# How to balance oxidation-reduction reaction?

Why do I have know the oxidation numbers of the reactants and products when I am balancing an acidic or basic solution? I know how to find the oxidation numbers and how to balance reactions usually but my answers keep coming out only partly correct and I think it is because I do not know how to integrate the oxidation numbers into the balancing process. could someone walk me through balancing an acidic and basic solution, detailing where the oxidation number play a role?

Acidic: MnO2 + HNO2----->MN2+ +NO3-

Basic:MN04- + I- -------->MNO2 + IO3-

I know the right answer for both but I can't get to it through balancing.

Acidic: MnO2 + HNO2----->MN2+ +NO3-
In this one Mn starts in the +4 OS and ends in the +2 OS (=reduction)  while N starts in the +3 OS and ends at +5 (=oxidation)
Best to separate oxidation and reduction halves. Then where needed, balance oxygen by adding water, balance H by adding H+ ions and balance charge by adding electrons.

MnO2----->Mn2+

MnO2----->Mn2+ + 2H2O

MnO2+ 4H+ ----->Mn2+ + 2H2O

MnO2+ 4H+ + 2e- ----->Mn2+ + 2H2O = reduction

HNO2 ----> NO3-

HNO2 + H2O ----> NO3-

HNO2  + H2O  ----> NO3- + 3H+

HNO2  + H2O  ----> NO3- + 3H+  + 2e- = oxidation

As electrons balance, just add the two equations

MnO2+ 4H+ + 2e- + HNO2  + H2O ----->Mn2+ + 2H2O + NO3- + 3H+  + 2e-

MnO2+ H+ + HNO2 ----->Mn2+ + H2O + NO3-

Basic:MN04- + I- -------->MNO2 + IO3-  Do the same and then add as many OH- ions to the side that has the H+ ions, and also add the same number OH- to the other side

Basic:MN04- + I- -------->MNO2 + IO3-. Mn changes from +7 to +4 (= reduction) while I- changes from -1 to +5 ( = oxidation)

MnO4- -------->MnO2

MnO4- -------->MnO2 + 2H2O

MnO4- + 4H+ -------->MnO2 + 2H2O

MnO4- + 4H++ 3e- -------->MnO2 + 2H2O (reduction)

I- ---> IO3-

I- + 3H2O ---> IO3-

I- + 3H2O ---> IO3- + 6H+

I- + 3H2O ---> IO3- + 6H+ + 6e-

Before adding the two balance the electrons by multiplyinn the reduction by 2

2MnO4- + 8H++ 6e- -------->2MnO2 + 4H2O

I- + 3H2O ---> IO3- + 6H+ + 6e-

2MnO4- + 8H++ 6e- + I- + 3H2O -------->2MnO2 + 4H2O + IO3- + 6H+ + 6e-

2MnO4- + 2H+ + I-  -------->2MnO2 + H2O + IO3-

For basic solution

2MnO4- + 2H+ + I- + 2OH- -------->2MnO2 + H2O + IO3- + 2OH-

H+ and OH- combine --> water

2MnO4- + 2H2O + I-  -------->2MnO2 + H2O + IO3- + 2OH-

2MnO4- + H2O + I-  -------->2MnO2  + IO3- + 2OH-