Balancing Redox Equations

A little background material courtesy of the College Board. Traditionally, this is the first topic in AP Chemistry that is totally new. However, the College Board has reworked their syllabus and has deemphasized balancing redox equations. In its place, there is more emphasis on the theory of an oxidation-reduction reaction. We are still going to take a look at them because it clarifies some of the more difficult aspects of redox.

Oxidation – Reduction Reactions Remember that although we talk about oxidation and reduction as if they were two completely separate processes, they always must be happening in unison. This is very evident when balancing redox equations. What this means, is that we are going to split each equation in half; a reduction half and an oxidation half. They will be balanced individually and then recombined at the end.

Before we work the problem Also, it makes quite a bit of difference if the reaction is taking place in either acidic or basic medium. How do you know which one, you ask ? Well, sometimes they will actually just come out and tell you. Other times, there is a known acidic or basic species in the problem. In other words, if the reaction is swimming in Sulfuric Acid (H2SO4) then we can be confident that it is in acidic medium.

Example Problem # 1 (acidic) Let’s walk through a redox problem. Each step will be explained as we go. Example: MnO4-1 + Fe+2  Fe+3 + Mn+2 (acidic) First we need to determine the oxidation states of each atom in the problem. Reactants: Mn  + 7 O  - 2 Fe  + 2 Products: Fe  + 3 Mn  + 2 We split this into a reduction and an oxidation, which is usually termed the reduction half-reaction and the oxidation half-reaction. We will do the reduction half-reaction first. we see that the Mn is being reduced.

Example Problem # 1 First Step So our half-reaction is going to look like. Reduction: MnO4-1  Mn+2 Mn goes from +7  +2 FIRST STEP: Since the atoms of Mn are balanced, we then balance the Oxygen by adding water. That means 4 water molecules on the product side. Our equation becomes MnO4-1  Mn+2 + 4 H2O Notice the water adds H+1 ions that are not present on the other side. No problem – we will take care of that in the next step.

Example Problem # 1 Second Step Here is what we have so far. MnO4-1  Mn+2 + 4 H2O SECOND STEP: Add Hydrogen Ions to the other side to balance the Hydrogen. We need to add 8 H+1 to the reactant side to balance the 8 Hydrogen atoms in water. 8 H+1 + MnO4-1  Mn+2 + 4 H2O

Example Problem # 1 Third Step Here is what we have so far. 8 H+1 + MnO4-1  Mn+2 + 4 H2O THIRD STEP: We need to balance according to charge. Total up the charges on each side and then add electrons so the charges balance. Remember that redox is about changing charges on atoms. The reactant side is a +7 and the product side is +2. So we need to add 5 e- to the reactant side. 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O

Example Problem # 1 Fourth Step Here is what we have so far 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O FOURTH STEP: Do a quick check to make sure that all atoms and charges are balanced. If they are, we go to the other half-reaction. This problem looks good to go, so on to the oxidation half-reaction.

Example Problem # 1 Oxidation Oxidation: Fe+2  Fe+3 Fe goes from +2 to +3 It is not unusual at all not to have to add any species nor to find Oxygen or Hydrogen on a half-reaction. We just happily go on our merry way and balance the charge, which in this case is by adding one electron to the product side to give us. Fe+2  Fe+3 + e-1

Example Problem # 1 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O Here are our two half-reactions 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O Fe+2  Fe+3 + e-1 NOTE: Notice that the electrons are always going to be found on opposite sides of the arrows. This makes sense if you think about it. Oxidation is loss of electrons, so we would expect to find electrons on the product side. Reduction is a gain of electrons, so they are picking up the electrons lost by the oxidized species.

Example Problem # 1 We now have our two half-reactions and we need to put them together. First we need to analyze them, but putting them next to each other. REDUCTION: 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O OXIDATION: Fe+2  Fe+3 + e-1 Let’s think about this. Does it seem right that we are only losing one electron but gaining 5 in this reaction ? Of course, not. We need to balance the charge. We do this by simply finding the lowest common multiple and make them equal. In this case the lowest common multiple is 5.

Example Problem # 1 The Reduction half-reaction gets multiplied by 1 The Oxidation half-reaction gets multiplied by 5. Here is what we have now. REDUCTION: 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O OXIDATION: 5 Fe+2  5 Fe+3 + 5 e-1 NOTE: Always check to see if the number of electrons being oxidized is equaling the number of electrons being reduced. THEY MUST EQUAL !!

Example Problem # 1 Final Step FINAL STEP: Combine the two half-reactions and cancel out anything that is found on both sides. The electrons will always be canceled out and often water will be as well. We now get the following. 8 H+1 + MnO4-1 + 5 Fe+2  Mn+2 + 5 Fe+3 + 4 H2O One last check to verify everything is balanced and we go out and celebrate our victory over redox.

Summary of Redox Steps for Acidic Medium 1. Determine the oxidation states of each atom. 2. Write out the oxidized and reduced species and set up their respective half-reactions 3. Balance the atoms (other than O and H). 4. Balance the Oxygen by adding water as needed. 5. Balance the Hydrogen by adding H+1 as needed. 6. Add the number of electrons needed to balance the equation according to charge. 7. Combine the two half reactions

Example Problem # 2 (basic) Know we will look at a problem when redox takes place in a basic solution. Here is our equation: NO2-1 (aq) + Al (s)  NH3 (g) + AlO2-1 (aq) Just like before, first we need to determine the oxidation states of each atom in the problem. Reactants: N  + 3 O  - 2 Al  0 Products: N  - 3 H  + 1 Al  + 3 O  - 2

Example Problem # 2 Once again, we split this into a reduction and an oxidation. Before we go any further, we need to talk about what are the different steps when balancing a reaction in basic solution as opposed to when it is in acidic solution. In an acidic medium, we work the problem, by adding water and hydrogen ions to make it balanced. If you are thinking that for basic solution, we will be adding water and hydroxide ions, you would be CORRECT. HOWEVER -- This is not as easy as it sounds. So a slightly different path is used to get this done. What is this you ask ?!! This might sound strange BUT…

Example Problem # 2 … when balancing a redox reaction in basic medium, we start by ‘pretending it is an acidic solution’. Of course, we will need to compensate for this later in the problem, but this method is much simpler. From our reaction we can see that the Nitrogen is the species being reduced. Reduction: NO2-1  NH3 N goes from + 3  - 3

Example Problem # 2 FIRST STEP: The Nitrogen atoms are balanced so we start off by adding water to balance the oxygen atoms. Just like we did when we were balancing in acidic medium. NO2-1  NH3 + 2 H2O Remember that when we balance any type of equation, we only balance one atom at a time and if while doing that, we UNBALANCE something else, that is fine. We will correct that later.

Example Problem # 2 7 H+1 + NO2-1  NH3 + 2 H2O NO2-1  NH3 + 2 H2O Here is what we have so far. NO2-1  NH3 + 2 H2O SECOND STEP: Add Hydrogen Ions to the other side to balance the Hydrogen. We need to add 7 H+1 to the reactant side to balance the 7 Hydrogen atoms on the reactant side. Make sure you take the Hydrogen atom in the NO2-1 species. 7 H+1 + NO2-1  NH3 + 2 H2O

Example Problem # 2 7 H+1 + NO2-1  NH3 + 2 H2O Look at where we are now: 7 H+1 + NO2-1  NH3 + 2 H2O There are Hydrogen ions present, which means an acidic solution. But we were told that the reaction is taking place in basic solution. Remember that we said that we will have to account for the addition of the Hydrogen Ions at a later time. WELL – THAT TIME IS NOW !!

Example Problem # 2 7 H+1 + NO2-1  NH3 + 2 H2O What we do is to add Hydroxide ions to any Hydrogen ions to form water. Of course, if we add OH-1 to the one side, then we have to add a similar number of OH-1 ions to the other side. Here is what we now have. 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1

Example Problem # 2 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1 This should look a bit better. Now we have a solution with hydroxide ions, which makes the statement of this reaction being in a basic solution a bit easier to see. We are not quite done yet. We still have to add the electrons to balance the charge. Also, remember that we have only done the reduction half-reaction. We still have to do the oxidation half-reaction.

Example Problem # 2 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1 The total of the charges on the reactant side is -1 while on the product side it is -7. 6 electrons need to be added to the reactant side. 6 e-1 + 7 H2O + HNO2-1  NH3 + 2 H2O + 7 OH-1 Now we are done with the reduction half-reaction.

Example Problem # 2 Oxidation: Al 0  + 3 Al  AlO2-1 We need to repeat the same steps over again, that we did for the reduction half-reaction. First we need to determine the oxidation states of each atom in the problem. Reactants: Al  0 Products: Al  + 3 O  - 2 We see that the Al is being reduced

Example Problem # 2 2 H2O + Al  AlO2-1 2 H2O + Al  AlO2-1 + 4 H+1 Here is our reaction: Al  AlO2-1 Since the Al is balanced, we balance the Oxygen with water to give us. 2 H2O + Al  AlO2-1 We add Hydrogen atoms to balance out those that we added when we inserted H2O 2 H2O + Al  AlO2-1 + 4 H+1

Example Problem # 2 2 H2O + Al  AlO2-1 + 4 H+1 Now we need to do our step to compensate for the basic medium. Let’s add the OH-1 on both sides as needed. The H+1 becomes H2O 4 OH-1 + 2 H2O + Al  AlO2-1 + 4 H2O And finally, balance according to charge. 4 OH-1 + 2 H2O + Al  AlO2-1 + 4 H2O + 3 e-1

Example Problem # 2 6 e-1 + 7 H2O + HNO2-1  NH3 + 2 H2O + 7 OH-1 Now we need to bring the two half-reactions together. 6 e-1 + 7 H2O + HNO2-1  NH3 + 2 H2O + 7 OH-1 4 OH-1 + 2 H2O + Al  AlO2-1 + 4 H2O + 3 e-1 We need to make the electrons being oxidized and the electrons being reduced equal to each other. So we multiply the reduction half-reaction by 2. The next slide has the updated equations

Example Problem # 2 6 e-1 + 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1 8 OH-1 + 4 H2O + 2 Al  2 AlO2-1 + 8 H2O + 6 e-1 Now we can start cancelling out species. Electrons First 6 e-1 from each side 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1 8 OH-1 + 4 H2O + 2 Al  2 AlO2-1 + 8 H2O

Example Problem # 2 7 H2O + NO2-1  NH3 + 2 H2O + 7 OH-1 8 OH-1 + 4 H2O + 2 Al  2 AlO2-1 + 8 H2O We can combine the reactions together now before we reduce anymore 8 OH-1 + 11 H2O + NO2-1 + 2 Al  NH3 + 10 H2O + 7 OH-1 + 2 AlO2-1 There are 11 water molecules on the reactant side and 10 on the right. As you can see, we can combine from the two equations when the same species is on the same side. This leaves 2 water molecules on the reactant side

Example Problem # 2 OH-1 + H2O + NO2-1 + 2 Al  NH3 + 2 AlO2-1 8 OH-1 + 11 H2O + NO2-1 + 2 Al  NH3 + 10 H2O + 7 OH-1 + 2 AlO2-1 We factor out the H2O and the OH-1 ion. This equation now becomes OH-1 + H2O + NO2-1 + 2 Al  NH3 + 2 AlO2-1 We check out everything to verify it is balanced according to all atoms as well as charge.

Example Problem # 2 OPTIONS You can combine the half-reactions before you start to cancel out on each side. Many people find this easier to do.

Summary of Redox Steps for Basic Medium 1. Determine the oxidation states of each atom. 2. Write out the oxidized and reduced species and set up their respective half-reactions 3. Balance the atoms (other than O and H). 4. Balance the Oxygen by adding water as needed. 5. Balance the Hydrogen by adding H+1 as needed. 6. Add Hydroxide Ion to each side to make the H+1 ion become water and add free OH-1 ions to the other side. 7. Add the number of electrons needed to balance the equation according to charge. 8. Combine the two half reactions