Lesson 4.5 Oxidation-Reduction Reactions
Suggested Reading
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Zumdahl Chapter 4 Section 4.9
Essential Question
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What are the characteristics of oxidation-reduction (redox) reactions.
Learning Objectives:
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Differentiate acid-base and precipitation reactions from redox reactions.
Define oxidation and reduction in terms of electron loss and gain.
Deduce the oxidation number of an element in a compound.
Deduce whether an element undergoes oxidation or reduction in
reactions using oxidation numbers.
Deduce simple oxidation and reduction half-equations given the
species involved in a redox reaction.
Deduce redox equations using half reactions.
Define the terms oxidizing agent and reducing agent.
Identify the oxidizing and reducing agents in redox equations.
Classify reactions by type.
In the two preceding lessons, we described precipitation reactions (reactions
involving a precipitate) and acid-base reactions (reactions involving proton
transfer). Here we will look at the third major class of reactions, oxidationreduction (redox) reactions, which are reactions involving a transfer of
electrons from one species to another.
As an example of a redox reaction, let us look at what happens when you dip
an ion nail in a blue solution of copper(II) sulfate. 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)
From this equation, you can obtain the net ionic equation, which shows that
iron metal reacts with copper(II) to produce iron(II) and copper metal.
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
The electron-transfer aspect of the reaction is apparent from this equations.
Note that each iron atom in the metal loses two electrons to form an iron(II)
ion, and each copper(II) 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 the 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, than an oxidationreduction number 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 using oxidation
number rules. In a redox reaction, one or more atoms change oxidation
number implying there has been a transfer of electrons.
Consider the combustion of calcium metal in oxygen gas.
2Ca(s) + O2(g) → 2CaO(s)
This is a redox reaction. To see this, you assign oxidation numbers to the
atoms in the equation and then note whether or not the atoms changed
oxidation number during the reaction.
Oxidation Number Rules
In redox reactions, oxidation numbers are assigned according to the following
rules.
The sum of the oxidation number 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.
Again, lets consider the combustion of calcium metal in oxygen gas. Lets
use the rules to assign oxidation numbers to the atoms in this reaction.
2Ca(s) + O2(g) → 2CaO(s)
Rule 1 states that the oxidation numbers of an atom in an element is
always 0, so the oxidation number for Ca in the metal and O in O2 have
oxidation numbers of zero.
Rule 2 states that the oxidation number for a monatomic ion is the same as
its charge. This rule applies to atoms that exist in a substances as
monatomic ions. Thus, the charge on Ca is +2.
Rule 4 states that the oxidation number of oxygen is usually -2. Thus the
oxidation number for oxygen in calcium oxide is -2.
0
0
+2 -2
2Ca(s) + O2(g) → 2CaO(s)
Most often, you will see the sign (+ or -) written to the left of the number if it
is a hypothetical oxidation number and to the right of the number if it is the
actual charge on an ion. However, this is not always the case.
Note that the sum of the oxidation numbers for calcium oxide equal zero
(2+(-2)=0). This is a requirement.
From the oxidation numbers assigned to the atoms in the reaction above,
you can see that the Ca and O atoms change in oxidation number during
the reaction. In effect, each calcium atom loses two electrons to form Ca2+
ions, and each O 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.
Oxidation-reduction reaction: a reaction in which electrons are
transferred between species or in which atoms change oxidation number.
Because calcium has gained in oxidation number due to the loss of
electrons, we say it has been oxidized. Oxygen on the other hand has
decreased in oxidation number due to a gain of electrons so we say it has
been reduced.
Use one of the acronyms below to help you remember this. They work!
Example: Assigning oxidation numbers
Use the rules to obtain the oxidation number of each atom in the following
compounds. a) HClO4 (perchloric acid)
Solution:
a) This is a neutral compound, so the sum of the oxidation numbers must
equal zero. Using this rule, we can write an equation that will help us to
determine the oxidation numbers of each elements (Note: There is no rule for
chlorine, so we will need to solve for the oxidation number of chlorine.
(oxidation # of H) + (oxidation # of Cl) + 4(oxidation # of O)=0
According to rules two and four the oxidation numbers for hydrogen and
oxygen are +1 and -2 respectively. Plugging this values into the equation
above will give you the oxidation number for Cl.
(1) + (oxidation # of Cl) + 4(-2)=0
oxidation # of Cl = 7
and, (1) + (7) + 4(-2)=0
Thus, the oxidation numbers are H = +1, Cl = +7, and O = -2
Describing Oxidation-Reduction Reactions
We use special terminology to describe redox reactions. To illustrate this
we will look gain at the reaction of iron with copper(II) sulfate. The net ionic
equation is
0
+2
+2
0
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
We can write this reaction in terms of two half-reactions.
A half-reaction is one of the two parts of an redox reaction, one part involves
a loss of electron (oxidation) and the other part involves a gain of electrons
(reduction)
The half reactions for the equation above are
Fe(s) → Fe2+(aq) + 2e- (electrons lost, oxidation half-reaction)
Cu2+(aq) + 2e- → Cu(s) (electrons gained, reduction half-reaction)
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 the equations above the oxidizing agent
is Cu2+ and the reducing agent is Fe(s).
The vocabulary reflects that fact that in a redox reaction, the oxidation and
reduction half-reactions must occur simultaneously if a redox reaction is to
occur.
Many important biochemical reactions, such those involved in cellular
respiration, occur by this mechanism.
Some Common Oxidation-Reduction Reactions
Many of the redox reactions we will discuss in this course fall into the following
simple kinds:
1. Combination reactions
2. Decomposition reactions
3. Displacement reactions
4. Combustion reactions
Lets look at examples of each (One of the free response questions on the AP
exam will ask you to write and predict products for several chemical reactions.
This lesson as well as the lessons on precipitation and acid-base reactions
should enable you to write most of the questions that are likely to show up on
the exam).
Combination Reactions
Watch the following YouTube Video
https://www.youtube.com/watch?v=Ftw7a5ccubs
A combination reaction is a reaction in which two substances combine to
form a third substance. Not all combination reactions are redox reactions.
However, the simplest cases are those in which two elements react to form a
compound; these are clearly redox reactions. Here are a couple of examples:
2Na(s) + Cl2(g) → 2NaCl(s)
CaO(s) + SO2(g) → CaSO3(s)
Combination reactions can involved elements or compounds (Can you name
all of the compounds above? There are both ionic and covalent compounds
so be sure to use the appropriate naming rules with each!). If you check the
oxidation numbers, you will see that that the first reaction is a redox reaction
while the second is not.
Decomposition Reactions
Watch the following YouTube Video:
https://www.youtube.com/watch?v=Wc0R7UlrlVA
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 (heat is added). Again, not all decomposition
reactions are redox reactions. Lets look at two examples. The first is the
reaction for the decomposition of ammonium dichromate seen in the video to
the right.
where delta represents that heat is being added to the reaction.
Check the oxidation numbers to determine which of the above reactions is an
redox reaction.
Displacement Reactions
Watch the following YouTube Video:
https://www.youtube.com/watch?v=UpjpQ34DZlQ
A displacement reaction (aka single-displacement reaction) is a reaction in
which an elements reacts with a compound , displacing an element from it.
Since these reactions involved an element and one of its compounds, these
must be redox 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)(aq) + 2Ag(s)
In this reaction 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)
In the reaction in the video, Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g), Zn
displaces hydrogen in the acid, producing zinc chloride and hydrogen gas.
The net ionic equation is
Zn(s) + 2H+(aq) → Zn+(aq) + H2(g)
Whether or not 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. The activity series to the right lists the elements in decreasing order
of their easy of losing electrons during reactions. The materials listing at the
top are the strongest reducing agents, because they lose electrons very easily
causing other substances to be reduced when they gain these lost electrons.
The materials at the bottom are the weakest reducing agents, since they do
not readily lose electrons. A free element reacts with the monatomic ion of
another element if the free element is above the other element in the
activity series. For example, consider the reaction
2K(s) + 2H+(aq) → 2K+(aq) + H2(g)
You would expect this reaction to proceed as written because potassium
metal is well above hydrogen in the activity series.
Combustion Reactions
Watch the following YouTube Video:
https://www.youtube.com/watch?v=gXcug7RqPgs
A combustion reaction is a reaction in which a substance reacts with
oxygen, usually with the rapid release of heat to produce a flame. The
products include one or more oxides. Organic compounds usually burn in
oxygen or air to yield carbon dioxide (this is what leads to global warming
since most of our fuels are made from organic compounds). If the compound
contains hydrogen (as most do) then what is also produced. For example,
butane burns in air as follows
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
In the "methane bubbles" video to the left, methane is allowed to burn in air
producing the plumes of flames observed.
Many metals burn in air air well. Although chunks of iron do not burn readily in
air, iron wool, which consists of fine strands of iron. does. The increased
surface area of the metal in iron wool allows oxygen fro air to react quickly
with it. The reaction is
4Fe(s) + 3O2(g) → 2Fe2O3(s)
The rusting of iron wool is essentially the slow burning of the wool in air.