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anonymous

  • 5 years ago

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?

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  1. anonymous
    • 5 years ago
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    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

  2. Preetha
    • 5 years ago
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    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

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