Transcript Oxidation-Reduction Chapter 17 Hein and Arena Version 1.1 Eugene Passer Chemistry Department 1 College Bronx Community © John Wiley and Sons, Inc.

Oxidation-Reduction
Chapter 17
Hein and Arena
Version 1.1
Eugene Passer
Chemistry Department
1 College
Bronx Community
© John Wiley and Sons, Inc.
Chapter Outline
17.1 Oxidation Number
17.2 Oxidation-Reduction
17.4 Balancing Ionic Redox
Redox Equations
17.5 Activity Series of Metals
17.3 Balancing OxidationReduction Equations
17.6 Electrolytic and Voltaic
Cells
2
Oxidation Number
3
The oxidation number (oxidation state) of an
atom represents the number of electrons lost,
gained, or unequally shared by an atom.
4
Oxidation numbers can be zero, positive
or negative.
5
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.
6
A positive oxidation number means the
atom has fewer electrons assigned to it
than in the neutral atom.
7
A negative oxidation number means the atom
has more electrons assigned to it than in the
neutral atom.
8
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.
9
In covalently bonded substances,
oxidation numbers are assigned by an
arbitrary system based on relative
electronegativities.
10
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
11
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
Number
Electronegativity
0
3.0
Oxidation
Number
Electronegativity
0
3.0
12
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 13
Many elements have multiple
oxidation numbers
N oxidation
number
N2
N2O
NO
N2O3
NO2
N2O5
NO-3
0
+1
+2
+3
+4
+5
+5
14
15
16
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.
17
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.
18
Determine the oxidation number for sulfur in sulfuric
acid.
24
C2O
Step 1
Step 2
Step 3
-2
4(-2) = -8
2C + (-8) = -2 (the charge on the ion)
Step 4 2C = +6
C = +3 (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.
19
Oxidation-Reduction
20
Oxidation-reduction (redox) is a chemical
process in which the oxidation number of an
element is changed.
21
Redox may involve the complete transfer of
electrons to form ionic bonds or a partial
transfer of electrons to form covalent bonds.
22
• 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.
23
• 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.
24
The reaction of zinc with sulfuric acid is a redox
reaction.
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
Hydrogen
Zinc
Hydrogen
Zinc
transfers
Hydrogen
accepts
Zinc
is the
is electrons
the
isreducing
oxidized.
electrons
is
oxidizing
reduced.
toagent.
hydrogen.
from
agent.
zinc.
The electron transfer is more clearly expressed
as
o
+
24
Zn + 2H + SO
 Zn
2+
24
o
2
+ SO + H
25
26
17.1
Balancing OxidationReduction Equations
27
Change-In-Oxidation
Number 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 below each element to avoid
confusing them with ionic charge.
Sn + HNO3 → SnO2 + NO2 + H2O
0
+1 +5 -2 +4
-2 +4 -2 +1
-2
oxidation number of tin increases
oxidation number of nitrogen decreases
29
Balance the equation
Sn + HNO3 → SnO2+ NO2+H2O
Step 2 Write two new equations, using only the
elements that change in oxidation number.
Then add electrons to bring the equations
into electrical balance.
Sno → Sn4+ + 4e4N5+ + 1e- → N4+
30
Balance the equation
Sn + HNO3 → SnO2+ NO2+H2O
Step 3 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.
Sno → Sn4+ + 4e-
44N5+ + 44e- → 44N4+
31
Balance the equation
Sn + HNO3 → SnO2+ NO2+H2O
Step 4 Transfer the coefficient in front of each
substance in the balanced oxidationreduction equations to the corresponding
substances in the original equation.
Sno → Sn4+ + 4e-
44N5+ + 44e- → 44N4+
Sn + 44HNO3 → SnO2 + 44NO2 + H2O
32
Balance the equation
Sn + HNO3 → SnO2+ NO2+H2O
Step 5 Balance the remaining elements that are
not oxidized or reduced to give the final
balanced equation.
2 2O
Sn + 4HNO3 → SnO2 + 4NO2 + 2H
33
Balancing Ionic
Redox Equations
34
In ionic redox equations both the
numbers of atoms and the charges on
both sides of the equation must be the
same.
35
The Ion-Electron Method
36
Step 1 Write the two half-reactions that
contain the elements being oxidized
and reduced.
Step 2 Balance the elements other than
hydrogen and oxygen.
37
Step 3 Balance hydrogen and oxygen.
Acidic solution:
• For reactions in acidic solution, use H+ and
H2O to balance oxygen and hydrogen.
• For each oxygen needed use one H2O.
• Then add H+ as needed to balance the hydrogen
atoms.
38
Step 3 Balance hydrogen and oxygen.
Basic solution:
• For reactions in alkaline solutions, first
balance as thought the reactions were in an
acid solution, using Steps 1-3.
• Then add as many OH- ions to each side of
the equation as there H+ ions in the equation.
• Combine OH- ions into water.
Example: 4H+ + 4OH- → 4H2O
• Rewrite the equation, canceling equal numbers
of water molecules that appear on opposite
39
side of the equation.
Step 4 Add electrons (e-) to each halfreaction to bring them into electrical
balance.
Step 5 Since the loss and gain of electrons
must be equal, multiply each halfreaction by the appropriate number
to make the number of electrons the
same in each half-reaction.
40
Step 6 Add the two half-reactions together,
canceling electrons and any other
identical substances that appear on
opposite sides of the equation.
41
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 1 Write two half-reactions, one containing the
element being oxidized and the other the
element being reduced (use the entire
molecule or ion).
S  S
2-
o
MnO  Mn
4
2+
Step 2 Balance elements other than oxygen and
hydrogen. (Step 2 is unnecessary, since
these elements are already balanced). 42
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 3 Balance O and H. The solution is acidic.
The oxidation requires neither O nor H, but
the reduction equation needs 4H2O on the
right and 8H+ 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.
43
Balance the equation
MnO + S  Mn
4
2-
2+
o
+ S (acidic solution)
Step 4 Balance each half-reaction electrically with
electrons:
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 + 8H + MnO  Mn
5e
-
+ 4H2O
44
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+
45
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 46
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 1 Write two half-reactions, one containing the
element being oxidized and the other the
element being reduced (use the entire
molecule or ion).
Fe(OH)2  Fe(OH)3
24
CrO
 Cr(OH)3
Step 2 Balance elements other oxygen and
hydrogen (Step 2 is unnecessary, since
these elements are already balanced).
47
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 3 Balance O and H as though the solution were
acidic. Use H2O and H+.
Fe(OH)2 + H2O  Fe(OH)3 + H
+
For each
unbalanced
O add
For is
one
each
unbalanced
H add
- toone
Since
the solution
basic,
add
1
OH
each
+
H2O to the other side Hof to
the the other side of the
side.
equation.
equation.
+
-
Fe(OH)2 + H2O + OH  Fe(OH)3 + H + OH
Combine H+ and OH- as H2O and rewrite,
canceling H2O on each side.
Fe(OH)2 + H2O + OH-  Fe(OH)3 + H2O
Fe(OH)2 + OH  Fe(OH)3
-
oxidation half-reaction
48
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 3 Balance O and H as though the solution were
acidic. Use H2O and H+.
CrO + 5H  Cr(OH)3 + H2O
24
+
For each unbalanced O For
add each
one unbalanced H add- one
the solution
is the
basic,
addside
5 OH
H2O to Since
the other
sideH+of tothe
other
of to
theeach
side.
equation.
equation.
2+
-
CrO4 + 5H + 5OH  Cr(OH)3 + H2O + 5OH
Combine 5H+ + 5OH- → 5H2O
CrO + 5H2O  Cr(OH)3 + H2O + 5OH
24
-
half-reaction
Rewrite,oxidation
canceling
1 H2O from each side:
CrO + 4H2O  Cr(OH)3 + 5OH
24
-
49
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 4 Balance each half-reaction electrically with
electrons:
balanced oxidation half-reaction
Fe(OH)2 + OH  Fe(OH)3 + e
-
-
net charge = -1 on each side
balanced reduction half-reaction
CrO + 4H2O + 3e  Cr(OH)3 + 5OH
24
-
-
net charge = -5 on each side
50
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 5 Equalize loss and gain of electrons.
Multiply the oxidation reaction by 3.
3Fe(OH)
Fe(OH)
OH  Fe(OH)
3Fe(OH)
2 2++3OH
3 3++e3e
-
-
-
It is not necessary to multiply the reduction
equation.
CrO + 4H2O + 3e  Cr(OH)3 + 5OH
24
-
-
51
Balance the equation
CrO + Fe(OH)2  Cr(OH)3 + Fe(OH)3 (basic solution)
24
Step 6 Add the two half-reactions together, canceling
3e- and 3OH- from each side of the equation.
3Fe(OH)2 + 3OH  3Fe(OH)3 + 3e
-
-
CrO + 4H2O + 3e  Cr(OH)3 + 5OH
24
-
-
CrO + 3Fe(OH)2 + 4H2O  Cr(OH)3 + 3Fe(OH)3 + 2OH
24
-
The charge on both sides of the balanced equation is -2 52
Activity Series
of Metals
53
activity series A listing of metallic
elements in descending order of
reactivity.
54
Sodium (Na) will displace
any element below it from
one of its compounds.
55
increasing activity
Mg(s) + PbS(s)  MgS(s) + 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 its compounds.
56
increasing activity
Ag(s) + CuCl2(s)  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.
57
Electrolytic and
Voltaic Cells
58
electrolytic cell An electrolysis apparatus in
electrolysis The process whereby electrical
which electrical energy from an outside
energy is used to bring about a chemical
source is used to produce a chemical
change.
change.
59
anode
positive
electrode.
cathodeThe
The
negative
electrode.
60
Electrolysis of
Hydrochloric Acid
61
In an electrolytic cell electrical energy from
the voltage source is used to bring about
nonspontaneous redox reactions.
62
Hydronium ions migrate
to the cathode and are
reduced.
H3O+ + 1e- → Ho + H2O
Ho + Ho → H2
Cathode Reaction
63
17.3
Chloride ions migrate
to the anode and are
oxidized.
Cl-→ Clo + eClo + Clo→ Cl2
Anode Reaction
64
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.
65
The Zinc-Copper Voltaic Cell
66
voltaic cell A cell that produces electrical
energy from a spontaneous chemical reaction.
(Also known as a galvanic cell).
67
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.
68
increasing activity
Zn(s) + CuSO4(s)  ZnSO4(s) + 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.
69
If this reaction is carried out in a voltaic
cell, an electric current is produced.
70
71
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)
72
Spontaneity
The
Spontaneous
Reactions
reactions
thatreactions
isofoccur
the zinc-copper
inoccur
crucial
electrolytic
in all
difference
cell
voltaic
cells
are
spontaneous.
cells.
are
between
nonspontaneous.
all voltaic and electrolytic cells.
73
Key Concepts
17.1 Oxidation Number
17.2 Oxidation-Reduction
17.4 Balancing Ionic Redox
Redox Equations
17.5 Activity Series of Metals
17.3 Balancing OxidationReduction Equations
17.6 Electrolytic and Voltaic
Cells
74