| Step 1 | Assign oxidation numbers to the elements in the equation to determine the oxidized and reduced substances. |
| Step 2 | Write the half-reactions for the oxidized and reduced substances. |
| Step 3 | Add electrons to the reactant side if the substance is reduced and to the product side for the oxidized substance. |
| Step 4 | Adjust the electrons in both equations so that the electrons lost equal the electrons gained. Do this by multiplying the reduced reaction by the number of electrons in the oxidized reaction and the oxidized reaction by the number of electrons in the reduced reaction. |
| Step 5 | Algebraically add the two half-reactions. |
| Step 6 | Place the coefficients into the original equation. |
| Step 7 | Balance all other elements, if any, other than hydrogen and oxygen. |
| Step 8 | Balance the hydrogen then balance the oxygen. |
| Step 9 | Double check your work. |
| equation to be balanced | KMnO 4 + HCl --> KCl + MnCl 2 + H 2O + Cl 2 |
| assigned charges | (+1+7-8) + (+1-1) --> (+1-1) + (+2-1) + (+2-2) + (0) KMnO 4 + HCl --> KCl + MnCl2 + H 2O + Cl 2 |
| half-reactions with charges & mass balanced | Mn +7 + 5e - Mn +1 2Cl - Cl 2 + 2e - |
| adjustment of coefficients to eliminate electrons | 2(Mn +7 + 5e - Mn +2) = 2Mn +7 + 10e - + 2Mn +2 5(2Cl - Cl 2 + 2e -) = 10Cl - + 5 Cl 2 + 10e - |
| coefficients placed in original equation | 2KmnO 4 10HCl --> KCl + 2MnCl 2 + H 2O + 5Cl 2 |
| other elements, hydrogen & oxygen, balanced | 2KmnO 4 16HCl --> 2KCl + 2MnCl 2 8H 2O + 5Cl 2 |
