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Newbie

Joined: 01 Oct 2017
Posts: 1
 Posted: Sun Oct 01, 2017 3:54 am    Post subject: Can somebody explain me how to do these chemical reactions? * Na2SO3 + KMnO4 + H2O = Na2SO4 + MnO2 + KOH * J2 + MnO2 + H2SO4 = MnSO4 + HJO4 + H2O Thanks

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Joined: 27 Sep 2017
Posts: 18
Location: Berlin, Germany
 Posted: Mon Oct 02, 2017 11:12 am    Post subject: Develop halfreactions MnO4 - to MnO2 MnO4 - + 2 H2O => MnO2 + 4 OH- Balance chargse MnO4 - + 2 H2O + 3 e- => MnO2 + 4 OH- Reduction SO3 2- to SO4 2- SO3 2- + 2 OH- => SO4 2- + H2O Balance of charge SO3 2- + 2OH - => SO4 2- + H2O + 2e- Oxidation Make electrons equal 2 MnO4 - + 2 H2O => + 6 e- =>2 MnO2 + 8 OH- 3 SO3 2- + 6 OH - => 3 SO4 2- + 3 H2O + 6 e- Addition and eliminate of electrons, water and OH- 2 MnO4 - + 3 SO3 2- => 3 SO4 2- + 2 MnO2 + 2 OH- + H2O K and Na spectator ions and not necessary in the calculation. I2 to HIO4 I2 + 8 H2O => 2 HIO4 + 14 H+ I2 + 8 H2O => 2 HIO4 + 14 H+ + 14 e- Oxidation MnO2 to Mn2+ MnO2 + 4 H+ => Mn2+ 2 H2O MnO2 + 4 H+ + 2 e- => Mn2+ 2 H2O times 7 7 MnO2 + 28 H+ + 14 e- => 7 Mn2+ + 14 H2O Reduction Addition, eliminate electrons , H+ and water. I2 + 7 MnO2 +14 H+ => 2 HIO4 + 6 H2O + 7 Mn2+ Sulfate is spectator ion
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