chemistry

Oxidation–Reduction Reactions
In the two preceding sections, we described precipitation reactions (reactions producing
a precipitate) and acid–base reactions (reactions involving proton transfer).
Here we discuss the third major class of reactions, oxidation–reduction reactions,
which are reactions involving a transfer of electrons from one species to another.
As a simple example of an oxidation–reduction reaction, let us look at what happens
when you dip an iron nail into a blue solution of copper(II) sulfate (Figure 4.11).
What you see is that the iron nail becomes coated with a reddish-brown tinge of metallic
copper. The molecular equation for this reaction is
Fe(s) CuSO4(aq) ±£ FeSO4(aq) Cu(s)
The net ionic equation is
Fe(s) Cu2 (aq) ±£ Fe2 (aq) Cu(s)
ZnS(s) 2H (aq) 2Cl (aq) Zn2 ±£ (aq) 2Cl (aq) H2S(g)
Experiments in General
Chemistry, 4B,“Ionic Reactions
in Aqueous Solutions.”
An alternate example is
provided in the Instructor’s
Resource Manual.
Video no. 24 (Series D),
“The Gold Penny.”
4.5 Oxidation–Reduction Reactions 145
The electron-transfer aspect of the reaction is apparent from this equation. Note that
each iron atom in the metal loses two electrons to form an iron(II) ion, and each
copper(II) ion gains two electrons to form a copper atom in the metal. The net effect
is that two electrons are transferred from each iron atom in the metal to each copper(
II) ion.
The concept of oxidation numbers was developed as a simple way of keeping
track of electrons in a reaction. Using oxidation numbers, you can determine whether
or not electrons have been transferred from one atom to another. If electrons have
been transferred, an oxidation–reduction reaction has occurred.
■ Oxidation Numbers
We define the oxidation number (or oxidation state) of an atom in a substance as
the actual charge of the atom if it exists as a monatomic ion, or a hypothetical charge
assigned to the atom in the substance by simple rules. An oxidation–reduction reaction
is one in which one or more atoms change oxidation number, implying that there
has been a transfer of electrons.
Consider the combustion of calcium metal in oxygen gas (Figure 4.12).
2Ca(s) O2(g) ±£ 2CaO(s)
This is an oxidation–reduction reaction. To see this, you assign oxidation numbers to
the atoms in the equation and then note that the atoms change oxidation number during
the reaction.
Since the oxidation number of an atom in an element is always zero, Ca and O
in O2 have oxidation numbers of zero. Another rule follows from the definition of
oxidation number: The oxidation number of an atom that exists in a substance as a
monatomic ion equals the charge on that ion. So the oxidation number of Ca in CaO
is 2 (the charge on Ca2 ), and the oxidation number of O in CaO is 2 (the charge
Figure 4.11
Reaction of iron with Cu2 (aq)
Left: Iron nail and copper(II) sulfate
solution, which has a blue
color. Center: Fe reacts with
Cu2 (aq) to yield Fe2 (aq) and
Cu(s). In the molecular view water
and the sulfate anion have been
omitted. Right: The copper metal
plates out on the nail.
Cu Fe2+ 2+
Fe
Cu
Video no. 27 (Series D),
“Complexes of Vanadium.”
Figure 4.12
The burning of calcium
metal in oxygen
The burning calcium emits a
red-orange flame.
O2–
O
Ca2+
Ca
146 CHAPTER 4 Chemical Reactions
on O2 ). To emphasize these oxidation numbers in an equation, we will write them
above the atomic symbols in the formulas.
From this, you see that the Ca and O atoms change in oxidation number during the
reaction. In effect, each calcium atom in the metal loses two electrons to form Ca2
ions, and each oxygen atom in O2 gains two electrons to form O2 ions. The net
result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidation–-
reduction reaction. In other words, an oxidation–reduction reaction (or redox reaction)
is a reaction in which electrons are transferred between species or in which
atoms change oxidation number.
Note that calcium has gained in oxidation number from 0 to 2. (Each calcium
atom loses two electrons.) We say that calcium has been oxidized. Oxygen, on the other
hand, has decreased in oxidation number from 0 to 2. (Each oxygen atom gains two
electrons.) We say that oxygen has been reduced. An oxidation–reduction reaction always
involves both oxidation (the loss of electrons) and reduction (the gain of electrons).
Formerly, the term oxidation meant “reaction with oxygen.” The current definition
greatly enlarges the meaning of this term. Consider the reaction of calcium metal
with chlorine gas (Figure 4.13); the reaction looks similar to the burning of calcium
in oxygen. The chemical equation is
In this reaction, the calcium atom is oxidized, because it increases in oxidation number
(from 0 to 2, as in the previous equation). Chlorine is reduced; it decreases in
oxidation number from 0 to 1. This is clearly an oxidation–reduction reaction that
does not involve oxygen.
■ Oxidation-Number Rules
So far, we have used two rules for obtaining oxidation numbers: (1) the oxidation
number of an atom in an element is zero, and (2) the oxidation number of an atom
in a monatomic ion equals the charge on the ion. These and several other rules for
assigning oxidation numbers are given in Table 4.5.
In molecular substances, we use these rules to give the approximate charges on
the atoms. Consider the molecule SO2. Oxygen atoms tend to attract electrons, pulling
them from other atoms (sulfur in the case of SO2). As a result, an oxygen atom in
SO2 takes on a negative charge relative to the sulfur atom. The magnitude of the
charge on an oxygen atom in a molecule is not a full 2 charge as in the O2 ion.
However, it is convenient to assign an oxidation number of 2 to oxygen in SO2 (and
in most other compounds of oxygen) to help us express the approximate charge distribution
in the molecule. Rule 3 in Table 4.5 says that an oxygen atom has an oxidation
number of 2 in most of its compounds.
Rules 4 and 5 are similar in that they tell you what to expect for the oxidation
number of certain elements in their compounds. Rule 4, for instance, says that hydrogen
has an oxidation number of 1 in most of its compounds.
Rule 6 states that the sum of the oxidation numbers of the atoms in a compound
is zero. This rule follows from the interpretation of oxidation numbers as (hypothetical)
charges on the atoms. Because any compound is electrically neutral, the sum of
the charges on its atoms must be zero. This rule is easily extended to ions: the sum
Ca(s) Cl2(g)
0 0 2 1
±£ CaCl2(s)
2Ca(s) O2(g)
0 0 2 2
±£ 2CaO(s)
Figure 4.13
The burning of calcium metal
in chlorine
The reaction appears similar to
the burning of calcium in oxygen.
4.5 Oxidation–Reduction Reactions 147
of the oxidation numbers (hypothetical charges) of the atoms in a polyatomic ion
equals the charge on the ion.
You can use Rule 6 to obtain the oxidation number of one atom in a compound
or ion, if you know the oxidation numbers of the other atoms in the compound or
ion. Consider the SO2 molecule. According to Rule 6,
(Oxidation number of S) 2 (oxidation number of O) 0
or
(Oxidation number of S) 2 ( 2) 0.
Therefore,
Oxidation number of S (in SO2) 2 ( 2) 4
The next example illustrates how to use the rules in Table 4.5 to assign oxidation
numbers.
EXAMPLE 4.7 Assigning Oxidation Numbers
Use the rules from Table 4.5 to obtain the oxidation number of the chlorine atom
in each of the following: (a) HClO4 (perchloric acid), (b) ClO3
(chlorate ion).
PROBLEM STRATEGY In each case, write the expression for the sum of the oxidation numbers, equating
this to zero for a compound or to the charge for an ion (Rule 6). Now, use Rules 2
to 5 to substitute oxidation numbers for particular atoms, such as 2 for oxygen and
1 for hydrogen, and solve for the unknown oxidation number (Cl in this example).
(continued)
Table 4.5
Rules for Assigning Oxidation Numbers
Rule Applies to Statement
1 Elements The oxidation number of an atom in an element is zero.
2 Monatomic ions The oxidation number of an atom in a monatomic ion
equals the charge on the ion.
3 Oxygen The oxidation number of oxygen is 2 in most of its
compounds. (An exception is O in H2O2 and other
peroxides, where the oxidation number is 1.)
4 Hydrogen The oxidation number of hydrogen is 1 in most of its
compounds. (The oxidation number of hydrogen is
1 in binary compounds with a metal, such as CaH2.)
5 Halogens The oxidation number of fluorine is 1 in all of its
compounds. Each of the other halogens (Cl, Br, I)
has an oxidation number of 1 in binary
compounds, except when the other element is
another halogen above it in the periodic table or
the other element is oxygen.
6 Compounds and ions The sum of the oxidation numbers of the atoms in a
compound is zero. The sum of the oxidation numbers
of the atoms in a polyatomic ion equals the charge on
the ion.
Two alternate examples are
provided in the Instructor’s
Resource Manual.
148 CHAPTER 4 Chemical Reactions
SOLUTION a. For perchloric acid, Rule 6 gives the equation
(Oxidation number of H) (oxidation number of Cl) 4
(oxidation number of O) 0
Using Rules 3 and 4, you obtain
( 1) (oxidation number of Cl) 4 ( 2) 0
Therefore,
Oxidation number of Cl (in HClO4) ( 1) 4 ( 2) 7
b. For the chlorate ion, Rule 6 gives the equation
(Oxidation number of Cl) 3 (oxidation number of O) 1
Using Rule 3, you obtain
(Oxidation number of Cl) 3 ( 2) 1
Therefore,
Oxidation number of Cl (in ClO3
) 1 3 ( 2) 5
ANSWER CHECK As most compounds do not have elements with very large positive or very large
negative oxidation numbers, you should always be on the alert for a possible
assignment mistake when you find oxidation states greater than 6 or less than
4. (From this example you see that a 7 oxidation state is possible; however, it
only occurs in a limited number of cases.)
EXERCISE 4.8 Obtain the oxidation numbers of the atoms in each of the following: (a) potassium
dichromate, K2Cr2O7, (b) permanganate ion, MnO4
.
See Problems 4.49, 4.50, 4.51, and 4.52.
■ Describing Oxidation–Reduction Reactions
We use special terminology to describe oxidation–reduction reactions. To illustrate
this, we will look again at the reaction of iron with copper(II) sulfate. The net ionic
equation is
We can write this reaction in terms of two half-reactions. A half-reaction is one of
two parts of an oxidation–reduction reaction, one part of which involves a loss of
electrons (or increase of oxidation number) and the other a gain of electrons (or
decrease of oxidation number). The half-reactions for the preceding equation are
Oxidation is the half-reaction in which there is a loss of electrons by a species
(or an increase of oxidation number of an atom). Reduction is the half-reaction
in which there is a gain of electrons by a species (or a decrease in the oxidation
number of an atom). Thus, the equation Fe(s) ±£ Fe2 (aq) 2e represents the
oxidation half-reaction, and the equation Cu2 (aq) 2e ±£ Cu(s) represents
the reduction half-reaction.
Fe(s) (electrons lost by Fe)
0 2
±£ Fe2 (aq) 2e
Cu2 (aq) 2e
2 0
±£ Cu(s) (electrons gained by Cu2 )
Fe(s) Cu2 (aq)
0 2 2 0
±£ Fe2 (aq) Cu(s)
4.5 Oxidation–Reduction Reactions 149
Recall that a species that is oxidized loses electrons (or contains an atom that
increases in oxidation number) and a species that is reduced gains electrons (or contains
an atom that decreases in oxidation number). An oxidizing agent is a species
that oxidizes another species; it is itself reduced. Similarly, a reducing agent is a
species that reduces another species; it is itself oxidized. In our example reaction, the
copper(II) ion is the oxidizing agent, whereas iron metal is the reducing agent.
The relationships among these terms are shown in the following diagram for the
reaction of iron with copper(II) ion.
■ Some Common Oxidation–Reduction Reactions
Many of the oxidation–reduction reactions can be described as:
1. Combination reactions
2. Decomposition reactions
3. Displacement reactions
4. Combustion reactions
We will describe examples of each of these in this section.
Combination Reactions A combination reaction is a reaction in which two
substances combine to form a third substance. Note that not all combination reactions
are oxidation–reduction reactions. However, the simplest cases are those in which two
elements react to form a compound; these are clearly oxidation–reduction reactions.
In Chapter 2, we discussed the reaction of sodium metal and chlorine gas, which is a
redox reaction (Figure 4.14).
2Na(s) Cl2(g) ±£ 2NaCl(s)
Fe(s)
0
reducing
agent
oxidation
oxidizing
agent
2 2
Cu2 (aq)
0
±£ Fe2 (aq) Cu(s)
reduction
Video Series C,“Zinc and Iodine.”
Na
NaCl
Cl2
Figure 4.14
Combination reaction
Left: Sodium metal and chlorine
gas. Right: The spectacular combination
reaction of sodium and
chlorine.
Video no. 19 (Series A),
“Ammonium Dichromate
Volcano.”
Video Series C,“Ammonium
Dichromate Volcano.”
150 CHAPTER 4 Chemical Reactions
Antimony and chlorine also combine in a fiery reaction.
2Sb 3Cl2 ±£ 2SbCl3
Some combination reactions involve compounds as reactants and are not oxidation–
reduction reactions. For example,
CaO(s) SO2(g) ±£ CaSO3(s)
(If you check the oxidation numbers, you will see that this is not an oxidation–
reduction reaction.)
Decomposition Reactions A decomposition reaction is a reaction in which a single
compound reacts to give two or more substances. Often these reactions occur when
the temperature is raised. In Chapter 1, we described the decomposition of mercury(II)
oxide into its elements when the compound is heated (Figure 4.15). This is an
oxidation–reduction reaction.
Another example is the preparation of oxygen by heating potassium chlorate with
manganese(IV) oxide as a catalyst.
In this reaction, a compound decomposes into another compound and an element; it
also is an oxidation–reduction reaction.
Not all decomposition reactions are of the oxidation–reduction type. For example,
calcium carbonate at high temperatures decomposes into calcium oxide and carbon
dioxide.
Is there a change in oxidation numbers? If not, this confirms that this is not an
oxidation–reduction reaction.
Displacement Reactions A displacement reaction (also called a single-replacement
reaction) is a reaction in which an element reacts with a compound, displacing an element
from it. Since these reactions involve an element and one of its compounds, these
must be oxidation–reduction reactions. An example is the reaction that occurs when you
dip a copper metal strip into a solution of silver nitrate.
Cu(s) 2AgNO3(aq) ±£ Cu(NO3)2(aq) 2Ag(s)
From the molecular equation, it appears that copper displaces silver in silver nitrate,
producing crystals of silver metal and a greenish-blue solution of copper(II) nitrate.
The net ionic equation, however, shows that the reaction involves the transfer of electrons
from copper metal to silver ion:
Cu(s) 2Ag (aq) ±£ Cu2 (aq) 2Ag(s)
When you dip a zinc metal strip into an acid solution, bubbles of hydrogen form
on the metal and escape from the solution (Figure 4.16).
Zn(s) 2HCl(aq) ±£ ZnCl2(aq) H2(g)
CaCO3(s) ±£ CaO(s) CO2(g)
2KClO3(s) ±±£ 2KCl(s) 3O2(g)
MnO2
2HgO(s) ±£ 2Hg(l) O2(g)
Figure 4.15
Decomposition reaction
The decomposition reaction
of mercury(II) oxide into its
elements,mercury and oxygen.
Demonstration of metal/metal
salt displacement reactions:
Summerlin and Ealy, Chemical
Demonstrations, Vol. 1
(Washington,D.C.: Am. Chem.
Soc., 1985), no. 96, p. 151.
O2–
Hg2+
Hg
O
4.5 Oxidation–Reduction Reactions 151
Zinc displaces hydrogen in the acid, producing zinc chloride solution and hydrogen
gas. The net ionic equation is
Zn(s) 2H (aq) ±£ Zn2 (aq) H2(g)
Whether a reaction occurs between a given element and a monatomic ion depends
on the relative ease with which the two species gain or lose electrons. Table 4.6 shows
the activity series of the elements, a listing of the elements in decreasing order of
their ease of losing electrons during reactions in aqueous solution. The metals listed
at the top are the strongest reducing agents (they lose electrons easily); those at the
bottom, the weakest. A free element reacts with the monatomic ion of another element
if the free element is above the other element in the activity series.
Consider this reaction:
2K(s) 2H (aq) ±£ 2K (aq) H2(g)
You would expect this reaction to proceed as written, because potassium metal (K) is
well above hydrogen in the activity series. In fact, potassium metal reacts violently
with water, which contains only a very small percentage of H ions. Imagine the reaction
of potassium metal with a strong acid like HC