How was it determined that $\ce{MnO4-}$ was reduction, and $\ce{H2O2}$ was oxidation?

Also, why wasn’t the equation $$\ce{MnO4- + 2H2O + 3e- -> MnO2 +4OH-}$$ used? Or the $$\ce{H2O2 + 2H+ +2e- -> 2H2O}$$? (These are from the half equation tables)



  • $\begingroup$ $\ce{KMnO4}$ is quite a powerful oxidizing agent, unless you add something really crazy like dioxygen difluoride to it, it's going to undergo reduction. $\endgroup$ – Pritt says Reinstate Monica Jun 1 '17 at 6:24

In the case of permanganate, it is in a very oxidised state, +7, and all the other stable oxidation states of manganese are lower (+6, +4, +2...) so it will typically reduce in redox reactions (i.e. it will be an oxidising agent). You can also know this from practical experience if you know that permanganate is used as a general oxidising reagent.

More generally, if you don't know whether a species is reducing or oxidising, try to build a picture of the entire reaction:

1) If you know what species it ends up as, it's easy: you only have to count electrons using standard charge rules. For instance, for MnO4- you know that oxygen atoms have a formal charge of 2-, so the manganese atom is formally Mn(7+); if you know that the product is Mn2+, it is reducing.

2) If you don't know that, look at what it is reacting with. Sometimes, one or two of the species will only oxidise or reduce (if it's the highest or lowest oxidation state you find in the table). For instance, MnO4- is in the highest oxidation state you'll find for manganese at the table (Mn7+), so you know it is reducing.

3) The trickiest case is when you have two species reacting that, depending on the circumstances, could both act as oxidising or reducing agents (i.e. species that appear at your tables both as reagents and as products). In that case, you have to use the standard reduction potentials to figure out.

About whether to use acidic or basic potentials (the tables with H+ or the tables with OH-), these are all equivalent and you are OK using either (as long as you are consistent!). Their difference in value comes from the water dissociation constant Kw, but if you write the semireactions and adjust the water, you end up with exactly the same results in both cases. If you know that your medium is acidic or basic it's generally a good idea to use the corresponding values (it saves you having to adjust the water), but it will still work as long as you don't mix them up and adjust the reaction (I can't stress that enough).


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