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1
Oxidation Number
2
The oxidation number (oxidation state) of an
atom represents the number of electrons lost,
gained, or unequally shared by an atom.
3
Oxidation numbers can be zero, positive,
negative or fractional.
4
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.
5
A positive oxidation number means the
atom has fewer electrons assigned to it
than in the neutral atom.
6
A negative oxidation number means the atom
has more electrons assigned to it than in the
neutral atom.
7
The oxidation number of an atom that
has gained or lost electrons to form an
ion is the same as the positive or
negative charge of the ion.
NaCl
The charge
oxidation
Sodium
has lost
The
on
number
sodium
is +1.
an electron
. of
sodium is +1.
The
Chlorine
oxidation
has
The charge on
number
gained an
of
chlorine is –1.
chlorine
electron.
is -1.
8
In covalently bonded substances,
oxidation numbers are assigned by an
arbitrary system based on relative
electronegativities.
9
For symmetrical 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.
Oxidation
Electronegativity
Number
2.1
0
Oxidation
Electronegativity
Number
2.1
0
10
When the covalent bond is between two
unlike atoms, the bonding electrons are
shared unequally because the more
electronegative element has a greater
attraction for them.
Oxidation
Number
Electronegativity
+1
2.1
there is a partial
after assignment
hydrogen
unequal
shared
electron
pair of
transfer of an
has one lesssharing
electron
than
electrons
electron to chlorine
neutral chlorine
Oxidation
Number
Electronegativity
-1
3.0
after
assignment
chlorine
both
shared
electrons
one to
more
electron
arehas
assigned
chlorine
than neutral chlorine 11
Many elements have multiple
oxidation numbers
N oxidation
number
N2
N2O
NO
N2O3
NO2
N2O5
NO-3
0
+1
+2
+3
+4
+5
+5
12
13
14
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.
15
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.
16
Determine the oxidation number for manganese in
potassium permanganate.
KMnO4
Step 1
+1
Step 2 1(+1) = +2
Step 3
-2
4(-2) = -8
+1 + Mn + (-8) = 0
Step 4 Mn = +7 (oxidation number for Mn)
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.
17
Determine the oxidation number for nitrogen in the
nitrate ion.
NO
3
Step 1
Step 2
Step 3
-2
3(-2) = -6
1N + (-6) = -1 (the charge on the ion)
Step 4 1N = +5
N = +5 (oxidation number for nitrogen)
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.
18
Determine the oxidation number for carbon in the
oxalate ion.
C 2O
Step 1
Step 2
Step 3
24
-2
4(-2) = -8
2C + (-8) = -2 (the charge on the ion)
Step 4 2C = +6
C = +3 (oxidation number for carbon)
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.
19
Determine the oxidation number for carbon in ethane.
C2 H6
Step 1
+1
Step 2
6(+1) = +6
Step 3
2C + (+6) = 0
Step 4
C = - 3 (oxidation number for carbon)
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.
20
Determine the oxidation number for carbon in propyne.
C3 H4
Step 1
+1
Step 2
4(+1) = +4
Step 3
3C + (+4) = 0
Step 4
3C = - 4 or C = - 4/3
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.
21
Oxidation-Reduction
22
Oxidation-reduction (redox) is a chemical
process in which the oxidation number of an
element is changed.
23
Redox may involve the complete transfer of
electrons to form ionic bonds or a partial
transfer of electrons to form covalent bonds.
24
• Oxidation occurs when the oxidation
number of an element increases as a
result of losing electrons. OIL
• Reduction occurs when the
oxidation number of an element
decreases as a result of gaining
electrons. RIG
• In a redox reaction oxidation and
reduction occur simultaneously.
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Leo the Lion says
GER
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Balancing OxidationReduction Equations
27
The Loop Method
28
Balance the equation
Sn + HNO3 → SnO2+ NO2+H2O
Step 1 Assign oxidation numbers to each element
to identify the elements being oxidized and
those being reduced. Write the oxidation
numbers above or below each element..
0
+1 +5 -2
+4 -2
+4 -2
+1 -2
Sn + HNO3 → SnO2 + NO2 + H2O
oxidation number of tin increases
oxidation number of nitrogen decreases
29
Balance the equation
Sn + HNO3
→ SnO + NO +H O
2
2
2
gain 1e-
0
+5
+4
+4
1 Sn + 4 HNO3 → SnO2 + NO2 + H2O
lost 4 e-
Step 2 Multiply the two equations by the smallest
whole numbers that will make the electrons
lost by oxidation equal to the number of
electrons gained by reduction.
30
Balance the equation
Sn + HNO3
→ SnO + NO +H O
2
2
2
gain 1e-
0
+5
+4
+4
1 Sn + 4 HNO3 → 1 SnO2 + 4 NO2 + 2 H2O
lost 4 e-
Step 3 Balance the remaining elements that are
not oxidized or reduced to give the final
balanced equation.
31
Balance the equation
C3H4 + O2 → CO2 + H2O
gain 2e-2  4e-  4  1
- 4/3
0
+4
-2
1 C3H4 + 4 O2 → 3 CO2 + 2 H2O
lost 16/3 e- 3  16e-  4  4
32
Balance the equation
MnO2  HCl  MnCl2  Cl2  H2O
lost 1e-
+4
2
+2
-1
+2
-1
0
Cl → 1 MnCl22 + 1 Cl2 + 2 H2O
1 MnO2 + 24 HCl
gain 2 e-
33
Balancing Practice
Ca  HCl  CaCl2  H2
ZnS  O2  SO2  ZnO
Cu  HNO3  H2SO4  CuSO4  H2O  NO
MnO2  HCl  MnCl2  Cl2  H2O
Fe  HNO3  Fe(NO3 )2  NO  H2O
34
Multiple Change Balancing
This one’s hard!
CrBr3  NaOH  Cl2  Na 2CrO4  NaBrO4  NaCl  H2O
35
Answer to Multiple Change
Oxidation numbers above the three elements
that change oxidation numbers in RED
+3 -1
0
2 CrBr3  64 NaOH  27 Cl2 
+6
+7
-1
2 Na 2CrO4  6 NaBrO 4  54 NaCl  32 H 2O
36
Activity Series
of Metals
37
38
activity series: A listing of metallic
elements in descending order of
reactivity.
39
Sodium (Na) will displace
any element below it from
one of its compounds.
40
41
increasing activity
Mg(s) + PbS(aq)  MgS(aq) + Pb(s)
K
Ba
Ca
Na
Mg
Al
Zn
Cr
Fe
Ni
Sn
Pb
H2
Cu
Magnesium is above lead in
the activity series.
Magnesium will displace lead
from one of its compounds.
42
increasing activity
Ag(s) + CuCl2(aq)  no reaction
Ba
Na
Mg
Al
Zn
Cr
Fe
Ni
Sn
Pb
H2
Cu
Ag
Hg
Silver is below copper in the
activity series.
Silver will not displace copper
from one of its compounds.
43
Electrolytic and
Voltaic Cells
44
electrolysis The process whereby electrical
energy is used to bring about a chemical
change.
electrolytic cell: An electrolysis apparatus
in which electrical energy from an outside
source is used to produce a chemical
change.
45
cathode The negative electrode.
anode The positive electrode.
46
Electrolysis of
Hydrochloric Acid
47
In an electrolytic cell electrical energy from
the voltage source is used to bring about
nonspontaneous redox reactions.
48
Hydronium ions migrate
to the cathode and are
reduced.
H3O+ + 1e- → Ho + H2O
Ho + Ho → H2
Cathode Reaction
49
Chloride ions migrate
to the anode and are
oxidized.
Cl-→ Clo + eClo + Clo→ Cl2
Anode Reaction
50
17.3
2HCl(aq)
electrolysis
H2(g) + Cl2(g)
The hydrogen and chlorine produced when
HCl is electrolyzed have more potential
energy than was present in the hydrochloric
acid before electrolysis.
51
The Zinc-Copper Voltaic Cell
52
voltaic cell: A cell that produces electrical
energy from a spontaneous chemical reaction.
(Also known as a galvanic cell).
53
When a piece of zinc is put in a copper(II)
sulfate solution, the zinc quickly becomes
coated with metallic copper. This occurs
because zinc is above copper in the
activity series.
54
increasing activity
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
K
Ba
Ca
Na
Mg
Al
Zn
Cr
Fe
Ni
Sn
Pb
H2
Cu
Zinc is above copper in the
activity series.
Zinc will displace copper from
one of its compounds.
55
If this reaction is carried out in a voltaic
cell, an electric current is produced.
56
57
loss of
electrons
anode
Zno(s) → Zn2+(aq) + 2e-
oxidation
gain of
electrons
cathode
Cu2+(aq) + 2e- → Cuo(s)
reduction
Net ionic reaction
Zno(s) + Cu2+(aq) → Zn2+(aq) + Cuo(s)
Overall equation
Zno(s) + CuSO4(aq) → ZnSO4(aq) + Cuo(s)
58
LeClanche Cell
• The LeClanche Cell was
described by Georges
LeClanche (1839-1882) in
1867. The two electrodes are
carbon and zinc, with a sal
ammoniac electrolyte. The
carbon electrode is mixed with
manganese peroxide. This
battery was used mainly for
intermittent service, such as
ringing electric bells.
59
60
Dry Cell battery
Anode:
Uses: Portable radios,
+2 + 2 e–
Zn
 Zn
toys,
flashlights.
+2 + 2 NH  Zn(NH ) +2
Zn
Advantages:
3
3 2
Inexpensive,
Removal
of NH3 safe, many
sizes.
Cathode:
Disadvantages:
High
+
–
2 NH4 + 2 e  2 NH3 + H2
current drain, NH3 builds
up
causing voltage drop,
H
2 + MnO2  MnO + H2O
short shelf life.
Removal of H2
61
62
Alkaline Battery
Uses: Same as dry cell.
Advantages: No voltage
drop, longer shelf life.
Disadvantage: Expensive
63
Super-iron battery
New type of Alkaline Battery
Uses: Same as dry
cell.
Advantages: works
well in high-drain-rate
electronics.
Disadvantages:
Expensive.
64
Lead Storage Battery
65
Mercury
Battery
Used in calculators,
watches, hearing aids,
cameras, and devices
where small size is
needed.
66
Lithium
Battery
Because it is light in
weight, and has a
large voltage (3.4 V
per cell) these are used
in pacemakers, cell phones,
laptops, camcorders.
67
Hydrogen Fuel Cell
68
Honda FCX Clarity Fuel Cell Vehicle
69
Downs cell for sodium production
70
Copper Electrolysis
71
Electrorefining of copper metal
72
Zinc strips help
protect the iron
hull of an oil
tanker from
oxidization. This
strip is attached
to the hull’s
interior surface.
73
Tarnish on silverware is
a coating of silver sulfide
(Ag2S). Tarnish begins
when silver atoms come
into contact with
hydrogen sulfide (H2S)
in the air. The silver ions
and sulfide ions combine
to form blackish silver
sulfide. Aluminum atoms
can help restore the
silver to its shiny self.
Directions at:
http://faculty.chemeketa.edu/lemme/CH%20122/handouts/Removing Silver Tarnish.pdf
74
Corrosion
75
Corrosion Prevention
76
Dental Voltaic Cell
Al → Al+3 + 3e–
O2 + 4 H+ + 4e– → 2 H2O
The short circuit between the Al foil and the
filling produces a current that is sensed by
the nerve of the tooth.
77
• In 1936, while excavating ruins of a 2000year-old village near Baghdad, workers
discovered mysterious small vase. A 6-inchhigh pot of bright yellow clay dating back two
millennia contained a cylinder of sheetcopper 5 inches by 1.5 inches. The edge of
the copper cylinder was soldered with a 6040 lead-tin alloy comparable to today's solder.
The bottom of the cylinder was capped with a
crimped-in copper disk and sealed with
bitumen or asphalt. Another insulating layer
of asphalt sealed the top and also held in
place an iron rod suspended into the center
of the copper cylinder. The rod showed
evidence of having been corroded with an
acidic agent.
78
The ancient battery in
the Baghdad Museum
79
The jar was found in Khujut Rabu just outside Baghdad and is composed
of a clay jar with a stopper made of asphalt. Sticking through the asphalt
is an iron rod surrounded by a copper cylinder. When filled with vinegar –
or any other electrolytic solution - the jar produces about 1.1 volts.
80
Atlantis Light-Bulb?
From Egypt Hieroglyphics
Filament
Light Blubs
Cord
81
http://www.world-mysteries.com/sar_lights_fd1.htm
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