To accompany Inquiry into
Chemistry
PowerPoint Presentation
prepared by Robert Schultz
robert.schultz@ei.educ.ab.ca
Chemistry 30 – Unit 2
Electrochemical Changes
Section 12.1 Characterizing
Oxidation Reduction
Chapter 12, Section 12.1
• Oxidation: Historically, the reaction of
a substance with O2
• Today, oxidation: loss of electrons
• Reduction: Historically, the reaction of
a metal ore to produce a smaller mass
of pure metal
• Today, reduction: gain of electrons
Chapter 12, Section 12.1
Memory Tools:
LEO the Lion says GER
gain
oxidation
lose
electrons
electrons
Oil Rig
oxidation
losing
is
reduction
reduction
is
gaining
Chapter 12, Section 12.1
• Oxidation and reduction cannot happen
individually – if one substance loses
electrons (oxidation), another
substance must gain electrons
(reduction)
• Oxidation/reduction reactions called
“redox reactions”
• Redox Song
Chapter 12, Section 12.1
• Consider the example:
Zn(s) + Cu(NO3)2(aq)
Zn(NO3)2(aq) + Cu(s)
This equation can be rewritten as a total
ionic equation:
Zn(s) + Cu2+(aq) + 2 NO3ˉ(aq)
Zn2+(aq) + 2 NO3ˉ(aq)
+ Cu(s)
spectators
and finally reduced to a net-ionic
equation:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Chapter 12, Section 12.1
electrons lost
oxidation
• Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
electrons gained
reduction
causes reduction
called “reducing agent”
causes oxidation
called “oxidizing agent”
Note: the oxidizing agent gets
reduced; the reducing agent gets
oxidized!!!
Chapter 12, Section 12.1
• The net-ionic equation:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
can be written as two half-reactions,
Cu2+(aq) + 2 e–
Zn(s)
Cu(s)
reduction
Zn2+(aq) + 2 e–
oxidation
Chapter 12, Section 12.1
• Half-Reactions WS
• Discuss questions 3 and 4, page 437
• Spontaneity of redox reactions – Nelson
Lab or Investigation 12.A, page 438
Chapter 12, Section 12.1
• Example: Cu(s), Pb(s), Ag(s) and Zn(s)
are reacted with solutions of
Cu(NO3)2(aq), Pb(NO3)2(aq), AgNO3(aq),
and Zn(NO3)2(aq)
• The experimental results are
summarized in the chart:
reducing agents
strongest oxidizing agent
Chapter 12, Section 12.1
Cu(NO
)2(aq) Pb(NO
)2(aq) AgNO
Cu2+3(aq)
Zn2+(aq)
Ag+(aq)
Pb2+3(aq)
3(aq) Zn(NO
3)2(aq)
Cu(s)

Pb(s)

Ag(s)


Zn(s)


oxidizing agents
 spontaneous;
strongest reducing agent






 non-spontaneous
Chapter 12, Section 12.1
• Generalizations to this point:
• Oxidizing agents: metal ions, e.g.
Ag+(aq), and non-metal elements, e.g.
Cl2(g)
• Reducing agents: non-metal ions, e.g.
Br –(aq), and metal elements, e.g. Fe(s)
Chapter 12, Section 12.1
• Results of experiment can be summarized
in a table of redox half-reactions:
Ag+(aq) + e–
Ag(s)
Cu2+(aq) + 2 e–
Cu(s)
Pb2+(aq) + 2 e–
Pb(s)
Zn(s)
“SOA” Zn2+(aq) + 2 e–
“SRA”
Where are the weakest OA and RA?
Examine data chart from previous
slide. How can chart of half-reactions
be used to predict whether or not a
reaction is spontaneous?
Chapter 12, Section 12.1
• Spontaneity Generalization:
OA
spontaneous
RA
RA
non-spontaneous
OA
Chapter 12, Section 12.1
Take the chart of half-reactions with
Zn, Cu, Pb, and Ag elements and ions
and combine with the two data charts
on the next page to produce a single
chart of half-reactions formatted in the
standard way
Chapter 12, Section 12.1
Chapter 12, Section 12.1
Ag+(aq) + e–
Cu2+(aq) + 2 e–
Pb2+(aq) + 2 e–
Zn2+(aq) + 2 e–
Ag(s)
Cu(s)
Pb(s)
Zn(s)
Chart 2:
Chart 1:
Cd2+(aq) + 2 e–
Cd(s)
Pb2+(aq) + 2 e–
Pb(s)
V2+(aq) + 2 e–
V(s)
Cd2+(aq) + 2 e–
Cd(s)
Be2+(aq) + 2 e–
Be(s)
Zn2+(aq) + 2 e–
Zn(s)
Ra2+(aq) + 2 e–
Ra(s)
V2+(aq) + 2 e–
V(s)
Chapter 12, Section 12.1
Ag+(aq) + e–
Cu2+(aq) + 2 e–
Pb2+(aq) + 2 e–
Cd2+(aq) + 2 e–
Zn2+(aq) + 2 e–
Ag(s)
Cu(s)
Pb(s)
Cd(s)
Zn(s)
V2+(aq) + 2 e–
V(s)
Be2+(aq) + 2 e–
Ra2+(aq) + 2 e–
Be(s)
Ra(s)
Chapter 12, Section 12.1
Chart on page 7
of Data Booklet
has been
prepared in this
manner
You can use the
chart and the
spontaneity
generalization
developed
earlier to
predict
spontaneity of
redox reactions
Chapter 12, Section 12.1
• Examples:
• #5a, 6a, and Review 6a page 440
purple box
bottom of page
Chapter 12, Section 12.2
• Writing balanced half-reactions in
acidic or neutral media:
• Step 1: balance elements other than O
or H by inspection
Must be
done in • Step 2: balance O’s using H2O(l)’s
this
• Step 3: balance H’s using H+(aq)’s
order!
• Step 4: balance total charge using
electrons (e–)
Chapter 12, Section 12.2
• Examples:
• Zn2+(aq) + 2 e–
Zn(s)
Step 4
neutral
media
• 2 Cl–(aq)
Step 1
Cl2(g) + 2 e–
Step 4
Chapter 12, Section 12.2
• Example:
acidic medium
• Cr2O72-(aq) + 14 H+(aq) + 6 e–
Step 3
2 Cr3+(aq) + 7 H2O(l)
Step 1
Step 2
Step 4
You’ll find this half-reaction on page 7
of your data booklet
Chapter 12, Section 12.2
• You are not required to write halfreactions in basic media, though you
should be able to use given halfreactions if they are in basic media
Chapter 12, Section 12.2
• Example: Practice Problem 1a, page
448
• Both of the required half-reactions are
in your data booklet
• Half-reactions, those you write
yourself, and those from your data
chart can be used to balance redox
reaction equations
MnO4–(aq) + 8 H+(aq) + 5 e–
5 x (Ag(s)
MnO4–(aq) + 8 H+(aq) + 5 Ag(s)
OA
RA
Mn2+(aq) + 4 H2O(l)
Ag+(aq) + e–)
Mn2+(aq) + 4 H2O(l) + 5 Ag+(aq)
Chapter 12, Section 12.2
• Practice Problem 1b, page 448
• Neither of these half-reactions are in
your data booklet you must write each
yourself
Hg(l) + 4 Cl–(aq)
HgCl42–(s) + 2 e–
2 x ( NO3–(aq) + 2 H+(aq) + e–
NO2(g) + H2O(l) )
Hg(l) + 4 Cl–(aq) + 2 NO3–(aq) + 4 H+(aq)
HgCl42–(s) + 2 NO2(g) + 2 H2O(l)
Chapter 12, Section 12.2
• Do Practice Problems 1c, 1d page 448
1c
AsH3(s) + 4 H2O(l)
4 x (Zn2+(aq) + 2 e–
H3AsO4(aq) + 8 H+(aq) + 8 e–
Zn(s))
AsH3(s) + 4 H2O(l) + 4 Zn2+(aq)
4 Zn(s) + H3AsO4(aq) + 8 H+(aq)
Data Booklet (basic)
1d
I2(s) + 6 H2O(l)
2 IO3–(aq) + 12 H+(aq) + 10 e–
5 x (OCl–(aq) + H2O(l) + 2 e–
I2(s) + 11 H2O(l) + 5 OCl–(aq)
Cl–(aq) + 2 OH−(aq))
2 IO3–(aq) + 12 H+(aq) + 5 Cl–(aq) + 10 OH−(aq)
Chapter 12, Section 12.2
• Half-reactions WS
• Do question 14 a,b,c (acidic conditions)
on page 475 (yes I do mean page 475)
• Time to work
Chapter 12, Section 12.2
• 14 a
• (both half-reactions are in the Data
Booklet)
MnO2(s) + 4 H+(aq) + 2 e–
2 Cl–(aq)
MnO2(s) + 4 H+(aq) + 2 Cl–(aq)
Mn2+(aq) + 2 H2O(l)
Cl2(g) + 2 e–
Mn2+(aq) + 2 H2O(l) + Cl2(g)
Chapter 12, Section 12.2
• 14 b
2 x (NO(g) + 3 H+(aq) + 3 e–
3 x (Sn(s)
2 NO(g) + 6 H+(aq) + 3 Sn(s)
NH2OH(aq))
Sn2+(aq) + 2 e–)
3 Sn2+(aq) + 2 NH2OH(aq)
Chapter 12, Section 12.2
• 14 c
3 x (Cd2+(aq) + 2 e–
2 x (V2+(aq) + 3 H2O(l)
3 Cd2+(aq) + 2 V2+(aq) + 6 H2O(l)
Cd(s))
VO3–(aq) + 6 H+(aq) + 3 e–)
3 Cd(s) + 2 VO3–(aq) + 12 H+(aq)
Chapter 12, Section 12.2
• Read Reducing Iron (to make steel),
pages 452-4
Chapter 12, Section 12.3
• Assigning oxidation numbers is a
method of electron bookkeeping –
they make it possible to see electron
gain and loss without writing halfreactions
• Oxidation Number Rules:
1. Pure elements have an oxidation
number of 0. e.g. in Cl2 each Cl atom
has an oxidation number of 0; in Fe
each Fe atom has an oxidation number
of 0.
Chapter 12, Section 12.3
2. The oxidation number of hydrogen in
compounds is +1. This is true in
molecular compounds and acids.
Exception: in metal hydrides, e.g. LiH,
H is -1
3. The oxidation number of oxygen in
compounds is -2. Exceptions:
peroxides, e.g. H2O2, where oxygen is
-1 and the compound OF2, where
oxygen is +2
Chapter 12, Section 12.3
4. In monatomic ions, e.g. Cl– and Fe3+,
the oxidation number equals the ion
charge
5. Oxidation numbers of atoms in
compounds add to zero. Oxidation
numbers of atoms in polyatomic ions
add to the ion charge
Note oxidation numbers can be
fractions
Do Oxidation State WS
Chapter 12, Section 12.3
• Uses for oxidation numbers:
1. Determining whether a substance is
oxidized or reduced – e.g. in the
reaction: (oxidation numbers in red)
0
5 Fe(s) + 2
+7 -2
MnO4–(aq)
+1
+ 16
H+(aq)
5
+2
Fe2+(aq)
+2
+2
Mn2+(aq)
+1-2
+ 8 H2O(l)
Fe changes from 0 to +2; it is oxidized
Mn in MnO4–(aq) changes from +7 to +2
in Mn2+; it is reduced
an increase in oxidation number is
oxidation; a decrease in oxidation
number is reduction
Chapter 12, Section 12.3
2. Determining the OA and RA in a redox
reaction. The OA is reduced (has a
decrease in oxidation number), the RA
is oxidized (has an increase in
oxidation number)
3. Determining whether or not a reaction
is redox. If it is, one* substance will
increase in oxidation number
(oxidation) and one* will decrease
(reduction). If no change in oxidation
numbers, the reaction is not redox.
Chapter 12, Section 12.3
• Do Practice Problem 11, page 461
+1 -1
+2 -2 +1
+3
2 Fe(OH)3(s) Redox
H2O2(aq) + 2 Fe(OH)2(s)
+3-1
+1 -2
PCl3(l) + 3 H2O(l)
-3 +1
0
2 C2H6(g) + 7 O2(g)
+4-2
+1 -2
3 NO2(g) + H2O(l)
-2+1
+1 +3-2
+1-1
H3PO3(aq) + 3 HCl(aq)
+4 -2
+1 -2
4 CO2(g) + 6 H2O(l)
+1+5-2
Not Redox
+2-2
2 HNO3(aq) + NO(g)
Redox
Redox
Chapter 12, Section 12.3
4. Determining the number of electrons
transferred in a redox reaction or
half-reaction. e.g.
+7 -2
+1
MnO4–(aq) + 8 H+(aq) + 5 e–
+2
+1 -2
Mn2+(aq) + 4 H2O(l)
Note that Mn changes from +7 to +2, a
change of 5 electrons. The number of
electrons transferred can be
determined without a half-reaction!
Chapter 12, Section 12.3
• Another example:
+6 -2
+1
Cr2O72-(aq) + 14 H+(aq) + 6 e–
+3
+1-2
2 Cr3+(aq) + 7 H2O(l)
• Note that Cr changes from +6 to +3, a
change of 3, yet the half-reaction has
6 e–. Why???
• There are 2 Cr’s, each changing by 3
2x3=6
Chapter 12, Section 12.3
5. Determining which substance in a
half-reaction is really functioning as
the OA (or RA) e.g.:
+7 -2
+1
MnO4–(aq) + 8 H+(aq) + 5 e–
+2
+1 -2
Mn2+(aq) + 4 H2O(l)
Note that the Mn in the MnO4−(aq) is
reduced
MnO4−(aq) is the OA
H+(aq) is a very necessary spectator
6. Balancing equations by the oxidation
number method
Chapter 12, Section 12.2
• Disproportionation Reaction – a redox
reaction where the same element, is
both oxidized and reduced (acts as
both OA and RA)
• Example:
+1-2+1
0
2 NaOH(aq) + Cl2(g)
+1 -1
+1 +1-2
+1 -2
NaCl(aq) + NaClO(aq) + H2O(l)
• Note that Cl in Cl2 goes from 0 to -1,
and from 0 to +1 in NaClO(aq)
Chapter 12, Section 12.3
• Another example:
+4-2
+1 -2
3 NO2(g) + H2O(l)
+1+5-2
+2-2
2 HNO3(aq) + NO(g)
• N is oxidized from +4 to +5 and reduced
from +4 to +2
Chapter 12, Section 12.3
• Redox reactions are balanced by
making electrons gained and lost equal
• This can be done using half-reactions or
by observing change in oxidation
number
• Try balancing the following equation by
inspection:
S8(s) + KMnO4(aq) + HCl(aq)
SO2(g) + MnCl2(s) + KCl(aq) + H2O(l)
Chapter 12, Section 12.3
• Try it using change in oxidation
number:
0
+1+7 -2
+1 -1
5 S8(s) + 32 KMnO4(aq) + 96HCl(aq)
5e–/Mn
4e–/S
32e– /S8
RA
+4 -2
+2 -1
+1 -1
40 SO2(g) + 32 MnCl2(s) + 32 KCl(aq) + 48 H2O(l)
5e– /KMnO4
OA
+1 -2
Insert coefficients in
equation to make
these numbers equal
Usually the other
coefficients will be easily
found
Chapter 12, Section 12.3
+1 -2
+1 -1
0
8 H2S(g) + 8 H2O2(aq)
2e−/S
1e−/O
2e− /H2S
2e− /H2O2
RA
+1 -2
S8(s) + 16 H2O(l)
OA
Because of the 8 in S8, it is
necessary to put 8 in front of H2S
and because of 2e−/2e− ratio, 8
must also go in front of the H2O2
Do WS 56 B,C
Chapter 12, Section 12.4
• Redox Titrations
Buret containing
titrant solution
Statement: (applies to all
titrations: “titration of sample
with titrant ”
Know this!
Erlenmeyer flask
containing sample
solution
Chapter 12, Section 12.4
• Oxidizing and reducing agents can be
titrated to find concentrations (just
like acids and bases)
• KMnO4(aq) is used as the titrant in
many redox titrations – acidified
MnO4–(aq) is a very strong OA in the
half-reaction:
MnO4–(aq) + 8 H+(aq) + 5 e–
dark
purple
Mn2+(aq) + 4 H2O(l)
colourless
Chapter 12, Section 12.4
• As long as some RA is left in the sample
solution, the purple colour of the
MnO4–(aq) disappears. Once the RA is
used up the purple colour remains.
Ideally the endpoint is a very light
purple
Chapter 12, Section 12.4
• Example Practice Problem 19, page 469
2 x (MnO4–(aq) + 8 H+(aq) + 5 e–
5 x (H2O2(l)
2 MnO4–(aq) + 16 H+(aq) + 5 H2O2(l)
6
O2(g) + 2 H+(aq) +2 e–)
5 O2(g) + 10 H+(aq) + 2 Mn2+(aq) + 8 H2O(l)
2 MnO4–(aq) + 6 H+(aq) + 5 H2O2(l)
n1
0.02045 mol/L
38.95 mL
Mn2+(aq) + 4 H2O(l))
n2
m=?
5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)
n1  0.02045 mol L  0.03895 L  7.965  10-4 mol
5
 1.991 10 3 mol
2
m  1.991 10 3 mol  34.02 g mol  0.06774 g
0.06774 g
%m 
 100%  5.276%
m
1.284 g
n2  7.965  10-4 mol 
Chapter 12, Section 12.4
• Example Practice Problem 22, page 469
The given equation:
Cr2O72-(aq) + Fe2+(aq)
Cr3+(aq) + Fe3+(aq)
is not complete because of the O’s
Both half-reactions are in Data Booklet:
Cr2O72-(aq) + 14 H+(aq) + 6 e–
6 x (Fe2+(aq)
Cr2O72-(aq) + 14 H+(aq) + 6 Fe2+(aq)
n2
n1
c=?
0.02043 mol/L
25.00 mL
35.55 mL
2 Cr3+(aq) + 7 H2O(l)
Fe3+(aq) + e–
6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
n1  0.02043 mol L  0.03555L  7.263  10 4 mol
Chapter 12, Section 12.4
n2  7.263  10 4 mol 
6
 4.358  10 3 mol
1
4.358  10 3 mol
c
 0.1743 mol L
0.02500 L
Chapter 12, Section 12.4
• Practice Problems 20 and 21, page 469
• Answers:
20.
0.1387%
m
m
*
checked
21.
0.0110 mol/L
Redox titration calculation worksheet
*
Detailed solution at end
Chapter 12, Section 12.4
• Vitamin C content of orange juice
experiment (Lab 12.C) Prelab
• Lab
Chapter 12, Section 12.4
Practice Problem 20, page 469
16 H+(aq) + 2 Cr2O72-(aq) + C2H5OH(l)
n1
0.05023 mol/L
32.35 mL
n2
m=?
4 Cr3+(aq) + 2 CO2(g) + 11 H2O(l)
n1  0.05023 mol L  0.03235 L  1.265  10 3 mol
1
 8.125  10 4 mol
2
m  8.125  10 4 mol  46.08 g mol  0.03745 g
n2  1.625  10 3 mol 
%m m 
0.03744 g
 100%  0.1387%
27.00 g
Chapter 12, Section 12.4