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Honors Chemistry
Julien
Name _________________________
Period _____ Date _______________
Oxidation and Reduction
Chapter 15
Conservation of Matter and Stoichiometry
3. The conservation of atoms in chemical reactions leads to the principle of conservation of
matter and the ability to calculate the mass of products and reactants. As a basis for
understanding this concept:
g.* Students know how to identify reactions that involve oxidation and reduction and how to balance
oxidation-reduction reactions.
I. Oxidation-Reduction Reactions.
Read Section 15.1 in your eText! I will identify what is oxidized and what is reduced in an
oxidation-reduction reaction.
Do the Identifying Oxidation-Reduction Reactions Tutorial
A. Oxidation-reduction reactions.
1. Oxidation-reduction reactions (ReDox)—
2. Oxidation—
3.
OIL RIG
Oxidation Is Loss of electrons
Reduction Is Gain of electrons
Reduction—
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Chapter 15 Notes from Basic Chemistry
4. When calcium reacts with sulfur, ions are formed from the loss and gain of electrons.
a. Ca(s) + S(s)  CaS
b. Calcium loses two electrons, so it is oxidized. Ca  Ca2+ + 2e–
c. Sulfur gains two electrons, so it is reduced. S + 2e–  S2–
5. The reaction of zinc with copper(II) sulfate can be rewritten as:
Zn(s) + Cu2+(aq) + SO42–(aq)  Zn2+(aq) + SO42–(aq) + Cu(s)
a. Zinc is oxidized because it loses two electrons: Zn(s)  Zn2+(aq) + 2e–
b. Copper is reduced because it gains two electrons: Cu2+(aq) + 2e–  Cu(s)
Problem 1: In the following equation, which reactant is oxidized and which is reduced?
2Al(s) + 3Sn2+(aq)  2Al3+(aq) + 3Sn(s)
You are now able to do Problems 15.2 and 15.4 in your first homework assignment.
II. Oxidation Numbers.
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Chapter 15 Notes from Basic Chemistry
Read Section 15.2 in your eText! I will assign and use oxidation numbers to identify
elements that are oxidized or reduced, and to balance an oxidation-reduction equation.
Do the Assigning Oxidation States Tutorial!
A. Rules for assigning oxidation numbers.
1. The sum of the oxidation numbers in a molecule is equal to zero (0), or for a polyatomic ion, the
sum of the oxidation numbers is equal to the charge.
2. The oxidation number of an element (monoatomic or diatomic) is zero (0).
3. The oxidation number of a monoatomic ion is equal to its charge.
4. In compounds, the oxidation numbers of Group 1A (1) elements is +1 and of Group 2A (2)
elements is +2.
5. In compounds, the oxidation number of fluorine is always –1. Other elements in Group 7A (7)
also are –1 except when combined with oxygen or fluorine.
6. In compounds, the oxidation number of oxygen is usually –2 except in OF2.
7. In compounds with nonmetals, the oxidation number of hydrogen is +1; in compounds with
metals, the oxidation number of hydrogen is –1.
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Chapter 15 Notes from Basic Chemistry
Problem 2: Assign oxidation numbers to the elements in each of the following:
a. H3PO4
b. MnO4–
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Chapter 15 Notes from Basic Chemistry
B. Using oxidation numbers to identify oxidation-reduction.
1. In oxidation, the loss of electrons increases the oxidation number so that it is higher (more
positive in the product than in the reactant.
2. In reduction, the gain of electrons decreases the oxidation number so that it is lower (more
negative) in the product than the reactant.
Do the Oxidation, Reduction, Oxidizing Agents, and Reduction Agents Tutorial!
C. Oxidizing and reducing agents.
1. Reducing agent—
2. Oxidizing agent—
3. Oxidation–Reduction terminology. (Memorize this.)
Problem 3: Use the oxidation numbers to identify the substance that is oxidized, the substance that is reduced,
the oxidizing agent, and the reducing agent.
Zn(s) + CuCl2 (aq) ® ZnCl2 (aq) + Cu(s)
D. Using oxidation numbers to balance oxidation-reduction equations.
1. Oxidation and reduction must occur simultaneously.
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Chapter 15 Notes from Basic Chemistry
2. The electrons lost by the substance that is oxidized must equal the electrons gained by the
substance that is reduced.
Guide to Balancing Equations Using Oxidation Numbers
STEP 1: Assign oxidation numbers to all the elements.
STEP 2: Identify the oxidized and reduced elements from the changes
in oxidation numbers.
STEP 3: Multiply the changes in oxidation numbers by small integers
to equalize the increase and decrease.
STEP 4: Balance the remaining elements by inspection.
Problem 4: Use oxidation numbers to balance the following equation:
Li(s) + AlCl3 (aq) ® LiCl(aq) + Al(s)
Step 1:
Step 2:
Step 3:
Step 4:
Problem 5: Use oxidation numbers to balance the equation for the oxidation-reduction reaction of iron(III)
oxide and carbon to form iron and carbon dioxide.
Fe2O3 (s) + C(s) ® Fe(s) + CO2 (g)
Step 1:
Step 2:
Step 3:
Step 4:
You are now able to do Problems 15.6, 15.8, 15.10, 15.12, 15.14, 15.16 and 15.20 in your first homework.
III. Balancing Oxidation-Reduction Equations Using Half-Reactions.
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Chapter 15 Notes from Basic Chemistry
Read Section 15.3 in your eText! I will balance oxidation-reduction equations using the
half-reaction method.
Do the Balancing Redox Reactions Tutorial!
A. The half-reaction method.
1. Half-reaction method—
2. Half-reactions are balanced by adding electrons to each half-reaction to give an overall zero
charge.
Guide to Balancing Redox Equations Using Half-Reactions
STEP 1:
Write two half-reactions for the equation.
STEP 2:
Balance elements other than H and O in each half-reaction. Add
H2O to the side that needs O and add H+ to the side that needs H.
STEP 3:
Balance each half-reaction for charge by adding electrons to the
side with more positive charge.
STEP 4:
Multiply each half-reaction by factors that equalize the loss and
gain of electrons.
STEP 5:
Add balanced half-reactions, cancel electrons, and combine H2O
and H+. Check balance of atoms and charge.
Problem 6: Use the half-reaction method to balance the equation:
Zn(s) + Fe3+ (aq) ® Zn2+ (aq) + Fe2+ (aq)
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
B. Balancing oxidation-reduction equations in acidic solution.
1. To balance half-reactions, balance O by adding H2O.
2. To balance half-reactions, balance H by adding H+.
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Chapter 15 Notes from Basic Chemistry
Problem 7: Use half-reactions to balance the following equation in acidic solution:
Cu2+ (aq) + SO2 (g) ® Cu(s) + SO4 2– (aq)
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
C. Balancing oxidation-reduction equations in basic solution.
1. Balance these reactions the same way as you would to balance the equation in acidic solution by
adding H+ to the side with more oxygen and H2O to the side with less oxygen.
2. To balance it for a basic solution, neutralize the H+ by adding the same amount of OH– to each
side that converts the amount of hydrogen ions to the same amount of water. (i.e. two hydrogen
ions would require two hydroxide ions to be added to each side.)
Problem 8: Use half-reactions to balance the following equation in basic solution:
N2O(g) + 2ClO– (aq) ® NO2 – (aq) + Cl– (aq)
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6: Add equal amounts of OH– to neutralize the acid (H+).
You are now able to do problem 15.26 in your second homework assignment.
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Chapter 15 Notes from Basic Chemistry
IV. Oxidation of Alcohols.
Read Section 15.4 in your eText! I will classify alcohols as primary, secondary, or tertiary;
write equations for the oxidation of alcohols.
A. Classification of alcohols.
1. Alcohols are classified by the number of carbon groups attached to the carbon atom bonded to
the hydroxyl (—OH) group.
2. Primary (1°) alcohol—
3. Secondary (2°) alcohol—
4. Tertiary (3°) alcohol—
H
CH 3
|
|
CH 3—C —OH
|
H
CH 3
|
CH 3— C —OH
|
H
CH 3— C —OH
|
CH 3
B. Oxidation of primary and secondary alcohols.
1. The oxidation of a primary alcohol removes two hydrogen atoms and produces water and an
aldehyde, often using KMnO4 or K2Cr2O7 as oxidizing agents.
O
OH
||
CH 3— C —H ¾ O¾® CH 3—C —H + H 2O
|
H
2. The oxidation of secondary alcohols produces a ketone when two hydrogen atoms are removed
to make a molecule of water.
|
O
OH
||
CH 3— C —CH 3 ¾ ¾® CH 3—C —CH 3 + H 2O
|
H
|
O
3. Tertiary alcohols don’t oxidize readily because there are no hydrogen atoms on the carbon
bonded to the —OH group.
Problem 9: Draw the condensed structural formula of the aldehyde or ketone formed by the oxidation of the
following alcohol:
CH3
CH3
|
|
CH 3—CH —CH —CH 2 —CH 2 —OH
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Chapter 15 Notes from Basic Chemistry
You are now able to do problems 15.28 and 15.30 in your second homework assignment.
V. Electrical Energy from Oxidation-Reduction Reactions.
Read Section 15.5 in your eText! I will write the half-reactions that occur at the anode and
cathode of a voltaic cell; write the shorthand cell notation.
Do the Toxicity of Alcohol Tutorial!
A. Voltaic cells.
1. Electrochemical cell—
2. Voltaic cell—
3. A simple voltaic cell can be made when a zinc anode is placed in a Cu2+ solution.
a. The zinc metal becomes coated with the red-brown copper and the blue color of the solution
fades to clear.
b. The half-reactions can be written as follows:
Zn(s)

Zn2+(aq) + 2e–
(oxidation)
Cu2+(aq) + 2e–

Cu(s)
(reduction)
c. The overall cell reaction is:
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
4. The transfer of electrons is direct from Zn to Cu2+ but the reaction can be divided into half-cells.
5. Electrons flow from one half-cell to the other when an external circuit connects half-cells.
a. Anode—
b. Cathode—
6. Placing a salt bridge, containing positive and negative ions, in the half-cell solutions completes
the circuit and allows electrons to flow.
Anode is where
Cathode is where
oxidation takes place
reduction takes place
electrons are produced
electrons are used up
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Chapter 15 Notes from Basic Chemistry
7. An electrical current is produced as electrons flow from the anode through the wire to the
cathode.
8. In the cell above, the loss of zinc decreases the mass of the zinc anode and the gain of copper
increases the mass of the copper cathode.
9. The two half cells can be depicted in the following equation:
Zn(s) | Zn2+ (aq) || Cu 2+ (aq) | Cu(s)
a. The components of the oxidation half-cell (anode) are written on the left side in this
shorthand notation.
b. The components of the reduction half-cell (cathode) are written on the right side in this
shorthand notation.
c. A single vertical line separates the solid zinc anode from the zinc(II) ion solution and the
copper(II) ion solution from the solid copper cathode.
d. A double vertical line separates the two half-cells.
e. Electrons flow from the anode to the cathode.
10. If there is no component in the half-reactions that can be used as an electrode, electrodes made of
graphite or platinum are used for the transfer of electrons.
a. If there are two ionic components in a cell, their symbols are separated by a comma.
b. A platinum electrode placed in an Sn2+ solution like Sn(NO3)2 would cause the Sn2+ to
oxidize into Sn4+ + 2e–.
c. Using a silver cathode in an Ag+ solution like AgNO3, a salt bridge, and a wire to connect the
circuit would begin a current.
d. The notation for the cell would be written as:
Pt(s) | Sn2+ (aq), Sn 4+ (aq) | Zn2+ (aq) || Ag+ (aq) | Ag(s)
Problem 10: Write the half-reactions and the overall cell reaction for the following notation of a voltaic cell:
Co(s) | Co2+(aq) || Cu2+(aq) | Cu(s)
B. Prevention of corrosion.
1. Metals used in buildings and ships can corrode unless painted.
2. Any scratch or bubble will allow corrosion to weaken the metal and cause the sinking of ships or
the collapse of buildings or bridges.
3. Metals that lose electrons easier than iron (Fe) will become an anode and rust cannot form.
4. Magnesium, zinc, or aluminum can be used to keep ships from rusting, to keep underground
pipes from corroding, and ship hulls from corroding.
C. Batteries.
1. Batteries are made of voltaic cells used to power your cell phone, car, and flashlight.
2. A car battery, or lead storage battery is made of six voltaic cells linked together.
a. Each cell has a lead plate anode and a lead(IV) oxide (PbO2) plate that acts as the cathode.
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Chapter 15 Notes from Basic Chemistry
b. Each half-cell has a sulfuric acid solution (H2SO4).
c. The half-cell reactions that produce electric current are as follows:
Anode (oxidation):
Pb(s) + SO42–(aq)  PbSO4(s) + 2e–
Cathode (reduction):
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–  PbSO4(s) + 2H2O(l)
Overall cell reaction:
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq)  2PbSO4(s) + 2H2O(l)
3. As the battery is used, PbSO4 builds up on the electrodes and the H2SO4 concentration goes
down.
4. As you drive, the alternator restores the Pb and PbO2 electrodes are rebuilt and the concentration
of H2SO4 increases.
D. Dry-cell batteries.
1. Dry-cell batteries (no liquid inside, just a paste) are used in calculators, watches, hearing aids,
and many toys.
2. Acidic dry cell batteries use an anode of zinc metal that contains a paste of MnO2, NH4Cl, ZnCl2,
H2O, and starch; they use a graphite cathode.
Anode (oxidation):
Cathode (reduction:
Overall cell reaction:
+ H2O(l)
Zn(s)  Zn2+(aq) + 2e–
2MnO2(s) + 2NH4+(aq) + 2e–  Mn2O3(s) + 2NH3(aq) + H2O(l)
Zn(s) + 2MnO2(s) + 2NH4+(aq)  Zn2+(aq) + Mn2O3(s) + 2NH3(aq)
3. Alkaline batteries have similar components to acidic batteries except for their use of NaOH or
KOH to replace the NH4Cl.
4. In alkaline batteries, the zinc anode oxidizes to ZnO2 and produces longer life and more power.
Anode (oxidation):
Cathode (reduction:
Overall cell reaction:
Zn(s) + 2OH–(aq)  ZnO(s) + H2O(l) + 2e–
2MnO2(s) + H2O(l) + 2e–  Mn2O3(s) + 2OH–(aq)
Zn(s) + 2MnO2(s)  ZnO(s) + Mn2O3(s)
E. Mercury and lithium batteries.
1. Mercury batteries use a zinc anode but the cathode is steel in a mixture of HgO, KOH, and
Zn(OH)2.
2. Reduced mercury is toxic and an environmental hazard, which should always be disposed of
properly.
3. Lithium batteries use lithium as the anode because it is less dense than most metals and they can
be very small.
Anode (oxidation):
Cathode (reduction:
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Zn(s) + 2OH–(aq)  ZnO(s) + H2O(l) + 2e–
HgO(s) + H2O(l) + 2e–  Hg(l) + 2OH–(aq)
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Chapter 15 Notes from Basic Chemistry
Overall cell reaction:
Zn(s) + HgO(s)  ZnO(s) + Hg(l)
F. Nickel-cadmium (NiCad) batteries.
1. NiCad batteries are popular because they are easily recharged.
2. The anode is cadmium and the cathode is made of solid nickel oxide, NiO(OH)
Anode (oxidation):
Cathode (reduction:
Overall cell reaction:
Cd(s) + 2OH–(aq)  Cd(OH)2(s) + 2e–
2NiO(OH)(s) + 2H2O(l) + 2e–  2Ni(OH)2(s) + 2OH–(aq)
Cd(s) + 2NiO(OH)(s) + 2H2O(l)  Cd(OH)2(s) + 2Ni(OH)2(s)
You are now able to do problems 15.32, 15.34 and 15.36 in your second homework assignment.
VI. Oxidation-Reduction Reactions That Require Electrical Energy.
Read Section 15.6 in your eText! I will describe the half-cell reactions and the overall
reactions that occur in electrolytic cells.
Do the Electrolysis Tutorial!
Do the Predict Spontaneous Reactions Using the Activity Series Tutorial!
A. Activity series for some metals.
1. More active metals lose electrons easier than the les active metals.
2. Metals listed below H2 on the activity chart will not react with H+ from acids.
3. A metal will oxidize spontaneously when it is combined with the ions of any metal below it on
the list.
4. If you place a metal strip from a metal near the top of the list in a solution of a metal ion below it
on the list, you will plate the metal from the solution on the metal strip.
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Chapter 15 Notes from Basic Chemistry
B. Electrolytic cells.
1. To force a nonspontaneous reaction, electric current is needed to make the reaction proceed in
the opposite direction.
2. Electrolysis—
3. Electrolytic cell—
C. Electrolysis of sodium chloride.
1. Electrodes are placed in a mixture of sodium and chloride ions and then connected to a battery.
2. Liquid sodium metal and chlorine gas is produced.
Anode (oxidation):
Cathode (reduction:
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2Cl–(l)  Cl2(g) + 2e–
2Na+(l) + 2e–  2Na(l)
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Chapter 15 Notes from Basic Chemistry
Overall cell reaction:
2Na+(l) + 2Cl–(l)  Cl2(g) + 2Na(l)
D. Electroplating.
1. The electroplating industry uses electrolytic cells to coat a metal with a thin layer of a metal such
as silver, platinum or gold.
2. Car bumpers are electroplated with chromium to prevent rusting.
3. Many homes have silver platters, bowls, and decorative items which contain a thin layer of silver
and can be replated easily so they look like new.
Problem 11: Why is energy needed to chrome plate the iron in Sample Problem 15.11? (Electrolysis is used to
chrome plate an iron hubcap by placing the hubcap in a Cr3+ solution.)
You are now able to do problem 15.40 in your second homework assignment.
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Chapter 15 Notes from Basic Chemistry