Steps for Balancing Redox Reactions

in Acidic Conditions

1. Assign oxidation numbers and find out what is being oxidized and what is being reduced.
2. Draw arrows connecting "like" elements on the left and right side of the chemical equation. Note that those elements MUST have different oxidation numbers on the left and right sides of the equation.
3. Write the number of electrons over your arrows that shows the differences in the oxidation states. You should also note here which 1/2-reaction is the oxidation and which one is the reduction.
4. Now adjust those electrons accordingly (and formulas) IF one of the formulas has more than one element in it.
   • Example: Cr₂O₇^2- is on the left side and Cr³⁺ is on the right side. That's an oxidation state for chromium of +6 in the dichromate and +3 for the chromium(III) ion. That is a difference of 3 electrons - you write +3e^- over the arrow. WAIT!!! There are 2 Cr's in the dichromate formula and you MUST account for them. So put a coefficient of 2 in front of the chromium(III) ion and CHANGE your total of electrons to +6e^- over the arrow. Dichromate reacts TWO chromiums at a time (not one) and therefore 6 electrons at a time.
5. Now multiply each of your 1/2-reactions by the appropriate scaler so as to make the total number of electrons transferred MATCH in each. Remember to ADJUST the amounts on BOTH sides of the equation. NOTE: If your doing a titration calculation, you can now stop and work the problem. Why?/How? Well because you now know the stoichiometric ratio of the two reactants. However, if the point of the problem is to FINISH balancing the equation then read on...
6. To balance out the oxygens, add however many H₂Os to the side that is short
7. Now balance the hydrogens with H⁺s to the side that is short.
8. Your done! Check out the charge balance now to make sure.

in Basic Conditions

• Balance the equation just like you do for acids (no OH^- will be in the equation at this point, only H₂O and H⁺ ).
• Now ADD as many OH^-s as there are H⁺s to BOTH sides of the equation.
• The OH^-s and the H⁺s will combine to give H₂O (as many as there were H⁺s and OH^-s). So cross out the OH^-s and the H⁺s and write in place of them that many H₂Os.
• Now cancel out any H₂Os from the two sides of the equation. H₂O should only be listed on one side or the other, not both.
• Your done! Check out the charge balance now to make sure.