 + Sn+2! Now try to balance it.

Redox reactions with poly atomic ions such as MnO4- or CrO4= in water solution need acid to react with the oxygens. This acid comes from H+ in acidic solution or H2O in neutral or basic solutions. The acids take part in the reaction and are consumed; that is the main reason that they are used to balance the equation, not just to make it easy. Fe+2 + MnO4- = Fe+3 + Mn+2 to remove 4O= you need 8H+. 8H+ + Fe+2 + MnO4- = Fe+3 + Mn+2 + 4H2O. The manganese is taken care of but we have 4 extra + charges on the left each Fe+2 adds a +charge to the right so add 4Fe+2 and 4 Fe+3 to give: 8H+ + 5Fe+2 + MnO4- = 5Fe+3 + Mn+2 + 4H2O. Balanced with no need for oxidation numbers or half reactions [although they can and do help]. A solid-state reaction, here there is no water or acid and I making up a possible reaction: FeO + KMnO4 = MnO2 + Fe2O3 + K2O. This balances easily using changes in oxidation numbers but lets do it chemically. First thing there are odd-even problems so fix them. 2FeO + 2KMnO4 = 2MnO2 + Fe2O3 + K2O; 10Os on the left 8 on the right. Each 2FeO adds one O so we need 4 more FeO! 6FeO +2KMnO4 = 2MnO2 + 3Fe2O3 +K2O.

The point of all this is that balancing equations really is about learning chemistry, identifying reaction conditions and determining all the products of a reaction; it is not learning tricks and rules. Do learn the various methods, oxidation number change, redox half reaction balancing, the chemical method I illustrated that is particularly useful in changes in functional groups in organic compounds.

  • 1
    $\begingroup$ You may find useful these links for text formatting ( not to be applied to titles ): notation , chem/math formula/equation formatting and upright vs italic $\endgroup$
    – Poutnik
    Dec 9, 2021 at 10:50
  • 1
    $\begingroup$ @jimchmst. $\ce{FeO}$ and $\ce{K2O}$ are not common substances at all. They are difficult to obtain and will not be available in the usual laboratory or by mixing aqueous solutions. $\endgroup$
    – Maurice
    Dec 9, 2021 at 21:39
  • $\begingroup$ FeO is far from uncommon. It's seen all the time as part of the scale formed on hot-rolled steel (the rest is mostly a different iron oxide, Fe3O4). $\endgroup$ Dec 9, 2021 at 21:46
  • $\begingroup$ I stated that I made up the reaction to demonstrate that an equation could be balanced easily if all the products and reactants were listed. That is a "possible" solid state [possibly in a melt] reaction and invoking acid or water is not appropriate in an attempt to balance. $\endgroup$
    – jimchmst
    Dec 9, 2021 at 23:25

The key trick to balance an equation that may involve, for example, a Mn compound undergoing a change in oxidation states from the left to the right side of the equation, is to focus on the fact that the same # of Mn atoms must appear, in total, on both side of the equation. The adjustment factor to the Mn compound, for this to occur, should be somewhat apparent (for example, going from 2 atoms to 3, requires 3 and 2 coefficients to equate each side at 6). Then use this information to assist in computing the required # of oxygen and say hydrogen atoms appearing outside of the Mn compound to arrive at a total balanced equation. This can be may easier, at times, by also noting that each side of the equation's total charges must also be equal (if the sum +/- charges is +1 on one side, it must match the other). Note, that an electron has a charge of -1 and two electrons -2, etc. Also, atom count and charge balance are both required (albeit, the latter often not emphasized as often naturally occurring) for a valid equation, in accord with the preservation of matter and energy.

Do your best to follow the teacher's rule, however, the last rule "The reaction is impossible to balance by manipulating coefficients" requires adjusting in practice.

If the actual resulting mathematics indicates answers for each coefficient that are unexpectedly large (irrational), present the results as a raw unbalanced equation. Clearly, some inputs are missing or not applicable for the reaction you are trying to balance (that is, to correctly explain).

Not surprising, there is a problem with universal rules for balancing chemical reactions, as all reactions are not transparent as to how precisely the observed products were actually formed.

This is evident, for example, when the experimentally estimated coefficients are decimal amounts. Evidently, the suggested reaction equation may a summary of several random occurring more elemental reactions (usually, these can be radical reactions, which are reversible and have temperature sensitive forward and backward reaction rates).

A larger problem arises when the reaction occurs in an open system with say the amount of oxygen from air placed into the system is not known. In case again of say fractional weights and some inputs unknown, how does one present results?

My recommendation is honest reporting, as an unbalanced equation, or otherwise, the ability to speculate on true reaction pathways could be lose under a perception of false certainty.

Per examination of the literature, this problem is also resolved by stating "+ other products". The latter is acceptable, in my opinion, as an honest statement of lack of knowledge as to underlying mechanics and/or product estimate accuracy issues.

Hope this helps!


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