A redox reaction is one in which both oxidation and reduction occur. Ass the name implies, oxidation often involves the gain of oxygen, but it doesn't have to. All that needed for oxidation is the loss of electrons, which will lead to an increase in the oxidation number of some atom in the species.
And because oxidation involves the loss of electrons, reduction must occur in tandem with oxidation. Reduction involves a gain of electrons and a lowering of the oxidation number. Many students use the pneumonic device OILRIG to remember the relationship: Oxidation is Loss (of electrons), Reduction is Gain.
The procedure for balancing a redox reaction is to separate the reaction into half-reactions, one involving oxidation and one involving reduction. Once separated, the following procedure is applied:
Balance the following skeletal reaction:
The two half-reactions will be
Let's tackle the Copper half-reaction first (because it is so simple). There are no oxygen or hydrogen atoms with which to content, so all that is needed is to add electrons ot balance the charge. This is accomplished by adding two electrons to the right side of the reaction:
The selenium half-reaction can now be tackled. Since the selenium is already balanced, we can move on to the oxygen. This is balanced by adding H2O to the oxygen deficient side.
This creates an imbalance of H atoms. This is corrected by adding H+ to the hydrogen deficient side.
finally, we address the charge imbalance by adding electrons to the side with the larger total charge.
The task now is to add the reactions together in a ratio that cancels the electrons. For this purprose, we must find the least-common-multiple between 2 (the number of electrons in the Cu half-reaction) and 4 (the number of electrons in the Se half-reaction). That least-common-multiple is (of course) 4. In order to generate this many electrons in both half-reactions, we must double the copper half-reaction before adding the two together.
Adding the two together, and canceling out the electrons yields:
It should be noted that not only are all of the atoms balanced in this reaction, but the total charge is the same on either side of the arrow!