Science, Tech, Math › Science How to Balance Redox Reactions Keeping atoms and charges in balance Share Flipboard Email Print This is a diagram that describes the half-reactions of a redox reaction or oxidation-reduction reaction. Cameron Garnham, Creative Commons License Science Chemistry Chemical Laws Basics Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated September 22, 2019 To balance redox reactions, you must assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge. The Half-Reaction Method First, separate the equation into two half-reactions: the oxidation portion, and the reduction portion. This is called the half-reaction method of balancing redox reactions, or the ion-electron method. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation. For this example, let's consider a redox reaction between KMnO4and HI in an acidic solution: MnO4- + I- → I2 + Mn2+ Separate the Reactions Separate the two half-reactions: I- → I2 MnO4- → Mn2+ Balance the Atoms To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add H. Balance the iodine atoms: 2 I- → I2 The Mn in the permanganate reaction is already balanced, so let's balance the oxygen: MnO4- → Mn2+ + 4 H2O Add H+ to balance the water molecules: MnO4- + 8 H+ → Mn2+ + 4 H2O The two half-reactions are now balanced for atoms: MnO4- + 8 H+ → Mn2+ + 4 H2O Balance the Charge Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions: 2 I- → I2 + 2e- 5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O Next, multiply the oxidation numbers so that the two half-reactions have the same number of electrons and can cancel each other out: 5(2I- → I2 +2e-) 2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O) Add the Half-Reactions Now add the two half-reactions: 10 I- → 5 I2 + 10 e- 16 H+ + 2 MnO4- + 10 e- → 2 Mn2+ + 8 H2O This yields the following equation: 10 I- + 10 e- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 10 e- + 8 H2O Simplify the overall equation by canceling out the electrons and H2O, H+, and OH- that may appear on both sides of the equation: 10 I- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 8 H2O Check Your Work Check your numbers to make certain that the mass and charge are balanced. In this example, the atoms are now stoichiometrically balanced with a +4 net charge on each side of the reaction. In summary: Step 1: Break reaction into half-reactions by ions.Step 2: Balance the half-reactions stoichiometrically by adding water, hydrogen ions (H+) and hydroxyl ions (OH-) to the half-reactions.Step 3: Balance the half-reactions charges by adding electrons to the half-reactions.Step 4: Multiply each half-reaction by a constant so both reactions have the same number of electrons.Step 5: Add the two half-reactions together. The electrons should cancel out, leaving a balanced complete redox reaction.