# Potassium Manganate and Acidified Potassium Dichromate colour changes [closed]

I would be obliged if somebody could explain the colour changes for the reduction of potassium manganate and acidified potassium dichromate. I understand that for potassium manganate, it is being reduced differently and this gives us a colour change. But I can’t understand the acid, base, neutral reactions.

I also keep on seeing this equation:

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

Unfortunately I cannot understand it, possibly because it’s an net equations and not the entire reaction.

## closed as unclear what you're asking by Mithoron, Todd Minehardt, Soumik Das, aventurin, andseliskMar 17 at 14:52

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Although you are asking for the color changes of the reduction of potassium manganate ($$\ce{K2MnO4}$$; a green-colored salt, (Wikipedia1)) by potassium dichromate ($$\ce{K2Cr2O7}$$; a red-orange-colored salt, (Wikipedia2)) in acidic medium, the equation you are showing is reduction half reaction of potassium permanganate ($$\ce{KMnO4}$$; a purple-colored salt, (Wikipedia3)) in acidic medium. According to Wikipedia1, potassium manganate is an intermediate in the industrial synthesis of potassium permanganate. Thus, color change for that specific reaction is green to purple (disregarding other interference such as color change of the oxidizing reagent).

The reduction half reaction of $$\ce{K2Cr2O7}$$ in acidic medium is: $$\ce{Cr2O7^2- + 14 H+ + 6e- <=> 2 Cr^3+ + 7 H2O} \qquad E^{\circ} = \pu{1.36 V}$$

The oxidation half reaction of $$\ce{K2MnO4}$$ is:

$$\ce{MnO4^2- <=> MnO4^- + e-} \qquad \qquad E^{\circ} = \pu{-0.558 V}$$

The total redox reaction of $$\ce{K2MnO4}$$ and $$\ce{K2Cr2O7}$$ in acidic medium is: $$\ce{Cr2O7^2- + 14 H+ + 6 MnO4^2- -> 2 Cr^3+ + 6 MnO4^- + 7 H2O} \qquad E^{\circ}_{\text{Rxn}} = \pu{0.802 V}$$

Since $$E^{\circ}_{\text{Rxn}} \gt 0$$, the reaction is spontaneous.