OxidationReduction
Chapter 16
Tro, 2nd ed.
1.1
1
Oxidation Number
The oxidation number (oxidation state) of an atom represents the
number of electrons lost, gained, or unequally shared by an
atom. (IT IS NOT THE SAME AS CHARGE!!)
Oxidation numbers can be zero, positive or negative.
An oxidation number of zero means the atom has the same
number of electrons assigned to it as there are in the free
neutral atom, WHICH ARE ALWAYS ZERO.
A positive oxidation number means the atom has fewer electrons
assigned to it than in the neutral atom.
A negative oxidation number means the atom has more electrons
assigned to it than in the neutral atom.
2
Rules for Assigning Oxidation Numbers
(similar to page 563)
Apply rules from top down; first one you reach has priority.
1. All elements in their free state have an oxidation number of zero.
(e.g., Na, Cu, H2, O2, Cl2, N2).
2. a. H is +1, except in metal hydrides, where it is -1 (e.g., NaH,
CaH2).
b. F is always -1, except when it is F2.
3. O is -2, except in peroxides, where it is -1, and in OF2, where it is
+2.
4. The metallic element in an ionic compound has a positive
oxidation number. (Group IA is always +1 and Group IIA is
always +2.) The nonmetallic element in an ionic compound has a
negative oxidation number. (Halogens -1, Group VIA -2, Group
VA -3.)
5. In covalent compounds the negative oxidation number is assigned
to the most electronegative atom
6. The algebraic sum of the oxidation numbers of the elements in a
compound is zero.
7. The algebraic sum of the oxidation numbers of the elements in a
polyatomic ion is equal to the charge of the ion.
3
Oxidation Number
For the seven diatomic elements which exist
as covalent molecules: each atom is
assigned an oxidation number of 0
because the bonding pair of electrons is
shared equally between two like atoms of
equal electronegativity.
H-H and Cl-Cl: both atoms in each molecule
have an oxidation number of 0.
Other elements: Cu, Ne, C, et., will all be
assigned a 0.
4
Oxidation Number
The oxidation number of an atom that
has gained or lost electrons to form a
monatomic ion is the same as the
positive or negative charge of the
ion.
NaCl
Na+ has an oxidation number of +1.
Cl- has an oxidation number of -1.
5
Oxidation Number
Given a compound like HCl, use the rules to assign
oxidation number. Rule 2 says H is +1 unless
with a metal: therefore H is +1.
Rule 6 says the sum of the oxidation numbers must
be zero, therefore Cl will be assigned a -1.
In binary compounds, halogens are often a -1, like
their monatomic ions would be. However, if a
halogen is part of a polyatomic ion, then its ON
must be determined.
HCl
+1,-1
6
Many elements have multiple
oxidation numbers
N oxidation
number
N2
N2 O
NO
N2O3
NO2
N2O5
NO-3
0
+1
+2
+3
+4
+5
+5
7
Notice that H and
group I metals have
O.N. of +1
Notice that group II
metals have O.N. of +2
Halogens are
typically assigned
O.N. of -1
Oxygen is usually -2
8
Rules for Determining the Oxidation Number
of an Element Within a Compound
Step 1 Write the oxidation number of each known
atom below the atom in the formula.
Step 2 Multiply each oxidation number by the
number of atoms of that element in the
compound.
Step 3 Write an expression indicating the sum of
all the oxidation numbers in the
compound. Remember: The sum of the
oxidation numbers in a compound must
equal zero.
9
Determine the oxidation number for sulfur in sulfuric
acid.
H2SO4
Step 1
+1
Step 2 2(+1) = +2
Step 3
-2
4(-2) = -8
+2 + S + (-8) = 0
Step 4 S = +6 (oxidation number for sulfur)
Write an
Multiply
expression
oxidation
indicating
numberthe
by sum
the
number
ofknown
all the
of
theeach
oxidation
number
of
each
atoms
oxidation
of that
numbers
element
in the
ininthe
compound.
compound.
atom below
the
atom
the
formula.
10
Determine the oxidation number for carbon in the
oxalate ion.
Step 1
Step 2
Step 3
24
C2O
-2
4(-2) = -8
2C + (-8) = -2 (the charge on the ion)
Multiply
oxidation
numberthe
by sum
the
number
of
theeach
oxidation
number
of
each
Write an
expression
indicating
ofknown
all the
atoms
of that
element
ininthe
compound.
atom below
the
atom
the
formula.
oxidation
numbers
in the
compound.
11
ASSIGNING OXIDATION
NUMBERS
PRACTICE: FeCl2, MnO2, F2, Al, KI,
IF5, H2O, OF2, SO2, SO42-, OH-, HCl,
ClO-, ClO4+2 and -1, +4 and -2, zero, zero, +1
and -1, +5 and -1, +1 and -2, +2
and -1, +4 and -2, +6 and -2, -2
and +1, +1 and -1, +1 and -2, +7
and -2
12
Oxidation-Reduction
Oxidation-reduction (redox) is a
chemical process in which the oxidation
number of an element is changed.
Redox may involve the complete transfer of
electrons to form ionic bonds or a partial
transfer of electrons to form covalent
bonds.
Loss of electrons = oxidation
Gain of electrons = reduction
LEO says GER
13
Oxidation-Reduction
Oxidation occurs when the oxidation
number of an element increases as a
result of losing electrons.
Reduction occurs when the oxidation
number of an element decreases as a
result of gaining electrons.
In a redox reaction oxidation and reduction
occur simultaneously, one cannot occur in
the absence of the other.
14
Oxidation-Reduction
Oxidizing agent The substance that causes
an increase in the oxidation state of
another substance.
The oxidizing agent is reduced in a redox
reaction.
Reducing agent The substance that causes
a decrease in the oxidation state of
another substance.
The reducing agent is oxidized in a redox
reaction.
15
Oxidation-Reduction
Electron-transfer reactions:
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
Zn: from atom to ion; gives up e-s to H
H: from ion to atom; accepts e-s from Zn
by electron transfer!
Zn atom becomes Zn2+ ion and is oxidized.
H+ ion becomes H2 atoms = reduced.
Zn is the reducing agent and H+ is the oxidizing
agent.
You must always have both oxidation and reduction
in the reaction.
16
SCALE OF OXIDATION NUMBERS AND MEANING
OF OXIDATION AND REDUCTION
17
17.1
Redox Reactions
Practice: find oxid # for all elements in these
reactions to see if electron-transfer occurs.
1. 2 Mg + O2  2 MgO
2. 2 KClO3  2 KCl + 3 O2
3. Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
4. HCl + NaOH  H2O + NaCl
5. CH4 + O2  CO2 + 2 H2O
1. YES
2. YES
3. YES (Single Replacement
reactions are ALWAYS redox.)
4. NO
5. YES
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Balancing Oxidation Reduction Equations:
The Half-Reaction Method
(acidic solution)
Step 1: Determine the oxidation number of every atom/ion in the reaction.
Figure out which is being oxidized and which reduced. Connect them with
arrows and determine the number of electrons lost or gained per atom.
Then write the two half-reactions that contain the elements being oxidized
and reduced.
Step 2: Balance the elements other than hydrogen and oxygen. (This is the
step that students forget!)
Step 3: Balance oxygen by adding H2O, then balance H by adding H+.
Step 4: Add electrons (e-) to each half-reaction to bring them into electrical
balance.
Step 5: Since the loss and gain of electrons must be equal, multiply each
half-reaction by the appropriate number (using least common
denominator) to make the number of electrons the same in each halfreaction.
Step 6: Add the two half-reactions together, canceling electrons and any
other identical substances (like H+ and H2O) that appear on opposite sides
of the equation. In ionic redox equations both the numbers of atoms and
the electrical charge on both sides of the equation must be the same.
19
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 1: MnO4- + S2-  Mn2+ + S
+7, -2
-2
+2
Mn reduced, gains 5 e-/Mn;
0
S oxidized, loses 2 e-s/S
S  S
2-
o
Oxidation halfreaction
MnO  Mn
4
2+
Reduction halfreaction
Step 2 Balance elements other than oxygen and
hydrogen. (Step 2 is unnecessary, since
these elements are already balanced).
20
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 3 The oxidation half-reaction requires neither
O nor H, but the reduction equation needs
4 H2O on the right and 8 H+ on the left.
S  S
2-
o
4H22O
O
8H
8H ++ MnO  Mn ++ 4H
++
4
8H+ balance the 8 hydrogens of
4 water molecules. MnO4.
2+
4 water molecules are necessary
to balance the 4 oxygens in
MnO4-4.
21
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 4 Balance each half-reaction electrically with electrons.
You have already determined the number of
electrons for each.
balanced oxidation half-reaction
-2o
S  S +
 2e
2e
net charge = -2 on each side
balanced reduction half-reaction
+
2+
4
net charge = +2 on each side
5e
5e + 8H + MnO  Mn
-
+ 4H2O
22
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 5 Equalize loss and gain of electrons. In this
case, multiply the oxidation equation by 5
and the reduction equation by 2.
5S
S  S
5S 2e
10e
2-
oo
-
-
10e
5e++16H
8H + 2MnO
MnO Mn
2Mn+ 4+H8H
2O2O
- -
+
- 44
2+ 2+
23
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 6 Add the two half-reactions together, canceling
the 10e- from each side, to obtain the
balanced equation.
5S  5S  10e
2-
o
-
10e + 16H + 2MnO  2Mn
-
+
4
2+
16H + 2MnO + 5S  2Mn
+
4
2-
2+
+ 8H2O
o
+ 5S + 8H2O
The charge on both sides of the balanced equation is +4
24
Balancing Redox Equations:
The Half-Reaction Method
(basic solution)
For reactions in alkaline (basic) solutions,
first balance as though the reactions were
in an acid solution, using Steps 1-6 as
before.
Add OH- to each side equal to the number of
H+ on one side. Where both OH- and H+
appear on same side of reaction, combine
them to make H2O.
Rewrite the equation, canceling equal
numbers of water molecules that appear
on opposite side of the equation.
25
PRACTICE WITH BASIC
SOLUTION
___Cl2 + ___IO3-  ___Cl- + ___IO4After steps 1-6:
Cl2 + IO3- + H2O  2 Cl- + IO4- + 2 H+
Add 2 OH- to both sides:
Cl2 + IO3- + H2O + 2 OH-  2 Cl- + IO4- + 2 H+ + 2 OH-
Combine H+ and OH- to make water:
Cl2 + IO3- + H2O + 2 OH-  2 Cl- + IO4- + 2 H2O
Cancel one water on each side for finished
equation:
Cl2 + IO3- + 2 OH-  2 Cl- + IO4- + H2O
26
AND THAT’S ALL WE NEED FROM CHP
16!
There is a lab on redox and practice
handouts and homework! Do it!
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