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How do i solve the following redox reaction question?
KMnO4 + Na2SO3 + H2O ---> MnO2 + Na2SO4 + KOH
Please use only H3O+ and NOT H+
THANKS:D
1 Answer
- ?Lv 78 years agoFavorite Answer
KMnO4 + Na2SO3 --> MnO2 + Na2SO4 + KOH
Mn in MnO4- = +7 and becomes +4 in MnO2, reduced, gained 3e-
S in SO3 -2 = +4 and becomes +6 in SO4 -2, oxidized lost 2e-
MnO4- + 3e- + 4H+ --> MnO2 + 2H2O
SO3 -2 + H2O --> SO4 -2 + 2e- + 2H+
balance electrons by multiplying equation 1 by 2 and equation 2 by 3
2MnO4- + 6e- + 8H+ --> 2MnO2 + 4H2O
3SO3 -2 + 3H2O --> 3SO4 -2 + 6e- + 6H+
combine and eliminate
2MnO4- + 6e- + 8H+ + 3SO3 -2 + 3H2O --> 2MnO2 + 4H2O + 3SO4 -2 + 6e- + 6H+
2MnO4- + 2H+ + 3SO3 -2 --> 2MnO2 + H2O + 3SO4 -2
to add/balance the OH- of KOH:
2KMnO4 + 2H2O + 3Na2SO3 --> 2MnO2 + H2O + 3Na2SO4 + 2KOH