Balancing Redox Reactions

The following procedure is guaranteed to produce an accurately balanced reaction equation for any electron-transfer (redox) equation that occurs in aqueous solution. In fact, it will even work for reaction equations in which no redox occurs, but then it's pretty tedious compared to balancing by inspection. The only thing you have to know to begin is the chemical identity of the major reactants and products of the reaction. Let's take the following example: a popular WWII German rocket propellant was nitric acid (HNO3) and hydrazine (N2H4).

1. Write a "skeleton" reaction equation.
   HNO₃ + N₂H₄ -----> N₂ + H₂O
2. Pick out an unusual element (non-hydrogen, non-oxygen unless that's all you have). Write a skeleton half-reaction involving only species of that element.
   HNO₃-----> N₂
3. Insert coefficients to balance the number of atoms of that element on each side of the half- reaction.
   2 HNO₃ -----> N₂
4. Insert species from the original reaction to take care of other non-H, non-O elements (if any). Balance the half-reaction with respect to the non-H, non-O element (if any).
   (Not applicable to this problem)
5. Insert enough H2O molecules on the side that's short of O atoms to balance the O atoms.
   2 HNO₃ -----> N₂ + 6 H₂O
6. Insert enough H+ ions on the side that's short of H atoms to balance the H atoms.
   2 HNO₃ + 10 H⁺ -----> N₂ + 6 H₂O
7. Total up the electrical charge on each side of the half-reaction equation.
   (+10 on left, 0 on right)
8. Add enough electrons to the side that's too positive (or not negative enough) to make the net charge balance.
   2 HNO₃ + 10 H⁺ + 10 e^- -----> N₂ + 6 H₂O
9. Check to make sure that atoms of each element are balanced and that charge is balanced in the half-reaction.
   2 N, 12 H, 6 O, 0 charge
10. Pick another unusual element in the original skeleton equation and write a skeleton half- reaction for it. Go through steps 3-9 again; that is, balance the unusual elements, then balance O using H₂O, then balance H using H+, then balance charge using e-.
    N₂H₄ -----> N₂
    N₂H₄ -----> N₂ + 4 H⁺
    N₂H₄ -----> N₂ + 4 H⁺ + 4 e^-
    2 N, 4 H, 0 charge
11. Now compare the two balanced half-reactions you've gotten. In the real beaker, one has to produce exactly as many electrons as the other uses up. So here you'll need to multiply one or both half-reactions by integers such that the final electron count is the same in both half-reactions.
    (multiply first reaction by 2 and second reaction by 5 to give 20 electrons)
12. Add the two half-reactions together, canceling the electrons and as many H+ and H2O as possible.
    4 HNO₃ + 20 H⁺ + 20 e^- -----> 2 N₂ + 12 H₂O
    5 N₂H₄ -----> 5 N₂ + 20 H⁺ + 20 e^-
    Total:
    4 HNO₃(l) + 5 N₂H₄(l) -----> 7 N₂(g) + 12 H₂O(g)
    The reactant which gains electrons is the oxidant (HNO3) and the reactant which loses electrons (N2H4) is the reductant.
13. If at this point all coefficients are multiples of 2, 3, 5, etc., divide the whole equation through appropriately to get the smallest possible set of integer coefficients.

SHORT SUMMARY: Separate reaction into unusual-element half-reactions, balance each half-reaction with respect to: (1) unusual element, (2) O using H2O, (3) H using H+, (4) charge using e-. Make e- donated equal e- accepted, and combine with appropriate cancellation.
In balancing these reactions you will always be given the reactants and products. The balanced equation cannot contain anything that was not in the skeleton reactionexcept that water, H⁺, and OH^- may be introduced as needed.