Balancing Redox Equations: A Step-by-Step Guide

Step 1: Assign Oxidation Numbers This follows the rules we learned in the last power point.

Step 2: Separating the Reaction into 2 Half-Reactions Let’s look at a specific reaction: Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq) You can clearly see the two half-reaction here: Sn2+  Sn4+ and 2Fe3+  2Fe2+

Try this: Cr2O72- + Cl- Cr3+ + Cl2

Answer Cr2O72-  Cr3+ Cl-  Cl2

Step 3: Add the appropriate number of electrons to the equation Ignore anything in the half-reaction that is not being oxidized or reduced. Think only about balancing the oxidation numbers - ignore any charges.

For Example… Cr2O72- --> Cr3+ Cr on the left has an oxidation number of 6 (each) The Cr on the right has an oxidation number of 3 The Cr is being (Oxidized? Reduced?).

Cr2O72-  Cr3+ In order to think about this properly, we need to balance the Cr’s, just as we would an ordinary chemical equation (again, ignore those O’s.) Cr2O72-  2Cr3+

Cr2O72- --> 2Cr3+ 12+ 6+ How many electrons do we need to add? We need 6+ to equal 3+ by adding negatives. But we need to do this twice, since there are 2 Cr’s on each side So we must add 6 electrons to the left. Cr2O72- + 6e-  2Cr3+

Step 4: Balance the Charges Cr2O72- + 6e-  2Cr3+ 12+ 6+ What is the charge on the left side of the equation? What is the charge on the right side of the equation? Remember: Charge is DIFFERENT than Oxidation Number

In Acids -2-6+14 = +6 Cr2O72- + 6e-  2Cr3+ 12+ 6+ add H+ ions to the more negative side to balance the charges Cr2O72- + 6e- + 14H+  2Cr3+ 12+ 6+ -2-6+14 = +6

Cr2O72- + 6e- --> 2Cr3+ + 14OH- In Bases Cr2O72- + 6e- 2Cr3+ 12+ 6+ add OH- ions to the more positive side to balance the charges Cr2O72- + 6e- --> 2Cr3+ + 14OH- 12+ 6+ -2-6= +6 -14

Now We Add H20 to Balance H’s and O’s In Acid Cr2O72- + 6e- + 14H+  2Cr3+ + 7H2O In Base 7H2O + Cr2O72- + 6e-  2Cr3+ + 14OH-

Now Check Are the charges the same on both sides? Are there the same numbers of atoms on both sides? If so, you’re done with that half Now do the same to the other half

The Other Half Cl-  Cl2 2Cl-  Cl2 2Cl-  Cl2 + 2e- Done!

Now Put the 2 Halves Back Together The trick here is: the number of electrons on the left has to equal the number of electrons on the right, so they can be canceled.

In this case, multiply the bottom equation by 3: 6Cl-  3Cl2 + 6e- Cr2O72- + 6e- + 14H+ 2Cr3+ +7H2O 2Cl-  Cl2 + 2e- In this case, multiply the bottom equation by 3: 6Cl-  3Cl2 + 6e-

Now, you can add the equations Cr2O72- + 6e- + 14H+ 2Cr3+ + 7H2O 6Cl-  3Cl2 + 6e- _________________________ Cr2O72- + 6Cl- + 14H+ 2Cr3+ + 3Cl2 + 7H2O

H2O + Cr(OH)3  CrO42- + 3e- + 5H+ Canceling You can cancel anything that’s the same on both sides, such as H+’s or waters. For example, add these halves: MnO4- + 8H+ + 5e-  Mn2+ + 4H2O And H2O + Cr(OH)3  CrO42- + 3e- + 5H+

First multiply the top equation by 3 3MnO4- + 24H+ + 15e-  3Mn2+ + 12H2O and the bottom equation by 5 5H2O + 5Cr(OH)3  5CrO42- + 15e- + 25H+ to even out the electrons.

Answer: Then add the equations up, canceling as you would in algebra: 3MnO4- + 24H+ + 15e-  3Mn2+ + 12H2O 5H2O + 5Cr(OH)3  5CrO42- + 15e- + 25H+ 3MnO4- + 5Cr(OH)3  3Mn2+ + 5CrO42- + 7H2O + H+

Summary Assign Oxidation numbers, determining what is being oxidized and reduced Split the equation into the 2 half-reactions that need to be balanced Balance each half reaction one at a time Balance the oxidation numbers by adding electrons Next balance the charge using H+ for acids solutions and OH- for basic solutions

Summary continued Add water to balance the Hydrogen’s and oxygen’s Put both halves together, making sure that you have the same number of electrons on each side so that they cancel each other out