Answer:
Well, purple permanganate, Mn(+VII) is REDUCED to colourless Mn2+ ion, i.e. a FIVE electron reduction (i.e. the difference in oxidation numbers, Mn(+VII)−Mn(+II), represents the NUMBER of electrons transferred..
MnO−4+8H++5e−⟶Mn2++4H2O(l)
And hydrogen peroxide, i.e. O(−I) is OXIDIZED to O(0) in dioxygen…
H2O2(l)⟶O2(g)↑+2H++2e−
And we add TWO of the reduction reaction, to FIVE of the oxidation reaction to retire the electrons…
2MnO−4+5H2O2(l)+16H++10e−⟶2Mn2++8H2O(l)+5O2(g)↑+10