# Why are we permitted to add things to balance redox equations?

Consider balancing redox half equations.

As a general rule, we can add electrons, water, hydrogen ions or hydroxide ions to balance the redox half equation. However, it is not entirely clear why we can just add something to an equation.

For electrons, it make sense that a change in oxidation state is occuring so the electrons must go somewhere, therefore balancing it by adding electrons to either side of the equation. However, for water, hydrogen, and hydroxide ions, why are we permitted to just add these. Does this mean we can just add anything?

• You're not really "adding" anything, just trying to describe actual reaction. Commented Apr 5 at 21:36
• οὐδὲν ἐξ οὐδενός, ex nihilo nihil, TANSTAAFL, matter is conserved. Unless there is a something to provide those "additions", such as water for H+ and OH- ions (and H2O should appear on the other side), the equation is not balanced. Commented Apr 5 at 22:24

By adding anything, you just remove one of several fundamental inequalities in the equation.

A chemical equation is formally nothing else but a very specific syntax for a set of linear algebraic equations with one degree of freedom. So you can deliberately choose one of stoichiometric coefficients and other ones are proportional to it. It is usually done in such a way that all coefficients are the smallest possible integers.

There is the mandatory requirement of equality of atom counts for all involved chemical elements, plus electric charges. No atom nor charge can appear or disappear during the reaction, but their combinations can be regrouped.

On the formal, accounting level, ignoring reality of chemistry, you can do the regrouping by any, even crazy way, involving components that are not even assumed to exist. $$\ce{F2 + 16 e- -> 2 F^8-}$$ is still formally correct.

Formal correctness is mandatory yet insufficient requirement for a correct chemical equation.

As there is another level, the level of chemistry, that must respect reality. There must be respected, what is obvious, observed, or assumed to happen.

Imagine as a redox half-reaction the well known reduction of a permanganate ion to manganese(II) ion in acidic solution:

\begin{align} \ce{MnO4-(aq) &-> Mn^2+(aq)}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &\ne 0\\ \ce{Ch}: -1 &\ne +2 \end{align}

Oxygen cannot disappear. As there is no formation of $$\ce{O2}$$ nor $$\ce{OH-}$$, there is probably formed water:

\begin{align} \ce{MnO4-(aq) &-> Mn^2+(aq) + 4 H2O}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &= 4\\ \ce{H}: 0 &\ne 8\\ \ce{Ch}: -1 &\ne +2 \end{align}

As hydrogen cannot be created and does not come from $$\ce{H2O}$$ nor $$\ce{OH-}$$, it comes probably from $$\ce{H+(aq)}$$ ions:

\begin{align} \ce{MnO4-(aq) + 8 H+(aq)&-> Mn^2+(aq) + 4 H2O}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &= 4\\ \ce{H}: 8 &= 8\\ \ce{Ch}: +7 &\ne +2 \end{align}

As the net charge cannot change, we have to add 5 electrons to the left:

\begin{align} \ce{MnO4-(aq) + 8 H+(aq) + 5 e- &-> Mn^2+(aq) + 4 H2O}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &= 4\\ \ce{H}: 8 &= 8\\ \ce{Ch}: +2 &= +2 \end{align}

• Enumeration is done, the formal level is satisfied.
• Chemistry level is satisfied too, as there is observed consuming of reactants and production of products, which all are known to exist.

IF we consider reduction of a permanganate ion to manganese dioxide in mildly alkalic solution:

\begin{align} \ce{MnO4-(aq) &-> MnO2(s)}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &\ne 2\\ \ce{Ch}: -1 &\ne 0 \end{align}

Oxygen must be passed somewhere and some negative charge as well, so there are probably formed $$\ce{OH-}$$ ions.

\begin{align} \ce{MnO4-(aq) &-> MnO2(s) + 2 OH-(aq)}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 4 &= 4\\ \ce{H}: 0 &\ne 2\\ \ce{Ch}: -1 &\ne -2 \end{align}

Hydrogen must come from $$\ce{H2O}$$, so does the half of oxygen in $$\ce{OH-}$$:

\begin{align} \ce{MnO4-(aq) + 2 H2O &-> MnO2(s) + 4 OH-(aq)}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 6 &= 6\\ \ce{H}: 4 &= 4\\ \ce{Ch}: -1 &\ne -4 \end{align}

3 electrons have to be added to the left side to balance charges:

\begin{align} \ce{MnO4-(aq) + 2 H2O + 3 e- &-> MnO2(s) + 4 OH-(aq)}\\ \ce{Mn}: 1 &= 1\\ \ce{O}: 6 &= 6\\ \ce{H}: 4 &= 4\\ \ce{Ch}: -4 &= -4 \end{align}

• Enumeration is done, the formal level is satisfied.
• Chemistry level is satisfied too, as there is observed consuming of reactants and production of products, which all are known to exist.

You can "add electrons, water, hydrogen ions or hydroxide ions to balance the redox half equation" only when the reaction occurs in water.

That means that lots of water molecules are always there surrounding the reactants — we just don't write them in the reaction formulas if they're "just spectating" and not actively participating in the reaction. But they're still there and they can participate in the reaction. And if you notice that a reaction you've written down seems to be unbalanced, so that electrons or (hydrogen or oxygen) atoms seem to appear or disappear during the reactions — well, given that the reaction occurs in water, the obvious place where those electrons or atoms could come from or go to is the water.

Also, given that water self-ionizes, hydronium and hydroxide ions will also always be present in the water solution. Their concentration is a lot lower than that of neutral water molecules (and depends on pH), but they're also a lot more reactive than neutral water molecules, so they can still bump into your reactants and react with them at an appreciable rate. So if your reaction seems to need one of those ions on the reactant side to balance it, well, it probably does. And it can still happen, because those ions are always present in water.

And, of course, if your reaction produces hydronium or hydroxide ions, those will just end up solvated in the water (and possibly eventually neutralized by recombining with the opposite type of ion to form neutral water molecules).

We add water molecules because most reactions tend to happen in aqueous solutions. As for whether to add $$H^+$$ or $$OH^-$$ ions depends on the pH of the reaction mixture.

For permanganate oxidations; $$\ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O}$$ We add $$H^+$$ because the reaction is taking place in an acidic medium while balancing the reaction. Similarly; $$\ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-}$$ We add $$OH^-$$ because the reaction is taking place in a basic medium while balancing the reaction. You can, technically add $$4H^+$$ ions in the above reaction on both sides and get; $$\ce{MnO4^- + 4H^+ + 3e^- -> MnO2 + 2H2O}$$ But this would imply that the reaction is taking place in an acidic medium.