1. Oxidation and Reduction
Introductory Chemistry: Concepts & Connections 4th Edition by Charles H. Corwin
Chapter 17
Oxidation and Reduction
Christopher G. Hamaker, Illinois State University, Normal IL
© 2005, Prentice Hall

2. Oxidation-Reduction Reactions
Oxidation-reduction reactions are reactions involving the transfer of electrons from one substance to another.
We have seen several “oxidation-reduction” reactions so far.
Whenever a metal and a nonmetal react, electrons are transferred.
2 Na(s) + Cl2(g) → 2 NaCl(s)
Combustion reactions also are examples of “oxidation-reduction” reactions.
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3. Example of Oxidation/Reduction
The rusting of iron is also an example of an oxidation-reduction reaction.
Iron metal reacts with oxygen in air to produce the ionic compound iron(III) oxide which is composed of Fe3+ and O2- ions.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
Iron loses electrons and is oxidized
Fe → Fe e-
Oxygen gains electrons and is reduced
O2 + 4 e- → 2 O2-
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4. Oxidation Numbers
The oxidation number describes how many electrons have been lost or gained by an atom.
Oxidation numbers are assigned according to seven rules:
A metal or a nonmetal in the free state has an oxidation number of 0.
A monoatomic ion has an oxidation number equal to its ionic charge.
A hydrogen atom is usually assigned an oxidation number of +1.
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5. Rules for Oxidation Numbers
An oxygen atom is usually assigned an oxidation number of -2.
For a molecular compound, the more electronegative element is assigned a negative oxidation number equal to its charge as an anion.
For an ionic compound, the sum of the oxidation numbers for each of the atoms in the compound is equal to 0.
For a polyatomic ion, the sum of the oxidation numbers for each of the atoms in the compound is equal to the ionic charge on the polyatomic ion.
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6. Assigning Oxidation Numbers
What is the oxidation number for magnesium metal, Mg?
Mg = 0 according to rule #1
What is the oxidation number for sulfur in the sulfide ion, S2-?
S = –2 (rule #2)
What is the oxidation number for barium and chloride in BaCl2?
Ba is present at Ba2+, so Ba = +2 (rules #2 and #6)
Cl is present as Cl-, so Cl = –1 (rules #2 and #6)
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7. Oxidation Numbers in Compounds
What are the oxidation numbers for each element in oxalic acid, H2C2O4?
H = +1 (rule #3)
O = -2 (rule #4)
To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals zero.
2(+1) + 2(ox no C) + 4(–2) = 0
2 + 2(ox no C) + (– 8) = 0
2(ox no C) = +6
C = +3
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8. Oxidation Numbers in Compounds
What are the oxidation numbers for each element in carbon tetrachloride, CCl4?
Cl = –1 (rule #5)
To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals zero.
(ox no C) + 4(–1) = 0
(ox no C) – 4 = 0 (ox no C) = +4
C = +4
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9. Oxidation Numbers in Polyatomic Ions
What are the oxidation numbers for chlorine and oxygen in the perchlorate ion, ClO4-?
O = –2 (rule #4)
To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals the charge on the ion (rule #7).
(ox no Cl) + 4(–2) = –1
(ox no Cl) – 8 = –1 (ox no Cl) = +7
Cl = +7
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10. Redox Reactions
Recall, a chemical reaction that involves the transfer of electrons is an oxidation-reduction reaction, or a redox reaction.
For example, iron metal is heated with sulfur to produce, iron(II) sulfide: Fe(s) + S(s) → FeS(s).
The sulfur changes from 0 to –2 and the iron changes from 0 to +2. ∆
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11. Oxidation and Reduction
The iron loses electrons and is oxidized.
Fe → Fe e-
The sulfur gains electrons and is reduced.
S + 2 e- → S2-
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12. Oxidizing & Reducing Agents
Oxidation is the loss of electrons and reduction is the gain of electrons.
An oxidizing agent is a substance that causes oxidation by accepting electrons. The oxidizing agent is reduced.
A reducing agent is a substance that causes reduction by donating electrons. The reducing agent is oxidized.
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13. Redox Reactions
In a redox reaction, one substance must be oxidized and one substance must be reduced.
The total number of electrons lost is equal to the total electrons gained.
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14. Redox Reactions
Identify the reducing agent, the oxidizing agent, and the oxidation and reduction in the following reaction:
CuS(s) + H2(g) → Cu(s) + H2S(g)
Cu is reduced from +2 to 0.
H is oxidized from 0 to +1. ∆
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15. 5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
Ionic Equations
Redox reactions in aqueous solution are most often shown in the ionic form.
Ionic equations readily show us the change in oxidation number.
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
We can easily tell that the oxidation number of iron changes from +2 to +3; iron is oxidized.
Manganese is reduced from +7 in MnO4- to +2 in Mn2+; manganese is reduced.
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16. Ionic Equations Continued
We can map the reaction to show the oxidation and reduction processes and to determine the oxidizing and reducing agents:
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17. Balancing Redox Reactions
When we balance redox reactions, the number of electrons lost must equal the number of electrons gained.
We will balance redox reactions using the oxidation number method which has 3 steps:
Inspect the reaction and the substances undergoing a change in oxidation number.
Write the oxidation number above each element.
Diagram the number of electrons lost by the oxidized substance and gained by the reduced substance.
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18. Oxidation Number Method
Balance each element in the equation using a coefficient. Remember, that the electrons lost must equal the electrons gained. If they are not the same, balance the electrons as follows:
In front of the oxidized substance, place a coefficient equal to the number of electrons gained by the reduced substance.
In front of the reduced substance, place a coefficient equal to the number of electrons lost by the oxidized substance.
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19. Oxidation Number Method
After balancing the equation, verify that the coefficients are correct.
Place a check mark above the symbol for each element to verify that the number of atoms is the same on both sides.
For ionic equations, verify that the total charge on the left side of the equation is the same as the total charge on the right side of the equation.
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20. Balancing a Redox Reaction
Balance the following redox reaction using the oxidation number method:
Fe2O3(l) + CO(g) → Fe(l) + CO2(g)
Since the total electrons gained and lost must be equal, we must find the lowest common multiple. For this reaction, it is 6.
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21. Balancing a Redox Reaction
Each iron gains 3 electrons, so place a 2 in front of the Fe. There are 2 iron atoms in Fe2O3, so no coefficient is necessary.
Each carbon loses 2 electrons, so place a 3 in front of CO and CO2.
Fe2O3(l) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)
Check to see that the number of each type of atom is the same on both sides:
There are 2 Fe atoms, 6 O atoms, and 3 C atoms on each side. √
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22. Balancing Redox Equations
An alternative method for balancing redox reactions is the half-reaction method.
A half-reaction shows the oxidation or reduction process of a redox reaction separately.
The steps are:
Write the half-reaction for both the oxidation and reduction processes.
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23. Half-Reaction Method Continued
Balance the atoms in each half-reaction using coefficients.
Balance all elements except oxygen and hydrogen.
Balance oxygen using H2O.
Balance hydrogen using H+.
For reactions in basic solution, add one OH- to each side for each H+ and combine H+ & OH- to H2O.
Balance the ionic charges using electrons.
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24. Half-Reaction Method Continued
Multiply each half-reaction by a whole number so that the total number of electrons in each is the same.
Add the two half-reactions together and cancel the identical species, including electrons.
After balancing, verify that the coefficients are correct by making sure there are the same number of each atom on each side of the reaction and that the overall charge is the same on both sides.
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25. Balancing a Redox Equation
Balance the following redox reaction using the half-reaction method:
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
The two unbalanced half-reactions are:
Fe2+ → Fe3+
MnO4- → Mn2+
We balance the two half-reactions as follows:
Fe2+ → Fe3+ + e-
5 e- + 8 H+ + MnO4- → Mn H2O
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26. Balancing a Redox Equation
Since Fe2+ loses 1 electron and MnO4- gains 5 electrons, we have to multiply the iron half-reaction by 5:
5 Fe2+ → 5 Fe e-
5 e- + 8 H+ + MnO4- → Mn H2O
We add the two half-reactions together and cancel out the 5 electrons on each side to get the balanced equation:
5 Fe H+ + MnO4- → 5 Fe2+ + Mn H2O
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27. Spontaneous Redox Reactions
Chemical reactions that occur without any input of energy are spontaneous.
The reaction of zinc metal with aqueous copper sulfate is spontaneous:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cu2+ has a greater tendency to gain electrons than Zn2+.
We can compare metals and arrange them in a series based on their ability to gain electrons.
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28. Reduction Potentials
The tendency for a substance to gain electrons is its reduction potential.
The strongest reducing agent is the most easily oxidized.
This is a table of reduction potentials for several metals.
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29. Spontaneous Reactions
A species can be reduced by any reducing agent lower in the table.
Any metal below H2 can react with acid and be oxidized.
A species can be oxidized by any oxidizing agent above it on the table.
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30. Predicting Spontaneous Reactions
A reaction will be spontaneous when the stronger oxidizing and reducing agents are the reactants and the weaker oxidizing and reducing agents are the products.
Predict whether the following reaction will be spontaneous:
Ni2+(aq) + Sn(s) → Ni(s) + Sn2+(aq)
stronger oxidizing agent
weaker oxidizing agent
stronger reducing agent
weaker reducing agent
The reaction is spontaneous as written.
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31. Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
Voltaic Cells
The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.
If we can physically separate the oxidation and reduction half-reactions, we can use the electrons from the redox reaction to do work. This is an electrochemical cell.
Lets look at the reaction of zinc metal with copper(II) sulfate:
Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
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32. Voltaic Cells
We place a zinc electrode in aqueous ZnSO4 and a copper electrode in aqueous CuSO4. The electrodes are connected by a wire to allow the flow of electrons.
A salt bridge is used to complete the circuit.
Zinc metal is oxidized and copper ions are reduced in each half cell.
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33. Voltaic Cells
Oxidation occurs at the anode of an electrochemical cell.
Reduction occurs at the cathode of an electrochemical cell.
Electrons flow through the wire from the anode to the cathode in a voltaic cell.
Negatively charged ions travel through the salt bridge away from the cathode and towards the anode in a voltaic cell.
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34. Electrolytic Cells
Electrolytic cells are electrochemical cells that do not operate spontaneously. The process is referred to as electrolysis.
A source of electricity is required to drive an electrolytic cell.
An example of an electrolysis reactions is the recharging of the battery in a cell phone.
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35. Conclusions
A redox reaction is a reaction involving the transfer of electrons from one substance to another.
The oxidation number describes how many electrons have been lost or gained by an atom.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
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36. Conclusions Continued
An oxidizing agent is a substance that causes oxidation by accepting electrons. The oxidizing agent is reduced.
A reducing agent is a substance that causes reduction by donating electrons. The reducing agent is oxidized.
In redox reactions, the number of electrons lost must equal the number of electrons gained.
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37. Conclusions Continued
There are two methods to balance redox reactions:
The Oxidation Number Method
The Half-Reaction Method
The tendency for a substance to gain electrons is its reduction potential.
The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.
We can physically separate oxidation and reduction half-reactions and use the electrons from the redox reaction to do work in an electrochemical cell.
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