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Balancing Redox Equations

Submitted by Anonymous (not verified) on Wed, 04/18/2007 - 16:00

How do you balacing redox equations, im scared that question 4 is gonna have one of them and im screwed if it does


LOL. I can imagine. It really isn't that bad. I have yet to write up a tutorial for it for the site. You have to atom balance and electron balance.

I'll have some of my students reply tomorrow and they can tell you there strategies.


Well, you will be given the full equation, and asked only for the molar coefficients, so you must first separate the overall equation into the 2 half reactions: oxidation and reduction. The best way to figure out which reaction is which is to go through the equation and label the oxidation state of each atom both before and after the reaction. (Note: in polyatomic ions, you must do whatever they are bonded with first, then the atoms within the polyatomic ion). Once this is done, you just have to look for an atom which has an oxidation number that is decreased, since that will be the atom that is reduced. The atom whose number is raised will be oxidized. Once you have the separated half reactions, you can begin to balance them separately. For example, if you have the oxidation half reaction Zn--> Zn+, your job is fairly easy, since you must only balance the charge by adding an electron to the right side. However, if you have something like MnO4-->Mn2+, then you must balance both the charges and the atoms. I generally start with atoms first, so I would add Oxygen (in the form of water) to the right side of the reaction. MnO4-->Mn2+ & 4H2O. However, this then creates the other problem that you need Hydrogen on the left. Therefore, you add H+ to balance the atoms: 8H+ & MnO4-->Mn2+ & 4H20. Now you take the overall charge of each side, and add electrons as necessary to balance the charge of the reaction. Therefore, since the left side is overall 8+, and the right is 2+, you add 6 electrons to the left side. Then both sides are 2+ total, and the reaction is balanced.

Once you do this to both the separate reactions, oxidation and reduction, you must combine them again. The only complication here lies in the need to balance the electrons in each half reaction, since the electrons on each side must be equal in the final reaction. Therefore, if you have unequal amounts of electrons transferred in each reaction, you must multiply by a certain factor of the electron's coefficients. For example, in the above example, the reduction of Manganese requires 6 electrons. If the accompanying oxidation half-reaction produces 8 electrons, then you must multiply each reaction by a certain number to get the least common multiple of 24.  You would multiply the whole manganese half-reaction by 4 and the whole oxidation by 3, and then combine them. Then, you must only strike off common elements, and you are finished. For example, if you have 5 waters on the left side of the final reaction, and 3 on the right, you would scratch off the 3 on the right, and subtract 3 from the total on the left.


1) Separate the equation into half-reactions
Oxidation (loss of e-)
Reduction (gain of e-)
2) Balance the atoms
3) Balance the charge
(Make sure the atoms and charge are balanced on BOTH half reactions)
4) After balancing the equations, you add the half-reactions together, and cancel the redundant elements and electrons
5) you have a balanced redox equation (if i missed anything someone change it)