Redo Equation:
5H202(aq) + 2MnO4-(aq) +6H(aq)+ --5o2(g) +2 Mn2+(aq)+8H20(l)

can anybody give me the half rxns for this equation and ID oxidizing and reducing agents?

Question

Redo Equation:
5H202(aq) + 2MnO4-(aq) +6H(aq)+ -->5o2(g) +2 Mn2+(aq)+8H20(l)

can anybody give me the half rxns for this equation and ID oxidizing and reducing agents?

anonymous · Answer

equation H202(aq)+MnO4{-}(aq)+H(aq){+}=O2+Mn+(aq)+H20(l) can be balanced in an infinite number of ways: this is a combination of two different reactions

preetha · Answer

You want to break it down into the oxidation half and the reduction half.  So MnO4 is clearly going to Mn2+, so that will be one half reaction.
That leaves the H2O2 going to H2O.
H202(aq)  -->o2(g) +H20(l) 
MnO4-(aq) -->Mn2+(aq)+H2O