* Na2SO3 + KMnO4 + H2O = Na2SO4 + MnO2 + KOH
* J2 + MnO2 + H2SO4 = MnSO4 + HJO4 + H2O
Thanks
Can somebody explain me how to do these chemical reactions?
Moderators: Xen, expert, ChenBeier
- ChenBeier
- Distinguished Member
- Posts: 1542
- Joined: Wed Sep 27, 2017 7:25 am
- Location: Berlin, Germany
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
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