redox reaction (idea)

How to Write a Redox Equation

There are two kinds of redox reactions: acidic and basic. The former is much more common than the latter. In order to write out an acidic redox equation, follow the following steps:

1. Write the half-reactions.
2. Balance the main elements.
3. Balance the electrons.
4. Balance your oxygens by using water (H₂O).
5. Balance your hydrogen atoms with hydrogen ions (H⁺).
6. Multiply one or both half-reactions by an integer to equalize the number of electrons present.
7. Cancel your electrons (e^-) H₂O's and H⁺s.
8. Combine the half-reactions into one complete equation.

If that sounded like gibberish to you, don't worry, you're not alone. The only way to really learn how to do acidic redox is by seeing examples, so I have presented one here:

You have the reaction NO₃^- + H₂S --> NO + S
1. First, write two half-reactions by breaking up the main one according to main elements, oxygen and hydrogen are not counted:
   
   NO₃^- --> NO
   
   and
   
   H₂S --> S
2. The next step, balancing the main elements, does not apply to this example because the main elements, those not oxygen and hydrogen, are already balanced. If this were not the case, you would balance them by placing a coefficient before the compounds with unequal elements. For example, let's say LiH --> Li₂O. To balance the main elements, you would add a "2" to the lithium hydride (LiH) and come up with 2LiH --> Li₂O.
3. Balancing electrons, the next step, is the hardest part to understand. In order to do it, you must remember that hydrogen has a +1 charge and oxygen has a -2 charge. You then set up an equation on both sides of the arrow to determine what the main elements' charges are. Look at the first half-reaction:
   
   NO₃ --> NO
   
   On the left side of the equation you get N-(2 x 3)=-1 (the "-1" is determined by the charge above the compound, in this case -1), or N-6=0. In order for this statement to be true, N must have a charge of +5 on the left side of the equation.
   
   Turning to the right side of this half-reaction, you would write the equation N-2=0 (NO has no charge so the sum is "0"). N must therefore have a charge of +2 in order to balance out the -2 oxygen ion.
   
   When we look at both of the charges together we get +5 on the left and +2 on the right. In order to balance this chemical formula, you must add 3e^- to the left; you then get a charge of +2 on both sides (you can't add charge to the +2 side to equalize the charges). The final result of the top half-reaction is:
   
   NO₃^- + 3e^- --> NO
   
   Now, remember the result of the first half-reaction while we turn our attention to the second half-reaction:
   
   H₂S --> S
   
   Keeping in mind what we learned for the last one, we set up two equations: the left side comes out to be 2+S=0 and the right S=0 (S has no superscript and therefore has a charge of 0). Solving the first equation, you come up with S=-2. In order to balance this half-reaction, 2e^- must be added to the right side, bringing both sides to a -2 charge. So, the final, electron-balanced half reactions come out to be:
   
   NO₃^- + 3e^- --> NO
   
   and
   H₂S --> 2e^- + S
4. Now, it's time to turn our attention to balancing oxygens in our half-reactions. Looking at the first half-reaction:
   
   NO₃^- + 3e^- --> NO
   
   we see that the left side has 3 oxygen ions while the right only has 1. To rectify this problem we must add 2H₂O to the right side. This results in an equation looking like this:
   
   NO₃^- + 3e^- --> 2H₂O + NO
   
   When we look at the second half-reaction we discover that there are no oxygen ions present, and therefore this step doesn't apply. So, at the end of this step our half-reactions look like this:
   
   NO₃^- + 3e^- --> 2H₂O + NO
   
   and
   
   H₂S --> 2e^- + S
   
   
5. At this point, you need to balance your hydrogen ions. Turning to the first half-reaction:
   
   NO₃^- + 3e^- --> 2H⁺ + NO
   
   you can see that the left side has 0 hydrogen ions while the right side has 4 (there are 2 hydrogens in a water molecule and we have 2 water molecules represented by 2H₂0). To make up this difference you need to add 4 hydrogen ions (H⁺) to the left side, ending up with a half-reaction looking like this:
   
   NO₃^- + 3e^- + 4H⁺--> 2H₂O + NO
   
   Shifting your focus to the next half-reaction:
   
   H₂S --> 2e^- + S
   
   you will be able to tell that the left side has 2 hydrogen ions while the right side has 0. Therefore, 2H⁺ must be added to the left side. At the end of this step, you come up with the following half-reactions:
   
   NO₃^- + 3e^- + 4H⁺ --> 2H₂O + NO
   
   and
   
   H₂S --> 2e^- + 2H⁺ + S
6. Now, the time has come to begin combining our half-reactions into one again. Lining up our half-reactions together,
   
   NO₃^- + 3e^- + 4H₂O --> 2H₂O + NO
   H₂S --> 2e^- + 2H⁺ + S
   
   You can see that there are 3 electrons on the top reaction and 2 on the bottom. The only way to combine the two is to canceling the electrons, and the only way to cancel electrons is to multiply one or both half-reactions by a number that allows both electron numbers to be equal. Thinking about it, you should realize that the top one should be multiplied by 2 and the bottom by 3. This leaves you with:
   
   2NO₃^- + 6e^- + 8H₂O --> 4H₂0 + 2NO
   3H₂S --> 6e^- + 6H⁺ + 3S
7. Now, it's time to cancel all the electrons in both half-reactions. At this time you also must simplify all the hydrogen ions and waters in the reactions; subtract the smaller number of water/ions from the larger. Doing this, you end up with:
   
   2NO₃^- + 2H⁺ --> 2NO + 4H₂O
   3H₂S --> 3S
8. Finally, we come to the end of our long process. All that remains is for the two half-reactions to be combined into one full reaction, which looks something like this:
   
   2NO₃^- + 3H₂S + 2H⁺ --> 2NO + 3S + 4H₂O