I'm trying to get a fool-proof way of finding the products of a redox reaction. I can do both the oxidation and reduction the half reactions. For example in the potassium permanganate with hydrogen peroxide I found the net ionic equation to be:

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

However, this troubles me. First, what happens to the $\ce{K+}$ that was attached to the permanganate ion? My guess is that it hangs around like a spectator ion to partner up with one of the products at the end, but it can't attach to the reduced manganese because that is also positively charged. Would it do this with the anion that was bound to the $\ce{H+}$ (e.g., $\ce{SO4^2-}$)? Would it attach to the $\ce{Mn^2+}$. Then, what does the $\ce{Mn^2+}$ cation bind to?

I get that only including the reacting species in the net ionic equation is supposed to make things easier, but it always seems to confuse me more. How do I decide what is important and what's not?

  • 1
    $\begingroup$ You're right. There are "spectator" anions and cations. $\endgroup$
    – MaxW
    Feb 17, 2016 at 23:38
  • 2
    $\begingroup$ Is this supposed to be a balanced equation? The charges don't balance. $\endgroup$
    – jerepierre
    Feb 18, 2016 at 0:51
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    $\begingroup$ @jerepierre you right - the charges should balance. The equation should be: $$\ce{5H2O2 + 2MnO4^{-} +6H+ -> 2Mn^{2}+ + 8H2O +5O2}$$ $\endgroup$
    – MaxW
    Feb 18, 2016 at 4:30

1 Answer 1


Spectator ions do indeed just ‘hang around’. They are unchanged during the reaction and after the reaction they just pair up with whatever they can find. Remember that redox (half) reactions are always charge-balanced (if you set it up correctly). Thus, including the spectator ions which also have a constant charge, the overall charge must be zero. Taking your equation $\text{(Redox1)}$ and its half reactions $\text{(Red1)}$ and $\text{(Ox1)}$:

$$\begin{align}\ce{MnO4- + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O}\tag{Red1}\\ \ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox1}\\ \ce{5 H2O2 + 2 MnO4- + 6H+ &-> 2Mn^2+ + 8 H2O + 5O2}\tag{Redox1}\end{align}$$

We added the permanganate as potassium permanganate so there should be a potassium ion. On the reactant sides of the equations $\text{(Red1)}$ and $\text{(Redox1)}$, it can happily hang around with permanganate but it seems a little out of place on the product side; see equations $\text{(Red2)}$ and $\text{(Redox2)}$:

$$\begin{align}\ce{KMnO4 + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O + K+}\tag{Red2}\\ \ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox2}\\ \ce{5 H2O2 + 2 KMnO4 + 6H+ &-> 2Mn^2+ + 8 H2O \\ &+ 5O2 + K+}\tag{Redox2}\end{align}$$

And obviously, that cannot be the end of the story. We need something for the product side — but we also need something to offset the protons. This is where the acid comes in. We generally write it simply as $\ce{H+}$, but there will always be a counterion. For example, we might have used sulfuric acid. In that case, instead of writing $\ce{H+}$ we can also include the counterion as $\ce{H2SO4}$, see the third set of equations:

$$\begin{align}\ce{2KMnO4 + 10 e- + 8 H2SO4 &-> 2 MnSO4 + 16 H2O\\ & + K2SO4 + 5 SO4^2-}\tag{Red3}\\ \ce{H2O2 + SO4^2- &-> O2 + 2e- + H2SO4}\tag{Ox3}\end{align}$$

As is, this seems kind of counter-intuitive since there are still naked sulfate ions. What balances those? It’s the electrons that we are transferring. Since the electrons cancel out when adding up the half-reactions, so do the sulfates, see $\text{(Redox3)}$:

$$\begin{multline}\ce{5 H2O2 + 2 KMnO4 + 3 H2SO4 -> \\ 2 MnSO4 + 8 H2O + 5 O2 + K2SO4}\quad\quad\tag{Redox3}\end{multline}$$

As you can see, only part of the sulfates pairs up with potassium; the remaining sulfates pair up with manganese. All in all, there are enough sulfates around for every species to be netural — as required by the law of conservation of charge.

Identifying spectator ions is rather easy: Simply, anything that doesn’t partake in the actual redox process can be determined a spectator ion. So if you realise that only the sulfite anion of sodium sulfite reacts, sodium is automatically a spectator ion and can be left out. Similarly, if you add nitric acid but no nitrogen oxides $\ce{NO_x}$ can be detected, the nitrate ion is a spectator ion.

I’ll admit, it gets a lot easier identifying spectator ions once you ‘get a feeling’. You can always imagine what a potential spectator ion could do in a redox reaction. For most, it’s either improbable (e.g. potassium being reduced) or would have a notable outcome (e.g. nitrate being reduced).


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