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What is the change in oxidation state of Mn in the reaction of MnO4- with H2O2 in acidic medium?
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What is the change in oxidation state of Mn in the reaction of MnO4- w...
Change in Oxidation State of Mn in the Reaction of MnO4- with H2O2 in Acidic Medium
The reaction between MnO4- and H2O2 in acidic medium can be represented as:
2MnO4- + 3H2O2 + 2H+ → 2Mn2+ + 4H2O + 3O2
Initial State
- The Mn atom in MnO4- has an oxidation state of +7.
- The O atoms in MnO4- have an oxidation state of -2.
- The H atoms in H2O2 have an oxidation state of +1.
- The O atoms in H2O2 have an oxidation state of -1.
Final State
- The Mn atom in Mn2+ has an oxidation state of +2.
- The O atoms in H2O have an oxidation state of -2.
- The O atoms in O2 have an oxidation state of 0.
Analysis
In the reaction, the Mn atom undergoes a reduction from an oxidation state of +7 to +2. This means that the Mn atom gains 5 electrons. The O atoms in MnO4- also undergo a reduction from an oxidation state of -2 to -2.5. The H atoms in H2O2 undergo an oxidation from an oxidation state of +1 to +0. The O atoms in H2O2 undergo an oxidation from an oxidation state of -1 to 0.
The reduction of MnO4- is due to the oxidation of H2O2. The H2O2 molecule donates electrons to the MnO4- ion, reducing it to Mn2+. The H2O2 molecule is oxidized to O2 in the process.
Conclusion
The change in the oxidation state of Mn in the reaction of MnO4- with H2O2 in acidic medium is from +7 to +2, indicating a reduction. This is due to the oxidation of H2O2, which donates electrons to MnO4-, reducing it to Mn2+. The H2O2 molecule is oxidized to O2 in the process.
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