Transcript 9.1 REDOX Introduction to Oxidation and Reduction

9.1 Introduction to Oxidation and
Reduction
Assessment Statements
9.1.1 Define oxidation and reduction (electron loss and
gain)
9.1.2 Deduce the oxidation number of an element in a
compound.
9.1.3 State the names of compounds using oxidation
numbers.
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers,
History
 Modern chemistry comes from John Dalton’s atomic
theory proposed in 1805
 Chemistry really begin when humans used fire
 Roman historian Pliny (27-79) said glass was first
discovered by accident by Phoenician sailors while
using fire to cook their food around 5000 BCE
 Saltpeter, KNO3, may have been used to balance
their cooking pots and the combination formed
glass
 Glass beads were found in Egypt and Mesopotamia
(Iraq) around 2500 BCE
 Heat made by burning charcoal was used to alloy
copper and tin during the Bronze Age, 4500 BCE
to 1200 BCE
 Copper & tin occur naturally, but this was the 1st
evidence for the use of heat to reduce metallic
ores, like iron oxides
 Evidence of iron oxides are found in India dating
1800 BCE, start of Iron Age
 Combustion wasn’t fully understood until after
oxygen was discovered by Joseph Priestley (17331804)
 Quickly realized that oxygen forms oxides so the
word oxidation was created to describe the
addition of oxygen
 When oxygen is removed “reduction” is used
 Now oxidation and reduction refer to transfer of
electrons
Assessment Statements
9.1.1 Define oxidation and reduction (electron loss
and gain)
REDOX
Zn
(S)
+ I2 (aq)  Zn2+ (aq) + 2I-(aq)
Lets look at this reaction:
Zinc atoms are becoming positively
charged.
Iodine atoms are becoming negatively
charged.
Half Reactions
Zn
(S)
 Zn2+ (aq) + 2e-
Zinc is losing electrons to become positively
charged  OXIDATION
I2 (aq) + 2e-  2I-(aq)
Iodine is gaining electrons to become
negatively charged  REDUCTION
LEO
LEO says
GER!
LEO
says
GER!
Loss of Electrons = Oxidation
Gain of Electrons = Reduction
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)
OIL RIG
Assessment Statements
9.1.2 Deduce the oxidation number of an element in
a compound.
Oxidation Numbers
 Oxidation/Reduction (ReDox)-- Whenever a substance
loses electrons and another substance gains electrons
 Oxidation Numbers--system used to keep track of
electron transfers
 Sometimes refered to as ‘ELECTRON ACCOUNTING’
 NOTE: Oxidation numbers are given a ‘+’ or ‘–’ THEN an
number (e.g. +7).
(do not say 7+, IB does not like this at all)
Oxidation Numbers Rules
Oxidation numbers always refer to
single atoms
The oxidation number of an
uncombined element is always 0
O2, H2, Ne
Zn
The oxidation number of Hydrogen
is usually +1 Hydrides are an
exception They are -1
HCl, H2SO4
The oxidation number of Oxygen is
usually -2 Peroxides are an exception
They are –1 In OF2 oxygen is a +2
H2O, NO2, et
Oxidation numbers of monatomic
ions follow the charge of the ion
O2-, Zn2+
The sum of oxidation numbers is
zero for a neutral compound. It is
the charge on a polyatomic ion
LiMnO4
SO42-
Rules for determining
Oxidation Number
1. Atoms of solid metals are neutral (i.e. 0)
(no charge has they have the same number of
protons and neutrons)

E.g. Fe (s) , Mg (s), Sn (s)
2. Natural molecules of gases are Neutral (i.e. 0)
E.g. H2, O2 , Cl2 , l2
GENERALLY if an element is in its natural
elemental state its Oxidation number is 0
3. Most elements will always produce the same
oxidation state (+, -)
This is true for
e.g. flourine will always be -1
most element in
period 1-7
But some can change………..
e.g. Hydrogen is +1 (H+1) except in hydride ion where
H is -1 (H-1)
Oxygen is -2 (O-2) but +1 in Hydrogen peroxide
(O+1)
4. The halogen group are always -1 except when
bonded to oxygen or a halogen higher in the group.
5. The total charge on a compound is 0. The number
of positives cancel out the negatives.
e.g. MgO (Magnesium is +2 and Oxygen is -2)
Practice Assigning Oxidation
Numbers
NO2
N2O5
HClO3
HNO3
Ca(NO3)2
KMnO4
Practice Assigning Oxidation
Numbers
Fe(OH)3
K2Cr2O7
CO32CNK3Fe(CN)6
Practice Assigning Oxidation
Numbers
NO2
N= +4, O = -2
N2O5
N = +5, O = -2
HClO3
H=+1, Cl=+5, O = -2
HNO3
H=+1, N = +5, O = -2
Ca(NO3)2
Ca=+2, N =+5, O= -2
KMnO4
K=+1, Mn=+7, O= -2
Practice Assigning Oxidation
Numbers
Fe(OH)3
Fe =+3, O=-2, H=+1
K2Cr2O7
K=+1, Cr=+6, O=-2
CO32-
C=+4, O =-2
CN-
C=+4, N=-5
K3Fe(CN)6
K=+1, Fe=+3, C=+4, N=-5
Assessment Statement
 9.1.3 State the names of compounds using
oxidation numbers.
Transition Elements
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
+1
+3
+2
+2
+2
+2
+2
+2
+2
+2
+3
+3
+3
+3
+3
+3
+3
+3
+4
+4
+4
+4
+4
+4
+4
+5
+5
+5
+5
+5
+6
+6
+6
+2
+7
These elements can have multiple oxidation states.
So when we talk to someone about, say, ‘Iron Oxide., we
also have to tell them is the iron’s oxidation number +2, +3
etc………….
Different Oxidation Numbers
The oxidation state of many transition metals can
be different
e.g. Fe can be +2 or +3
FeCl2 (ferrous Chloride, iron (II) Chloride)
FeCl3 (ferric Chloride, iron (III) Chloride)
These compounds have very different chemical and
physical properties so it is important that we
highlight which one we are talking about.
Activity
 Give the oxidation number of the stated element in
each of the following species:
 Iron in Fe
2+
 Silicon in Si02
 Sulphur in SCl4
 Nitrogen in N2O
 Carbon in CH4OH
 Vandium in VO
2+
 Bromine in BrO Iodine in IF6-
 Cromium in Cr2O7
2-
Assessment Statements
9.1.4 Deduce whether an element undergoes
oxidation or reduction in reactions using oxidation
numbers,
Using Oxidation Numbers
 An increase in the oxidation number indicates
that an atom has lost electrons and therefore
oxidized.
 A decrease in the oxidation number indicates
that an atom has gained electrons and therefore
reduced
 Example
Zn
+
CuSO4  ZnSO4 + Cu
0
+2+6-2
+2+6-2
0
Zn: 0  + 2 
Oxidized
Cu: +2  0
 Reduced
Exercise
For each of the following reactions find the
element oxidized and the element reduced
Cl2
0
+
KBr
+1-1

KCl +
+1-1
Br increases from –1 to 0 -- oxidized
Cl decreases from 0 to –1 -- Reduced
K remains unchanged at +1
Br2
0
Exercise
For each of the following reactions find the element
oxidized and the element reduced
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
0
+1+5-2
+2 +5-2
+4 –2 +1-2
Cu increases from 0 to +2. It is oxidized
Only part of the N in nitric acid changes from +5
to +4. It is reduced
The nitrogen that ends up in copper nitrate
remains unchanged
Exercise
For each of the following reactions find the
element oxidized and the element reduced
HNO3 +
I2
 HIO3 +
NO2
1 +5 -2
0
+1+5-2
+4-2

N is reduced from +5 to +4. It is reduced

I is increased from 0 to +5 It is oxidized

The hydrogen and oxygen remain unchanged.
Exercise
For each of the following reactions find the
element oxidized and the element reduced
Cl2 + KBr

Cu + HNO3 
HNO3 + I2 
KCl + Br2
Cu(NO3)2 + NO2 + H2O
HIO3
+
NO2
Activity
 Write the half equations for each.
 Work out which is being oxidised or reduced.
2 H2 (g) + O2 (g)  2 H2O (l)
Ca (S) + H2O(l)  CaO (S) + H2 (g)
Mg (S) + CuO (S) → Mg0 (S) + Cu (S)
Zn (S) + Cu S04 (aq) → ZnS04 (aq) + Cu (S)
Assignment
 Oxidation Numbers/Naming Review packet
Balancing Redox Reactions
 Many Redox rxns are complex and difficult to
balance .
 A systematic approach to balancing these reaction
is required.
Balancing Redox Equations 1
1.
2.
Divide the equation into 2
half reactions—one for
oxidation, one for
reduction.
Balance each half
reaction
1.
Balance elements
other than H and O
2.
Balance O by adding
H2O as needed
3.
Balance H by adding
H+ as needed.(acidic
solution)
4.
Balance charge by
adding e- as needed.
3.
Multiply half reactions by
integers so that the # of
e- lost in one reaction = #
of e- gained in the other
reaction.
4.
Add the two half
reactions. Simplify by
canceling species that
appear on both sides of
the arrow.
MnO41- (aq) + C2O42- (aq)  Mn2+(aq) + CO2 (aq)
MnO4 + 5e-  Mn+2
C2O4  2CO2 + 2e-
(Mn7+  Mn2+ )
(C+3  2C+4 )
8H+ + MnO4 + 5e-  Mn+2 + 4H2O
C2O4  2CO2 + 2e-
16H+ + 2MnO4 + 10e-  2Mn+2 + 8H2O
5C2O4  10CO2 + 10e16H+ + 2MnO4 + 5C2O4  2Mn+2 + 8H2O + 10CO2
5. Check your work. Make sure that both the
atoms and charges balance
Balancing Redox Equations 1
1.
2.
Divide the equation into 2
half reactions—one for
oxidation, one for
reduction.
Cr2O72- (aq) + Cl1- (aq)  Cr 3+ (aq) + Cl2 (g)
Balance each half
reaction
1.
Balance elements
other than H and O
2.
Balance O by adding
H2O as needed
3.
Balance H by adding
H+ as needed.(acidic
solution)
4.
Balance charge by
adding e- as needed.
3.
Multiply half reactions by
integers so that the # of
e- lost in one reaction = #
of e- gained in the other
reaction.
4.
Add the two half
reactions. Simplify by
canceling species that
appear on both sides of
the arrow.
5. Check your work. Make sure that both the
atoms and charges balance
Balancing Redox Equations 1
1.
2.
Divide the equation into 2
half reactions—one for
oxidation, one for
reduction.
Cu (s) + NO31- (aq)  Cu 2+ (aq) + NO2 (aq)
Balance each half
reaction
1.
Balance elements
other than H and O
2.
Balance O by adding
H2O as needed
3.
Balance H by adding
H+ as needed.(acidic
solution)
4.
Balance charge by
adding e- as needed.
3.
Multiply half reactions by
integers so that the # of
e- lost in one reaction = #
of e- gained in the other
reaction.
4.
Add the two half
reactions. Simplify by
canceling species that
appear on both sides of
the arrow.
5. Check your work. Make sure that both the
atoms and charges balance
Balancing Redox Equations 1
1.
2.
Divide the equation into 2
half reactions—one for
Mn 2+ (aq) + NaBiO3 (s)
oxidation, one for
 Bi 3+ (aq) + MnO4 1- (aq) + Na
reduction.
1+
(aq)
Balance each half
reaction
1.
Balance elements
other than H and O
2.
Balance O by adding
H2O as needed
3.
Balance H by adding
H+ as needed.(acidic
solution)
4.
Balance charge by
adding e- as needed.
3.
Multiply half reactions by
integers so that the # of
e- lost in one reaction = #
of e- gained in the other
reaction.
4.
Add the two half
reactions. Simplify by
canceling species that
appear on both sides of
the arrow.
5. Check your work. Make sure that both the
atoms and charges balance
Plenary: TOK
Are Oxidation Numbers ‘Real’?
Lets use the analogy of money.
Different parts of the world have a different belief of
money.
Capitalists  the value of money is a measure of work.
Communist  everyone is equal and money is distributed
equally.
Monarchist  it is a measure of nations worth and
means little to the population.
BUT is money REAL or just a value depending on a
collective belief??????
TOK
 CHARLES SANDERS PIERCE (philosopher) came
up with the concept of ‘PRAGMATIC TRUTH’ in
which he states that it is more convenient for a
society to believe (truth is more an attribute of a
society rather than an attribute of the physical
world).
TOK: Are Oxidation Numbers
Useful?
 They able to to give us some answers of a




chemical change.
But is it just a matter of ‘electron accounting’
that helps us balance equations.
It tells us what is more energetically favourable.
They are useful………but so many exceptions.
BUT is it actually what is happening………..