2MnO4- + 5H2S+ 6H+ ----> 2Mn2+ + 5S+ 8H2O. The oxidation state of manganese changes from what to what? How many electrons are transferred in the reaction?

Answer 1

First, remember that O has an oxidation of 2- (except in peroxides when it is 1), and H has an oxidation number of 1+ (except in hydrides when it is 1-). In this problem you don't need to worry about H.

Looking at MnO₄^-

We know that O has an oxidation number of 2-, so what does Mn have to be so that the overall charge is 1-?

There are 4 oxygens for a total of -8 so Mn would have to be +7 in MnO₄^-.

Looking at Mn²⁺, we see that Mn has an oxidation number of 2+

So, Mn went from 7+ to 2+ which means it gained 5 electrons