# How does KMnO₄ affect electrolysis?

I'm just blindly experimenting. I tried to dissolve some potassium permanganate in water before performing electrolysis.

I can't see much difference. How is $\ce{KMnO4}$ dissolved in water? How does it react during electrolysis?

I added salt and the electrolysis was so violent that the copper started to burn (probably reacting with something).

A lot of questions have been posed on electrolysis and electrochemistry on this forum recently, with many recurring themes. I will explain electrolysis in depth in order for a true understanding to come out of this; if you're in a hurry then skip to the last three paragraphs.

One of the most important things about learning electrochemistry is the abstract property of voltage. Voltage can be considered a driving force, which denotes which direction charges are going to flow in a system. A high positive voltage means that charges most definitely wish to flow in that direction, and a high negative voltage means that the charges wish to flow strongly in the opposite direction. If we really want charges to flow in the direction of negative voltage, we must supply energy to the system in order for that to happen—in other words, we're supplying energy to move a charged particle against an electric field (work must therefore be done to do this, as with mass in a gravitational field).

And so it is with $$\ce{H2O}$$! The reaction corresponding to our notion of electrolysis is:

$$\ce{2H2O -> O2 + 2H2}$$

At least two of the intermediate species in electrolysis are going to be $$\ce{OH-}$$ and $$\ce{H+}$$, and so the energy requirement for this process takes into account the energy required to move an electron from $$\ce{OH-}$$ to the anode, and also to move an electron from the cathode to $$\ce{H+}$$.The voltage requirement for this process is $$\pu{-1.23V}$$. In reality the voltage requirement is closer to $$\pu{-2V}$$ taking into account overpotential (the electrochemical equivalent to activation energy).

So this negative voltage tells us that this process will not happen spontaneously, and we will need to supply a voltage of $$\pu{2V}$$ in order for this to happen. So far so good: we never observe electrolysis in a standard beaker of water just sitting in the lab!

Your question asked what happened to the $$\ce{KMnO4}$$ during electrolysis. $$\ce{Mn}$$ is quite happy in a $$\ce{Mn^7+}$$ and $$\ce{Mn^4+}$$ oxidation state. Therefore in solution, the following reaction can happen:

$$\ce{Mn^7+ + 3e- -> Mn^4+}$$

Note that the above is a simplification, $$\ce{Mn^7+}$$ and $$\ce{Mn^4+}$$ don't occur simply as isolated ions in solution. In our case, these oxidation states of manganese are tied up in molecules:

$$\ce{MnO4^- -> MnO2}$$

But the above isn't balanced, so we ask ourselves what other likely molecules are going to be produced in order to balance our oxygens. It's common for these redox systems to produce $$\ce{H2O}$$, so we add $$\ce{H2O}$$ on the right to balance the oxygens, and add $$\ce{H+}$$ on the left so we have enough hydrogens. Thus:

$$\ce{MnO4^- + 4 H+ -> MnO2 + 2 H2O}$$

The voltage requirement for this reaction is $$\pu{+1.68 V}$$, so this means that the electron transfer happens spontaneously. In other words, this reaction is more likely to occur than the electrolysis of water at a less than neutral pH. If you start your electrolysis of water using distilled water (neutral pH implied) then the reduction of $$\ce{Mn^7+}$$ doesn't happen initially. It's only when the electrolysis starts and $$\ce{H+}$$ is generated that the reduction of $$\ce{Mn^7+}$$ can happen; this is then favoured over electrolysis on thermodynamic grounds.

As a last point, it's unlikely your copper electrode was burning, but these can get very hot if the voltage is high enough and hence melt. Turn the voltage down as only $$\pu{2V}$$ is needed for water electrolysis.

Experimental hint: try using $$\ce{NaOH}$$ and $$\ce{KMnO4}$$ in combination and see if there is any change in your observations. Then try increasing the amount of $$\ce{NaOH}$$ relative to $$\ce{KMnO4}$$ to see if you observe any changes which would suggest my theory. Note that other manganese reactions (pdf via The Internet Archive) can still happen at lower voltages, which does complicate things.

To answer your question directly, nothing additional is observed using $$\ce{KMnO4}$$ as it reacts instead of behaving as a charge carrier for electrolysis. A quality of a good electrolyte is that it does not interfere with your reaction of interest.