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Balancing Redox Reactions: The Half-Reaction Method

Balanced chemical equations accurately describe the quantities of reactants and products in chemical reactions. They serve as the basis of stoichiometry by showing how atoms and mass are conserved during reactions. Oxidation and reduction reactions need to be bal-anced just as all other reactions do, but they are often quite compli-cated. Balancing redox reactions involves balancing not only atoms but also positive and negative charges. Balancing either the atoms or the charges is not hard, but the process becomes difficult when both must be balanced simultaneously. Chemists have developed a way to balance redox reactions in an organized and straightforward process.

Redox reactions that are difficult to balance can be managed more easily if the processes of oxidation and reduction are con-sidered separately before they are considered together. In the half-reaction method, or ion-electron method, the redox reaction is split into two hypothetical parts called half-reactions. The oxidation half-reaction deals with all the substances that become oxidized and all the electrons they lose. The reduction half-reaction involves all the substances that become reduced and all the electrons they gain.

The half-reaction method consists of eight steps that help bal-ance reactions in an organized fashion. This method can reduce the task of balancing even complicated reactions, such as the one between nitric acid and copper (I) oxide, to a series of manageable procedures.

HNO3 + Cu2O Cu(NO3)2 + NO + H2O

Step 1. Assign oxidation numbers to all atoms, and then deter-mine which atoms are oxidized (LEO) and which ones are reduced (GER). Since the oxidation numbers of hydrogen and oxygen remain unchanged in this particular reaction, ignore them for now and focus on the other atoms. As previously discussed, ion charges are written with the charge designation after the numeral (3–), and the oxidation numbers are written with the charge designation preced-ing the numeral (+5). This will help avoid confusion when dealing with reactions containing ions.

In this reaction, the oxidation number of nitrogen changes from +5 in nitric acid to +2 in nitrogen (II) oxide, so nitrogen is reduced (GER: gains 3 electrons). The oxidation number of copper changes from +1 in copper (I) oxide to +2 in copper (II) nitrate, so copper is oxidized (LEO: loses 1 electron).

Step 2. Write a half-reaction for the oxidation process and one for the reduction process. Labeling each half-reaction with either LEO or GER will help keep them separate as they are balanced in the following steps. The half-reactions include only those substances that were either oxidized or reduced. Since the oxidation number of copper went from +1 to +2, it lost an electron; thus, it was oxidized (LEO). All the compounds with copper atoms in them should be in-cluded in the oxidation half-reaction.

LEO: Cu2O Cu(NO3)2

HNO Cu O Cu(NO + NO + H O3 2 3 2

+2+ + + ++ →

5 1 2 5

2)

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1 Oxidation-Reduction

The reduction half-reaction includes only those substances con-taining the atoms that are reduced. The oxidation number of nitro-gen went from +5 to +2 by gaining three electrons (GER). Notice that even though copper (II) nitrate contains nitrogen, it is not included in the reduction half-reaction because the nitrogen in copper (II) ni-trate is not reduced (the oxidation number remains +5).

GER: HNO3 NO

Step 3. Balance the atoms that were reduced or oxidized in the two half-reactions by inspection. A coefficient of 2 was added to the right side of the oxidation half-reaction to balance the copper atoms. In the reduction half-reaction the nitrogen atoms are balanced al-ready.

LEO: Cu2O 2Cu(NO3)2 2 Cu = 2 Cu

GER: HNO3 NO 1 N = 1 N

Step 4. Balance all other atoms by inspection. Most redox reac-tions (at least all the ones that will be considered in this text) take place in acidic solutions where there is an unlimited supply of hydro-gen ions and water molecules. When needed, these can be added to one side of the equation or the other. If additional oxygen is needed to balance the reaction, add water. If additional hydrogen is needed, add hydrogen ions. In the oxidation (LEO) half-reaction, the addi-tion of 4 to the HNO3 on the left side is needed to balance the nitro-gen atoms.

LEO: 4HNO3 + Cu2O 2Cu(NO3)2

The extra oxygen on the left side can be balanced by the addition of one water molecule to the right-hand side. The extra hydrogens can be balanced by the addition of two hydrogen ions also to the right-hand side.

LEO: 4HNO3 + Cu2O 2Cu(NO3)2 + H2O + 2H+

The balancing of the reduction (GER) half-reaction involves bal-ancing the hydrogen and the oxygen atoms and is performed using water and hydrogen ions.

GER: HNO3 + 3H+ NO + 2H2O

Step 5. Balance the charges in each half-reaction by adding elec-trons until the total charge is the same on both sides. Note: the total charge does not have to equal zero.

LEO: 4HNO3 + Cu2O 2Cu(NO3)2 + H2O + 2H+ + 2e–

In the above oxidation half-reaction, the total charge of the left side is zero, and the right side is +2 from the previously added hydro-gen ions. Therefore, two electrons must be added to the right side to make its total charge zero.

GER: HNO3 + 3H+ + 3e– NO + 2H2O

In the reduction half-reaction, the total charge of the right side is zero, and the left side is +3; therefore, three electrons must be added to the left side to equalize the charges. Now both half-reactions are balanced, and the total charges of each side are equal.

Step 6. Multiply each half-reaction by an appropriate whole number so that the number of electrons produced by the oxidation half-reaction equals the number used by the reduction half-reaction.

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2 Chapter Seventeen

In this case, multiply all coefficients in the oxidation half-reaction by three.

LEO: 3[4HNO3 + Cu2O 2Cu(NO3)2 + H2O + 2H+ + 2e–]

12HNO3 + 3Cu2O 6Cu(NO3)2 + 3H2O + 6H+ + 6e–

To make both half-reactions show the same number of electrons, multiply all the coefficients in the reduction half-reaction by two.

GER: 2(HNO3 + 3H+ + 3e– NO + 2H2O)

2HNO3 + 6H+ + 6e– 2NO + 4H2O

Step 7. Add the oxidation and reduction half-reactions together. Add the reactants from both half-reactions together and put the re-sult on the left-hand side. The products from both half-reactions are then totaled and placed on the right-hand side.

12HNO3 + 3Cu2O 6Cu(NO3)2 + 3H2O + 6H+ + 6e–

2HNO3 + 6H+ + 6e– 2NO + 4H2O+

Step 8. Cancel any quantities that appear on both sides of the overall reaction.

Since there are six hydrogen ions and six electrons on both sides, they cancel out.

14HNO3 + 3Cu2O 6Cu(NO3)2 + 2NO + 7H2O

The reaction is balanced. The result is the same as the one that would have been obtained by the balance-by-inspection method, but the extra steps help to keep you on the right track throughout the en-tire balancing process. Balancing redox reactions may seem difficult at first, but do not despair; practice makes perfect.

ExamplE problEm 17-4 Sulfite ions and permanganate ions can react to form sul-

fate ions and manganese ions. Balance this reaction, using the half-reaction method.

SO32– + MnO4

– SO42– + Mn2+

Solution Step 1. Assign oxidation numbers and determine which

atoms are reduced and which are oxidized.

The sulfur lost 2 electrons (LEO), so it was oxidized; the manganese gained 5 electrons (GER), so it was reduced.

Step 2. Identify half-reactions.

14HNO3 + 3Cu2O + 6H+ + 6e– 6Cu(NO3)2 + 2NO + 7H2O + 6H+ + 6e–

14HNO3 + 3Cu2O + 6H+ + 6e– 6Cu(NO3)2 + 2NO + 7H2O + 6H+ + 6e–

SO + MnO SO + Mn+4 +7 +6 +2

32

4 42 2+− − −→

LEO: SO SO

GER: MnO Mn

+4 +6

+7 +2

32

42

42+

− −

−

→

→

Steps to Balancing Redox Reactions by the Half-Reaction Method

Step 1: Assign oxidation numbers.•Step 2: Write half-reactions.•Step 3: Balance reduced/oxidized atoms •

in each half-reaction.

Step 4: Balance all other atoms.•Step 5: Balance charges in each half-•

reaction.

Step 6: Equalize electrons in each half-•reaction.

Step 7: Add half-reactions together.•Step 8: Cancel quantities.•

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3 Oxidation-Reduction

Step 3. Balance the reduced/oxidized atoms in each of the half-reactions. In these half-reactions, the atoms are al-ready balanced.

Step 4. Balance the remaining atoms in the half-reactions, using water and H+ where needed.

LEO: SO32– + H2O SO4

2– + 2H+

GER: MnO4– + 8H+ Mn2+ + 4H2O

Step 5. Balance the charges in each half-reaction. There is a –2 charge on the left side of the oxidation half-reaction and zero on the right side (the +2 of the hydrogen ions will cancel out the –2 on the sulfate ion). Remember that the charges do not have to equal zero, but they do need to be equal. There-fore, 2 electrons can be added to the right side to equalize the charges.

LEO: SO32– + H2O SO4

2– + 2H+ + 2e–

The right side of the reduction half-reaction has a +2 charge; the left side has a +7 charge—the sum of a –1 from the permanganate ion and +8 from the 8 hydrogen ions. Therefore, to equalize the charge of +2 on the right side, 5 electrons must be added to the left side.

GER: MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Step 6. Multiply all of the coefficients in both half-reactions so that the electrons will cancel out. In this case the oxida-tion half-reaction should be multiplied by 5, and the re-duction half-reaction by 2.

LEO: 5SO32– + 5H2O 5SO4

2– + 10H+ + 10e–

GER: 2MnO4– + 16H+ + 10e– 2Mn2+ + 8H2O

Step 7. Add the half-reactions.

5SO32– + 5H2O 5SO4

2– + 10H++ 10e–

2MnO4– + 16H+ + 10e– 2Mn2+ + 8H2O

+

Step 8. Cancel. All electrons, 10 hydrogen ions, and 5 water molecules will cancel from both sides, resulting in the final reaction.

5SO32– + 2MnO4

– + 6H+ 5SO42– + 2Mn2+ + 3H2O

5SO32– + 2MnO4

– + 5H2O + 16H+ + 10e– 5SO42– + 2Mn2+ + 8H2O +10H+ + 10e–

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