Assigning Oxidation Numbers
We will hold off for the time being the formal
definition of an oxidation number.
The oxidation number (state) of a free or
uncombined element is 0.
 Fe, Na, O2
The oxidation number of a monatomic ion is
equal to its charge.
 K+, Al3+, Ca2+, Cu+
8-1
The oxidation number of hydrogen is +1 when
it is covalently bonded.
 HCl, H2O2, H2O, H2SO4
The oxidation number of hydrogen is -1 when
it is ionically bonded with a metal.
 NaH, CaH2, KH
 These are examples of hydrides and are
named sodium hydride, calcium hydride,
and potassium hydride.
8-2
The oxidation number of oxygen is -2 in most
of its covalent compounds.
 One exception is when oxygen is bonded
to fluorine as in OF2.
In OF2, the oxidation number of oxygen is
+2.
Fluorine being the most electronegative
element is always assigned an oxidation
number of -1.
8-3
 Another exception is when the oxidation
number for oxygen is -1 when it is a
peroxide.
O22- can be found covalently bonded as it
is in H2O2 (hydrogen peroxide).
O22- can be found ionically bonded as it
is in K2O2 (potassium peroxide).
Something to remember is that you do
not reduce the subscripts in a metallic
peroxide (K2O2 ).
8-4
 A second exception is when oxygen has
an oxidation number of -½.
This occurs when oxygen is the anion,
O2-, and is called a superoxide.
LiO2 is such a compound and is called
lithium superoxide.
For binary compounds, the element with the
greater electronegativity is assigned a
negative oxidation number equal to its
charge.
8-5
The algebraic sum of the oxidation numbers
in a neutral compound must equal zero.
+3 -3
+3 -1
+2 -2
+1 -1
-3 +3
-3 +1
 AlCl3
H2O2
NH3
The algebraic sum of the oxidation numbers
in a polyatomic ion must equal the charge on
the ion.
+3 -2
+1 -2
-3 +4
-3 +1
+5 -8
+5 -2
 H3O+
NH4+
PO438-6
Oxidation Numbers
An oxidation number indicates the charge
that is assigned to an atom for a particular
electron environment.
 An electron environment consists of the
number of neighboring atoms and their
electronegativities.
 From species to species, the oxidation
number of the same atom can vary.
8-7
 As an example, compare the oxidation
numbers for chlorine in the following two
acids.
1 7 -8
1 3 -4
1
1
-2
HClO4
-2
HClO2
The Cl atom in the HClO4 molecule finds
itself covalently bonded to two more
atoms more electronegative than itself.
The difference in oxidation numbers for
chlorine is found in the Lewis structures.
8-8
The Lewis structure for HClO4 and ClO4- is
-1
O
O
O Cl O H
O Cl O
O
O
Comparing these two structures, one sees
that the hydrogen forms a coordinate
covalent bond.
The black electron to the left of chlorine is the
electron resulting in the -1 charge.
8-9
A coordinate covalent bond is not really that
different from it’s first cousin, the covalent
bond, except that the oxygen provides both
electrons.
-1
O
O
O Cl O H
O Cl O
O
O
The electron pair is assigned to the more
electronegative atom which in this case are
the oxygens.
8 - 10
O
O Cl O H
O
O
O Cl O
-1
O
There is a total of 32 valence electrons in the
perchlorate polyatomic ion with only 24
coming from the oxygens.
The remaining 8 electrons must come from
the “extra” electron the ion had to begin with
(because its charge is -1) and 7 from the Cl.
8 - 11
After assigning Cl’s 7 valence electrons to the
O’s, the apparent charge on Cl is +7.
This line of reasoning is used to show Cl’s
oxidation number in ClO3- (+5), ClO2- (+3), and
ClO- (+1).
The polyatomic ions ClO4-, ClO3-, ClO2-, and
ClO- also illustrate the Law of Multiple
Proportions.
The Law of Multiple Proportions states that
when two different elements form more than
one compound their mass ratios are
represented small whole numbers.
8 - 12
What about H2S and HNO3?
 There is a total of 8 valence e- in
hydrosulfuric acid with only 6 coming
from the sulfur.
After assigning 2 e- to the more
electronegative S, its oxidation number
is -2 and that of each H is 1.
 Nitric acid has 24 valence e- with 18
coming from the O’s giving the apparent
charge of 1 to H and 5 to N.
8 - 13
Oxidation State Method
The Oxidation States Method is a means of
balancing the more simpler redox reactions.
The steps for a variation on this method are
as follows:
 Assign oxidation numbers to all of the
atoms in the equation.
 Identify the substance reduced
(oxidizing agent) and determine the
number of electrons gained.
8 - 14
 Identify the substance oxidized
(reducing agent) and determine the
number of electrons given off.
 Use coefficients to balance the atoms
and electrons in the species reduced.
 Use coefficients to balance the atoms
and electrons in the species oxidized.
 Use coefficients to balance the
remaining atoms by inspection.
8 - 15
Oxidizing And Reducing Agents
The terminology for oxidation-reduction
(redox) can be somewhat confusing.
The oxidizing agent is the species (atom or
ion) being reduced.
In the following example, elemental chlorine
is said to be reduced.
 Cl2 → Cl-
8 - 16
 For a species to be reduced, its
oxidation is reduced to a smaller
number.
 Another way of saying the same thing is
that the oxidation number of chlorine
becomes less positive or more negative.
 Oxidation-reduction always occur
together.
 Because chlorine is reduced, it is called
an oxidizing agent.
8 - 17
 When chlorine is reduced, it is bringing
about the oxidizing of another species.
 Hence, chlorine is called an oxidizing
agent.
 The balanced equation for chlorine is
given by:
0
-2
-2
0
-1
-1
Cl2 + 2e- → 2Cl-
8 - 18
 The previous equation shows the
conservation of mass and charge.
In the following example, sodium is said to be
oxidized.
 Na → Na+ + e For a species to be oxidized, its
oxidation number is increased to a
larger number.
 Another way of saying the same thing is
that the oxidation number of sodium
becomes more positive or less negative.
8 - 19
 Because sodium is oxidized, it is called a
reducing agent.
 When sodium is oxidized, it is bringing
about the reduction of another species.
 Hence, sodium is called an reducing
agent.
 The balanced equation for sodium is
given by:
0
1
-1
Na → Na+ + e8 - 20
 The previous equation shows the
conservation of mass and charge.
8 - 21
Summary Of Double Agents
In summary,
 An oxidizing agent is reduced.
 A reducing agent is oxidized.
Atoms in Group I have a weak attraction for
their valence and are easily oxidized.
Atoms in Group VII have a strong attraction
for their valence electrons and are easily
reduced.
8 - 22
Single Replacement Reactions
A single replacement reaction is one in which
a free element becomes an ion and an ion in
solution becomes a neutral atom.
For example,
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
Zn becomes Zn2+ and H+ becomes H2.
8 - 23
Balancing the net ionic equation could be
done by inspection but we will use the
ion-electron method.
Zn + H2SO4 → ZnSO4 + H2
Zn + 2H+ + SO42- → Zn2+ + SO42- + H2
Zn + 2H+ + SO42- → Zn2+ + SO42- + H2
Zn → Zn2+ + 2e2H+ + 2e- → H2
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
8 - 24
Alkali metals (Group I) and alkaline
earth metals (Group II) are very active metals.
 These metals are easily oxidized and are
good reducing agents.
 Such active metals react with water
(H2O(l)) or steam (H2O(g)) to replace the
hydrogen from the water forming a
metallic hydroxide and hydrogen gas.
 Sometimes water will be written as HOH
to show that the first hydrogen is the one
that gets replaced.
8 - 25
Use the ion-electron method to write the
balanced net ionic equation for the reaction
between sodium and water.
Na + HOH → NaOH + H2
Na + HOH → Na+ + OH- + H2
Na → Na+ + eH+ + e- → H2
8 - 26
2Na → 2Na+ + 2e2H+ + 2e- → H2
2Na(s) + 2H+(aq) → 2Na+(aq) + H2(g)
Very unreactive metals such as Ag, Pt, and
Au, are not found in nature chemically
combined.
8 - 27
When Is There A Reaction?
One example of a single replacement reaction
is when one metal tries replacing another.
For example, what are the products of the
following reaction?
 Cu + AgNO3 →
To determine the products, one can use the
values from Standard Reduction Potential
Table.
8 - 28
 From the title, one expects to find the
half-reactions to be written as
reductions.
From the table, we find that:
Ag+ + e- → Ag
E°red = 0.80 V
Cu2+ + 2e- → Cu
E°red = 0.34 V
These E° values are measured with
respect to the SHE (Standard Hydrogen
Electrode).
8 - 29
Hydrogen is arbitrarily assigned an
E°red = 0 V (where V is volts).
This is covered in much more detail in
Electrochemistry.
For now, TP (Think Positive)!
Because E°red(Ag) > E°red(Cu) or E°red(Ag)
is more positive than E°red(Cu), the single
replacement takes place.
8 - 30
The more formal approach gives the same
result.
If the reaction was to take place, it would be:
Cu + AgNO3 → Ag + Cu(NO3)2
(unbalanced)
The two half-reactions would be:
Cu → Cu2+ + 2e- E°ox = -0.34 V
2Ag+ + 2e- → 2Ag
E°red = 0.80 V
Cu(s) + 2Ag+(aq) → Ag(s) + Cu2+(aq) E° = 0.46 V
8 - 31
Important fall out from this problem:
 Notice that the half-reaction for copper
is reversed and the sign changes to
negative.
This is necessary because Cu is required
to be oxidized in the reaction, not
reduced.
You must multiply the Ag+ half-reaction
by 2 so the electrons cancel out in the
net ionic equation.
8 - 32
A very important note is that you do not
multiply the E° by 2 because voltage is
an intensive property.
For the single replacement to occur,
E° must be positive!
8 - 33
When Is There A No Reaction?
Not all single replacements occur!
For example, what are the products of the
following reaction?
 Ag + Cu(NO3)2 →
From the table, we find that:
Ag+ + e- → Ag
E°red = 0.80 V
Cu2+ + 2e- → Cu
E°red = 0.34 V
8 - 34
Because E°red(Ag) > E°red(Cu) or E°red(Ag)
is more positive than E°red(Cu), the single
replacement does not take place!
This reaction is significantly different.
Ag → Ag+ + e-
Eox = -0.80 V
Cu → Cu2+ + 2e-
Eox = -0.34 V
TP (Think Positive) still works!
The oxidation potential of copper is
more positive than the silver.
8 - 35
But the problem is that copper needs to
be reduced in this problem!
The more formal approach gives the same
result.
If the reaction was to take place, it would be:
Ag + Cu(NO3)2 → Cu + AgNO3
The two half-reactions would be:
8 - 36
Cu2+ + 2e- → Cu
E°red = 0.34 V
2Ag → 2Ag+ + 2e- E°ox = -0.80 V
E° = -0.46 V
Because E° < 0, the reaction does not occur!
8 - 37
When Is There A Reaction?
Another example of a single replacement
reaction is when one halogen tries replacing
another.
For example, what are the products of the
following reaction?
 Cl2 + NaBr →
To determine the products, one can use the
values from Standard Reduction Potential
Table.
8 - 38
 From the table, we find that:
Cl2 + 2e- → 2Cl-
E°red = 1.36 V
Br2 + 2e- → 2Br
E°red = 1.07 V
Because E°red(Cl2) > E°red(Br2) or
E°red(Cl2) is more positive than E°red(Br2),
the single replacement takes place.
8 - 39
The more formal approach gives the same
result.
If the reaction was to take place, it would be:
Cl2 + NaBr → Br2 + NaCl
(unbalanced)
The two half-reactions would be:
Cl2 + 2e- → 2Cl2Br- → Br2 + 2e-
E°red = 1.36 V
E°red = -1.07 V
Cl2(g) + 2Br-(aq) → Br2(l) + 2Cl-(aq) E° = 0.46 V
8 - 40
When Is There A No Reaction?
For example, what are the products of the
following reaction?
 I2 + LiF →
From the table, we find that:
F2 + 2e- → 2F-
E°red = 2.87 V
I2 + 2e- → 2I-
E°red = 0.53 V
8 - 41
Because E°red(F2) > E°red(I2) or E°red(F2)
is more positive than E°red(I2), the single
replacement does not take place!
This reaction is significantly different.
F2 + 2e- → 2F-
E°red = 2.87 V
I2 + 2e- → 2I-
E°red = 0.53 V
TP (Think Positive) still works!
The oxidation potential of fluorine is
more positive than the iodine.
8 - 42
But the problem is that fluorine needs to
be oxidized in this problem!
The more formal approach gives the same
result.
If the reaction was to take place, it would be:
I2(s) + 2LiF(aq) → F2(g) + 2LiI(aq)
The two half-reactions would be:
8 - 43
2F- → F2 + 2eI2 + 2e- → 2I-
E°ox = -2.87 V
E°red = 0.53 V
E° = -2.34 V
Because E° < 0, the reaction does not occur!
8 - 44
Summarizing SRP Rxns For Halogens
The Standard Reduction Potentials for
nonmetals measures the ability of a nonmetal
to accept valence electrons.
The more reactive nonmetals have a greater
tendency to gain an electron or be reduced.
 Fluorine with its highest
electronegativity (4.0) is the most
reactive halogen and is the strongest
oxidizing agent.
8 - 45
 Iodine has the smallest tendency to gain
electrons because it has the largest
radius and smaller electronegativity.
8 - 46
Balance Rxns Using Oxidation Numbers
When the equations become somewhat more
difficult to balance by inspection, use the
following rules.
 Assign oxidation numbers to all
elements.
 Identify the element that is reduced and
write its half-reaction.
 Identify the element that is oxidized and
write its half-reaction.
8 - 47
 Balance and add the half-reactions so
the electrons cancel.
 Use the resulting coefficients as
coefficients in the original equation.
 Balance the remaining atoms by
inspection.
8 - 48
More Interesting Equations To Balance
Balance the following reaction using the
oxidation-number method.
HNO3 + H2S → NO + S + H2O
1 5 -6
1 -2
1 5 -2
1 -2
2 -2
2 -2
0
1 -2
HNO3 + H2S → NO + S + H2O
8 - 49
-2
0
S → S + 2e5
2
N + 3e- → N
-2
5
0
3S → 3S + 6e2
2N + 6e- → 2N
-2
5
0
2
3S + 2N → 3S + 2N
8 - 50
The sum of the reactant oxidation numbers is
+4.
The sum of the product oxidation numbers is
+4.
Using the coefficients of the net equation in
the original equation gives:
2HNO3 + 3H2S → 2NO + 3S + H2O
2HNO3(aq) + 3H2S(aq) → 2NO(g) + 3S(s) +
4H2O(l)
8 - 51
Another Interesting Problem
Potassium dichromate and hydrochloric acid
react to produce potassium chloride,
chromium(III) chloride, water and chlorine.
Write the balanced equation for this reaction.
2 12 -14
1 -1
1 -1
3 -3
1 -2
0
1
1 -1
1 -1
3 -1
1 -2
0
6 -2
K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2
8 - 52
Cl is being oxidized. Cr is being reduced.
Cl- → Cl2
Cr2 → Cr3+
2Cl- → Cl2
Cr2 → 2Cr3+
2Cl- → Cl2 + 2e-
Cr2 + 6e- → 2Cr3+
6Cl- → 3Cl2 + 6eCr2 + 6e- → 2Cr3+
6Cl- + Cr2 → 3Cl2 + 2Cr3+
8 - 53
Using the coefficients from the net equation
gives,
K2Cr2O7 + 6HCl → KCl + 2CrCl3 + H2O + 3Cl2
and balancing by inspection gives:
K2Cr2O7(aq) + 14HCl(aq) → 2KCl(aq) +
2CrCl3(aq) + 7H2O(l) + 3Cl2(g)
8 - 54
Ion-Electron Method
The ion-electron method is a means of
balancing the more difficult redox reactions
that take place in an acidic or basic solutions.
The steps for this method are as follows:
 Assign oxidation numbers to all of the
atoms in the equation.
 Split the given equation into two
half-reactions (one for oxidation and one
for reduction).
8 - 55
 Balance atoms other than hydrogen and
oxygen.
 Balance the number of electrons
involved in the transfer.
 Balance oxygen by adding water to the
side that needs oxygen.
 Balance hydrogen by adding H+ to the
side that needs hydrogen.
 Add the two half-reactions canceling
electrons and eligible atoms.
8 - 56
Redox in an Acidic Solution
Balance the following reaction in an acidified
(an acid included as a reactant) solution.
MnO4- + Fe2+ → Mn2+ + Fe3+
7 -8
-2
MnO4- → Mn2+
MnO4- + 5e- → Mn2+
8 - 57
MnO4- + 5e- → Mn2+ + 4H2O
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Fe2+ → Fe3+
Fe2+ → Fe3+ + e5Fe2+ → 5Fe3+ + 5eMnO4- + 8H+ + 5e- → Mn2+ + 4H2O
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) →
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
8 - 58
Redox in a Basic Solution
Balance the following reaction in a basic
solution.
I- + MnO4- → IO3- + MnO2
-1
7 -8
5 -6
4 -4
-1
-2
-2
-2
I- + MnO4- → IO3- + MnO2
8 - 59
I- → IO3I- → IO3- + 6eI- + 3H2O → IO3- + 6eI- + 3H2O → IO3- + 6e- + 6H+
MnO4- → MnO2
MnO4- + 3e- → MnO2
MnO4- + 3e- → MnO2 + 2H2O
MnO4- + 4H+ + 3e- → MnO2 + 2H2O
8 - 60
I- + 3H2O → IO3- + 6e- + 6H+
2
1
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
I- + 2MnO4- + 2H+ → IO3- + 2MnO2 + H2O
When you balance a reaction in a basic solution,
at this point you must add nOH- to both sides of
the equation to get rid of the H+. n represents the
coefficient of the H+.
When you add 2OH- to 2H+, you get 2H2O.
8 - 61
.
I- + 2MnO4- + 2H+ → IO3- + 2MnO2 + H2O
2OH-
2OH-
1
I- + 2MnO4- + 2H2O → IO3- + 2MnO2 + H2O + 2OHI-(aq) + 2MnO4-(aq) + H2O(l) →
IO3-(aq) + 2MnO2(s) + 2OH-(aq)
8 - 62