Question

Could someone explain to me how to identify the atom oxidized and the atom reduce for this equation?

Answer 1

An old chesnut...`Mn(VII+)` is REDUCED to `Mn(II+)`...we use the method of half-equations...

Explanation:

Oxidations by permanganate ion are particularly useful inasmuch as the oxidized form `Mn(VII+)` is INTENSELY purple-coloured...and the reduced form, `Mn(II+)` is almost colourless, i.e. `Mn^(2+)` has a `d^5` electronic state, and electronic transitions are spin disallowed...and thus...

`underbrace(MnO_4^(-) +8H^+ + 5e^(-))_"purple"rarr underbrace(Mn^(2+) +4H_2O)_"colourless"`

And nitrite, `N(+III)`, is oxidized to nitrate, `NO_3^(-)`...`N(+V)`, a two-electron oxidation...

`NO_2^(-) +H_2O(l) rarr NO_3^(-)+2H^+ +2e^(-)`

And so we take TWO of the former equations, and FIVE of the latter to eliminate the electrons, virtual particles of convenience...

`2MnO_4^(-)+5NO_2^(-) +5H_2O(l) +16H^+ + 10e^(-)rarr 2Mn^(2+) +8H_2O+5NO_3^(-)+10H^+ +10e^(-)`

And we cancel away....

`2MnO_4^(-)+5NO_2^(-) +6H^+ rarr 2Mn^(2+) +5NO_3^(-)+3H_2O`

The which is balanced with respect to mass and charge...as is ABSOLUTELY required... Capisce? Of course the sulfate ions, and the potassium ions are along for the ride, and these are not involved in the redox process. If you like you can add the gegenions as an exercise...