When l asked my teacher why the reaction between $\ce{S2O8^2-}$ and $\ce{Co^3+}$ did not occur, he said both were oxidising agents so their reaction would not be redox and thus the reaction would not occur. So am l right to say then that all reactions that are not redox will not occur ?

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    $\begingroup$ no, e.g. common neutralization reactions are not redox $\endgroup$
    – 234ff
    Commented Aug 3, 2020 at 3:12
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    $\begingroup$ No!! $\ce{}$ $\endgroup$ Commented Aug 3, 2020 at 3:56
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    $\begingroup$ You say the reaction, but which one? You need to propose products and stochiometry, then one can say yes or no. $\endgroup$
    – Karl
    Commented Aug 3, 2020 at 7:39

2 Answers 2


I think some of the responses are not getting to the crux of the problem. The issue is not about whether other reaction types can occur. There are several classes of reactions, of course, such as redox reactions (which involve changes in oxidation state), acid-base reactions (exact definition depends on how you define acid and base, but generally these don't involve oxidation state changes), and more.

I would also say that, as quoted, this explanation provided to you is exceedingly poorly phrased. So let's explain what's the matter.

The issue is that for any redox-type reaction to occur, you need a reducing agent and an oxidising agent. Changes in oxidation states are essentially additions or subtractions of electrons. If an atom gains electrons, its oxidation state is lowered, and vice versa. But in a chemical reaction, you cannot spontaneously create electrons out of thin air. So for atom X to gain electrons, those electrons have to come from somewhere else: in other words, there has to be a simultaneous loss of electrons from atom Y. In this situation one would say that X is reduced and Y is oxidised.

One cannot have a reaction in which both X and Y are gaining electrons at the same time, because electrons can't be magically pulled out of a hat. Likewise, one cannot have a reaction in which both X and Y lose electrons at the same time, because those electrons would have nowhere to go to, and you can't just vanish them away into thin air.

Now, both $\ce{S2O8^2-}$ as well as $\ce{Co^3+}$ are species which have pretty high oxidation states. Consequently, they don't have any electrons to give away, or at the very least you'd need some serious firepower to get them to give up the electrons that they do have. This means that there's very little chance of any electron transfer occurring between them, as neither is keen on giving up their electrons to the other.

This doesn't preclude other types of reaction which don't involve electron transfer from occurring, of course. Ruling these out would require a whole new explanation, so is beyond the scope of this answer. But I get the feeling that that isn't the point of your question.


We cannot say something to be a truly oxidising or truly reducing agent. Although, some compounds can be observed to display solely reducing activity or solely oxidising property in many reactions, there may be reagents which cause them to reverse their activity.

Eg. Xenon fluorides are considered oxidising agents in majority of reactions. It can oxidise bromate ion to perbromate, $\ce{Ce(III)}$ to $\ce{Ce(IV)}$, $\ce{Fe^2+}$ to $\ce{Fe^3+}$ etc. But it would be incorrect to name it as an absolute oxidising agent. In the reaction,

$\ce{O2F2 + XeF4 -> XeF6 + O2}$

The $\ce{XeF4}$ is a reducing agent, while $\ce{O2F2}$ is the oxidizing agent. Thus the attributes interchange.

And there are several reactions which are not redox. Neutralization, as in the comment is an example.

  • $\begingroup$ Then why does the reaction between S2O8 2− and Co 3+ not occur? $\endgroup$
    – Rabbit
    Commented Aug 3, 2020 at 4:13
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    $\begingroup$ Nobody knows why a reaction does not happen. Why does charcoal not react with oxygen at room temperature ? Why does potassium permanganate not react with water at room temperature ? It is probably a question of activation energy. That's what we say when we don't know the answer. $\endgroup$
    – Maurice
    Commented Aug 3, 2020 at 9:12
  • $\begingroup$ @Rabbit Which reaction between $\ce{S2O8^2-}$ and $\ce{Co^3+}$ should happen ? It is like asking why a wooden log does not burn in petrol vapour, or why vater vapour mixed with oxygen does not ignite. Reactions are like business, there must be a seller and a buyer. If there are 2 sellers and no buyer, there is no business. And vice versa. Both above ions are buyers ( of electrons ), but there is no electron seller. $\endgroup$
    – Poutnik
    Commented Aug 12, 2020 at 15:28

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