I'm assuming you are referring to balancing redox reactions in acidic or basic conditions.
When you are balancing equations, you want to make sure that the same amount of atoms are on both sides of the equation right? This applies to hydrogen as well, and when you are balancing redox reactions that involve Oxygen and/or hydrogen in the species being oxidized/reduced, there are a few ways we can go about doing this.
In acidic solutions, we know that H+ is readily available and so if we need to have more hydrogen on one side of a reduction or oxidation, we can just add H+ to that side.
1) staring with
in acidic solution
2) add 2 H2O to the right side in order to balance out the 2 O atoms in MnO2:
3) now add 4H+ to the left side in order to balance out the 4H atoms in 2H2O:
4) Check the final equation and now you see that the number of atoms and total charge on each side is equal.
In basic solutions, we know that OH- ions are readily available. For every hydrogen you need on one side of a reaction to balance out H, you add one H2O to that side and one OH- on the other side. Since the difference between OH- and H2O is one H atom, doing this is the equivalent of adding one H atom to the side of the reaction that you need H atoms.
1) starting with
in basic solution
2) add one H2O to the right side to balance out O in BrO-:
3) To balance out H, add 2HO- to the right side and 2H2O on the left side:
4) The last step results in the left side having 4 more H than before, and now both sides are balanced. However, we can cancel out the H2O on the right, which gives us:
5) Check the final equation and you will see that The O, H, and total charge on each side is now equal.