Chapter 22
Oxidation – Reduction Rxns
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Section 22.1
The Meaning of
Oxidation and Reduction
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OBJECTIVES:
– Define the terms “oxidation” and
“reduction” in terms of electron loss or
gain.
– State the characteristics of a RedOx
Rxn, and recognize oxidizing and
reducing agents.
Oxygen in RedOx Reactions
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“Oxidation” has traditionally meant the
combination of elements (or compounds) with
oxygen to form oxides.
Oxidation is most closely associated with
combustion (burning).
“Reduction” traditionally meant the loss of
oxygen from a compound.
When ores (especially iron) have their oxygen
removed they are “reduced” back to their
original elements
Reduction and oxidation occur
simultaneously and the process is termed a
“RedOx Reaction”
Electron Transfer in RedOx Reactions
There are 2 mnemonics for helping you
memorize the definitions for reduction
and oxidation:
OIL RIG = Oxidation Is Loss of Electrons
Reduction Is Gain of Electrons
LEO goes GER =
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
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Example of a RedOx Reaction
Mg + S  MgS
In this example elemental magnesium and sulfur
combine to produce the compound magnesium
sulfide (a synthesis reaction).
Mg  Mg
2+
1-
+ 2e
The magnesium atom becomes a magnesium cation by
losing two electrons. The two electrons are needed on
the product side to balance the charge of this OXIDATION
HALF REACTION.
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1-
2e
+ S  S
2-
The sulfur atom gains the two electrons lost
by the magnesium atom to become a sulfide
anion. This is the REDUCTION HALF REACTION.
Mg + S  Mg
2+
2-
+ S
The over all balanced reaction is the sum of the two half
reactions. NO electrons can show, and it must balance
by both mass and charge!
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More Good Stuff
The oxidizing reagent is itself reduced
(and thereby causes oxidation).
 The reducing agent is itself oxidized
(and thereby causes reduction).
 In the preceding example, Mg was
oxidized, so it is the reducing agent.
The S was reduced, so it is the oxidizing
agent.
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Table 22.1 Processes Leading
to Oxidation and Reduction
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Oxidation can occur by:
Loss of electrons
Shift of electrons away from a covalent
bond
Gain of Oxygen
Increase in oxidation number (charge)
Loss of hydrogen (biology definition)
Table 22.1 Processes Leading
to Oxidation and Reduction
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Reduction can occur by:
Gain of electrons
Shift of electrons toward an atom in a
covalent compound
Loss of Oxygen
Decrease in Oxidation Number (charge)
Gain of hydrogen (biology definition)
Some Important Facts
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On the NYS Regents Exam the answer to any
RedOx question will be one of the reactants,
never the products!
If you are asked to identify a RedOx reaction
from a list of reactions, the answer can simply
be found by identifying which reaction has an
element on one side the arrow, and then a
compound containing that element on the
other side of the arrow! Simple!!
Synthesis and Single Replacement reactions
are always RedOx reactions.
Decomposition reactions can be RedOx, but
Double Replacement are rarely RedOx.
Corrosion Facts
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Iron reacts with atmospheric oxygen gas to
form iron (III) oxide under dry conditions.
Under moist conditions, especially in the
presence of salts, iron forms iron
hydroxide(s).
Iron corrodes all the way through and forms
pits or holes as it rusts.
Not all metals corrode - Gold and platinum
(Noble Metals) are examples.
Some metals, such as silver or aluminum,
form thin layers of oxide (rust) that protect
them from further damage.
How Can We Prevent Corrosion?
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Rust free metal surfaces can be cleaned and
varnished or painted to keep molecular
oxygen from reacting with the metal.
A “sacrifice metal” can be placed in contact
with the metal we are trying to protect. In this
case, the sacrifice metal rusts faster than its
partner, and supplies electrons to it such that
it will not be easily oxidized. Typically Mg or
Zn are used. If Zn is the sacrifice metal, we
say the metal we are trying to protect has
been “galvanized”.
One More Method
If the metal object we are trying to
protect is really valuable, we can use a
D.C. current source (a battery) to supply
a feed of electrons to the metal we are
trying to protect.
 This process is usually done on
expensive automobiles, underground
pipelines, storage tanks, etc.
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Section 22.2
Oxidation Numbers
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OBJECTIVES:
– Determine the oxidation number of an
atom in pure substances.
– Define RedOx in terms of changes in
oxidation number
Rules for Assigning Oxidation Numbers
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The oxidation number of a monatomic ion is
equal in magnitude and sign to its ionic charge.
Ex: The oxidation number of bromide is -1.
The oxidation number of hydrogen is +1 except
in hydrides where it is -1. Ex: In NaH hydrogen
is -1.
The oxidation number of oxygen is normally -2,
except in peroxides where it is -1. Ex: In H2O2,
the oxygen is -1.
The oxidation number of any uncombined
element (including diatomic elements) is 0. Ex:
The charges on S, Na, and N2 are all 0.
Rules for Assigning Oxidation Numbers
The sum of all the oxidation numbers in
an electrically neutral compound is 0.
Ex: In SO2, S = +4 and O = -2
(S) + 2(O) = 0 or (+4) + 2(-2) = 0
 The sum of all the oxidation numbers in
a polyatomic ion is equal to the charge
on that ion.
Ex: In CO32-, C = +4 and O = -2
(C) + 3(O) = -2 or (+4) + 3(-2) = -2
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Try These!
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What are the charges (oxidation numbers) of all the elements in
these examples?
S2O3
S = +3, O = -2
2(S) + 3(O) = 0 or 2(+3) + 3(-2) = 0
KMnO4
K = +1, Mn = +7, O = -2
K + Mn + 4(O) = 0 or (+1) + (+7) + 4(-2) = 0
N2
N = 0 (remember, diatomic elements are all zero!)
Cr2O72Cr = +6, O = -2
2(Cr) + 7(O) = -2
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or
2(+6) + 7(-2) = -2
Write the Net Ionic Rxn and Identify what is
Reduced and Oxidized in this Equation:
2AgNO3(aq) + Cu(s)  2Ag(s) + Cu(NO3)2(aq)
For this single replacement rxn, the balanced complete ionic rxn is:
2Ag1+(aq) + Cu0(s) + 2NO31-(aq)  2Ag0(s) +
Cu2+(aq) + 2NO31-(aq)
The spectator ion must be 2NO31-(aq), so that leaves the net ionic rxn as:
2Ag1+(aq) + Cu0(s)  2Ag0(s) + Cu2+(aq)
(Note that this rxn is balanced by both mass and charge!)
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2Ag1+(aq) + Cu0(s)  2Ag0(s) + Cu2+(aq)
The half rxns are:
2Ag1+(aq) + 2e1-  2Ag0(s)
Reduction
and
Cu0(s)  Cu2+(aq) + 2e1-
Oxidation
Thus, 2Ag1+(aq) is reduced (making it the
oxidizing agent), and Cu0(s) is oxidized (making it
the reducing agent). Notice that all the answers
to a potential NYS Regents question on this rxn
are on the reactant (left) side of the reaction!
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Section 22.3
Balancing RedOx Rxns
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OBJECTIVES:
– Balance RedOx rxns by either
“inspection” or the “half-rxn” method
Balancing by “Inspection”
Most chemical rxns can be balanced by
simply examining the unbalanced
reaction, picking out one particular
species that is unbalanced, and
iteratively changing the coefficients until
balance is achieved
 This is the method you have used all
along, and it will be more than sufficient
to balance any RedOx reaction you may
find on the NYS Regents exam
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Balancing by the “Half-Reaction” Method
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This method is far more useful than what is
stated on pg. 663 – 665 of your text book
Not only can you balance simple Regents
level rxns, you can begin to think about far
more involved IB course rxns
The basic idea is to: (1) break up the rxn into
its Red and Ox parts, (2) multiple each half by
what ever factor is needed to make the
number of electrons in both half-rxns equal,
and the (3) adding the two half-rxns back
together while canceling any species that is
found to be in common
Example of Half-Rxn Method
Li(s) + Al(NO3)3(aq)  LiNO3(aq) + Al(s)
Write the complete ionic reaction:
Li(s) + Al3+ (aq) + 3(NO3)1-(aq)  Li1+ (aq) + NO31-(aq) + Al(s)
Eliminate the spectator ions. Write the net ionic reaction:
Li(s) + Al3+ (aq)  Li1+ (aq) + Al(s)
Now write the two half-rxns:
3e1- + Al3+ (aq)  Al(s)
Li(s)  Li1+ (aq) + 1e123
Reduction
Oxidation
As you can see, the number of electrons is not the same,
so we will multiple the oxidation half by 3:
3e1- + Al3+ (aq)  Al(s)
Reduction
3Li(s)  3Li1+ (aq) + 3e1Oxidation
Now add the two half-reactions back together to get:
3Li(s) + Al3+ (aq)  3Li1+ (aq) + Al(s)
Put the spectator ions back in and you get:
3Li(s) + Al(NO3)3(aq)  3LiNO3(aq) + Al(s)
Of course you could have done this by simple
“inspection”, but the technique we just used can be
applied to solve far more difficult problems!
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