Download Chapter 17

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
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.
Chapter 17
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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 → Fe3+ + 3 e-
• Oxygen gains electrons and is reduced
– O2 + 4 e- → 2 O2-
Chapter 17
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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:
1. A metal or a nonmetal in the free state has an
oxidation number of 0.
2. A monoatomic ion has an oxidation number equal to
its ionic charge.
3. A hydrogen atom is usually assigned an oxidation
number of +1.
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Rules for Oxidation Numbers
4. An oxygen atom is usually assigned an oxidation
number of -2.
5. For a molecular compound, the more electronegative
element is assigned a negative oxidation number
equal to its charge as an anion.
6. For an ionic compound, the sum of the oxidation
numbers for each of the atoms in the compound is
equal to 0.
7. 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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5
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)
Chapter 17
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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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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
Chapter 17
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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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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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Oxidation and Reduction
• The iron loses electrons and is oxidized.
– Fe → Fe2+ + 2 e-
• The sulfur gains electrons and is reduced.
– S + 2 e- → S2-
Chapter 17
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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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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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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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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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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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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:
1. Inspect the reaction and the substances
undergoing a change in oxidation number.
a) Write the oxidation number above each element.
b) Diagram the number of electrons lost by the oxidized
substance and gained by the reduced substance.
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Oxidation Number Method
2. 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:
a) In front of the oxidized substance, place a coefficient
equal to the number of electrons gained by the
reduced substance.
b) In front of the reduced substance, place a coefficient
equal to the number of electrons lost by the oxidized
substance.
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Oxidation Number Method
3. After balancing the equation, verify that the
coefficients are correct.
a) Place a check mark above the symbol for each
element to verify that the number of atoms is the
same on both sides.
b) 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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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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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.
Chapter 17
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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:
1. Write the half-reaction for both the oxidation and
reduction processes.
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Half-Reaction Method Continued
2. Balance the atoms in each half-reaction using
coefficients.
a) Balance all elements except oxygen and hydrogen.
b) Balance oxygen using H2O.
c) Balance hydrogen using H+.
d) For reactions in basic solution, add one OH- to each
side for each H+ and combine H+ & OH- to H2O.
e) Balance the ionic charges using electrons.
Chapter 17
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Half-Reaction Method Continued
3. Multiply each half-reaction by a whole number
so that the total number of electrons in each is
the same.
4. Add the two half-reactions together and cancel
the identical species, including electrons.
5. 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.
Chapter 17
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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- → Mn2+ + 4 H2O
Chapter 17
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Balancing a Redox Equation
• Since Fe2+ loses 1 electron and MnO4- gains 5
electrons, we have to multiply the iron halfreaction by 5:
5 Fe2+ → 5 Fe3+ + 5 e5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O
• We add the two half-reactions together and cancel
out the 5 electrons on each side to get the
balanced equation:
5 Fe2+ + 8 H+ + MnO4- → 5 Fe2+ + Mn2+ + 4 H2O
Chapter 17
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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.
Chapter 17
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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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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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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
stronger
oxidizing agent reducing agent
weaker
reducing agent
weaker
oxidizing agent
• The reaction is spontaneous as written.
Chapter 17
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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)
Chapter 17
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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.
Chapter 17
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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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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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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.
Chapter 17
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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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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. Chapter 17
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