# Regarding the oxidation of Manganese(II) ion by Peroxodisulphate ion

The following information is mentioned in my textbook:

In the laboratory, a manganese(II) ion salt is oxidized by peroxodisulphate to permanganate.

$$\ce{2Mn^2+ + 5S_2O_8^2- + 8H2O -> 2MnO4- + 10SO_4^2- + 16H^+}$$

The manganese ion does get oxidized, but as the oxidation number of sulphur in $$\ce{S_2O_8^2-}$$ and $$\ce{SO_4^2-}$$ is the same, i.e., equal to +6, $$\ce{S_2O_8^2-}$$ doesn't seem to undergo reduction. For a redox reaction, both oxidation and reduction are necessary to occur simultaneously. But in this equation, only oxidation seems to be taking place.

I think it might have to do with the peroxy-bond in the peroxodisulphate ion, but I am not entirely certain. Is it because - oxidation state of two oxygens (of $$\ce{S_2O_8^2-}$$) on the reactant side is -1, while all the oxygen atoms on the product side have their oxidation state equal to -2, so there is a net reduction? Could anyone please confirm if my understanding is correct?

• Review how the oxidation number is defined and you have your answer. // simple hint: Do you need any electrons to convert peroxodisulphate to sulphate ? Are peroxocompounds generally oxidation agents ? How is managed oxidation e.g. by hydrogenperoxide in oxidation number context ? Aug 4 '21 at 7:49
• @Poutnik Thank you so much, I see it now; $\ce{S_2O_8^2^- + 2e^- -> 2SO_4^2-}$ - gain of two electrons, so reduction is indeed taking place
– Pal
Aug 4 '21 at 7:53

It seems OP understands Poutnik's rationalization on the reduction step, so this question does not need an answer. However, I feel it is not completed because OP's statement of:

[...] but as the oxidation number of sulphur in $$\ce{S2O8^2−}$$ and $$\ce{SO4^2−}$$ is the same, i.e., equal to +6, [...]

This statement is not quite true if you just do the calculations and conclude from that as OP initially did. For benefit of other novice chemistry readers, I think it is better to explain what's going on in this situation in detail.

Suppose the oxidation number of sulfur in peroxodisulphate ion $$(\ce{S2O8^2−})$$ is $$\alpha$$. Then if you do usual calculations, it is like:

$$2 \cdot \alpha + 8(-2) = -2 \ \Rightarrow \ \therefore \alpha = \frac{16 - 2}{2} = +7$$

This means there is a reduction happening. However, this calculation is also erroneous. Let's look at the Lewis structure of the peroxodisulphate ion, which is a particularly interesting one:

The trick here is that you should be able to notice that $$\ce{S2O8^2−}$$ ion has two oxygen atoms that are bonded via a single bond, which is known as a peroxide linkage. This linkage will influence the average oxidation state of oxygen in $$\ce{S2O8^2−}$$ ion. The two oxygen atoms in peroxide linkage will have a $$−1$$ oxidation state, similar to the same oxidation state that oxygen has in all peroxides. The rest of oxygens in the anion have usual $$−2$$ oxidation state, which implies that the average oxidation state of oxygen in $$\ce{S2O8^2−}$$ ion is: $$\frac{6(-2) + 2(-1)}{8} = -\frac74$$ Thus, if we redo the calculations:

$$2 \cdot \alpha + 8(-\frac74) = -2 \ \Rightarrow \ \therefore \alpha = \frac{14 -2}{2} = +6$$

This did not imply a reduction when we look at the oxidation number of sulfur in $$\ce{SO4^2−}$$ ion. If it is $$\beta$$ in $$\ce{SO4^2−}$$, then, $$\beta + 4(-2) = -2 \ \Rightarrow \ \therefore \beta = 8 - 2 = +6$$

Yet, if you know the Lewis structure of the peroxodisulphate ion, you would realize there is something going on for oxygen atoms in $$\ce{S2O8^2−}$$ ion, which cannot be explained by these calculations. Therefore, it is always better to conclude by the facts found using the redox half reactions. Consequently, I admit that I also prefer to do usual redox half reactions if I find no change in oxidation states doing usual calculations, because that gives no place to make erroneous calculations. Accordingly, for the reduction half reaction:

$$\ce{S2O8^2− + 2 e- <=> 2SO4^2- } \tag1$$

This is the reduction half reaction. The equation $$(1)$$ clearly indicates the $$\ce{S2O8^2−}$$ ion has undergone a reduction during the complete reaction. The corresponding oxidation half reaction would be:

$$\ce{Mn^2+ + 4H2O <=> MnO4- + 8H+ + 5 e-} \tag2$$

If you combine the equations $$(1)$$ and $$(2)$$ in order to cancel the electrons, you get:

$$\ce{2Mn^2+ + 8H2O + 5 S2O8^2- -> 2MnO4- + 16H+ + 10 SO4^2-} \tag3$$ This is the given equation. When you consider the Standard Electrode (Reduction) Potentials $$(E^\circ)$$ in aqueous solutions at $$\pu{25 ^\circ C}$$, you would find for reaction $$(1)$$ it is $$\pu{2.01 V}$$ and for reaction $$(2)$$ it is $$\pu{-1.51 V}$$ (oxidation potential). Thus, for the completed reaction $$(3)$$ at standard conditions, $$E^\circ_\mathrm{Rxn} = 2.01 + (-1.51) = \pu{+0.50 V}$$, implying it is spontaneous at standard conditions.

• There were 4-5 minor grammar errors (probably typos or keyboard autocorrect) which I edited. Please rollback if the changes are not appropriate.
– TRC
Aug 5 '21 at 13:53