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Sunteiks
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Joined: 01 Oct 2017
Posts: 1
PostPosted: Sun Oct 01, 2017 3:54 am    Post subject: Can somebody explain me how to do these chemical reactions? Reply with quoteFind all posts by Sunteiks

* Na2SO3 + KMnO4 + H2O = Na2SO4 + MnO2 + KOH
* J2 + MnO2 + H2SO4 = MnSO4 + HJO4 + H2O

Thanks Wink
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ChenBeier
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Joined: 27 Sep 2017
Posts: 18
Location: Berlin, Germany
PostPosted: Mon Oct 02, 2017 11:12 am    Post subject: Reply with quoteFind all posts by ChenBeier

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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