20.
Researched Information on Chlorophyll
Ɣ
the structure of the main
form of chlorophyll
a ring structure with a long tail
Ɣ
its general role in
electron transfer
reactions
As chlorophyll absorbs light energy, one of its electrons moves
from a lower to a higher energy state. A series of electron-transfer
steps follows, ending in an electron being transferred to carbon
dioxide. Once carbon dioxide has the extra electron, it can
participate in further reactions, resulting in the formation of glucose.
Ɣ
close relatives containing
iron and cobalt
The part of chlorophyll that surrounds the magnesium is called
porphyrin. Other compounds containing porphyrin (but with
different metal ions) are hemoglobin (containing iron) and
vitamin B12 (containing cobalt).
Ɣ
a technological
application inspired by
the chlorophyll molecule
As many porphyrins are brightly coloured, they have been used as
natural dyes. Similar synthetic dyes have been manufactured and
can be used medically to kill certain undesirable cells when
activated by light.
13.3 OXIDATION STATES
Practice
(Page 585)
1. (a) +4
(b) +7
(c) +6
(d) +6
(e) –1
(f) –1
2. (a) +1
(b) +2
(c) +4
(d) –3
(e) –2
(f) +5
(g) 0
(h) –3
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3. (a) 0
(b) 0
(c) +4
(d) +2
4. +5 –2
–1
+1
+4 –2
0
+1 –2
2 ClO3–(aq) + 2 Cl–(aq) + 4 H+(aq) o 2 ClO2(g) + Cl2(g) + 2 H2O(l)
5. methane o methanol o methanal o methanoic acid o carbon dioxide
CH3OH
CH2O
HCOOH
CO2
CH4
–4
–2
–4; –2; 0; +2; +4
0
+2
+4
Lab Exercise 13.B: Oxidation States of Vanadium
(Page 586)
Purpose
The purpose of this lab exercise is to use the concept of oxidation states to investigate some redox
chemistry of vanadium compounds.
Problem
What are the oxidation numbers and changes in oxidation number for vanadium ions?
Analysis
(a) (1) VO3–(aq) [vanadate(V) ion from the dissociation of ammonium vanadate(V)]
(2) VO3–(aq) o VO2+(aq) o V3+(aq) o V2+(aq)
(3) V2+(aq) o V3+(aq)
(4) VO3–(aq) o VO2+(aq)
(5) VO2+(aq) (no change)
(6) V2+(aq) o V3+(aq) o VO2+(aq) o VO3–(aq)
(b) (1) dissociation; no change
(2) vanadium is being reduced from +5 to +4 to +3 to +2
(3) vanadium is being oxidized from +2 to +3
(4) vanadium is being reduced from +5 to +4
(5) no change in oxidation number
(6) vanadium is being oxidized from +2 to +3 to +4 to +5
(c) (3) The V2+(aq) is slowly changed to V3+(aq). This is an oxidation reaction probably caused
by the reaction of oxygen in the air with V2+(aq).
(4) The VO3–(aq) is likely reduced to VO2+(aq), which is blue. At the same time, iodide ions
are likely being oxidized to iodine, which is yellow-brown in aqueous solution. The
combination of blue and yellow-brown would produce a very dark coloured mixture.
(5) Iodide ions are not able to reduce VO2+(aq).
(6) Permanganate ions are able to successively oxidize V2+(aq) to V3+(aq) to VO2+(aq) and
finally, to VO3–(aq).
Web Activity: Case Study—Catalytic Converters
(Page 588)
Main components:
A three-way catalytic converter contains a reduction catalyst and an oxidation catalyst (usually
platinum, rhodium, and/or palladium), each contained in a honeycomb structure.
Main reactions:
Nitrogen in nitrogen oxides is reduced to nitrogen gas, and carbon in carbon monoxide and
volatile organic compounds (VOCs) is oxidized to carbon dioxide.
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+2 –2
0
0
+4 –2
2 NO(g) o N2(g) + O2(g)
+2 –2
0
or
+4 –2
2 CO(g) + O2(g) o 2 CO2(g)
–1 +1
0
0
0
2 NO2(g) o N2(g) + 2 O2(g)
+4 –2
+1 –2
CxHy(g) + n O2(g) o x CO2(g) + m H2O(g)
(VOC)
Practice
(Pages 588–589)
6.
–2 +1 –2 +1
+7 –2
+1
0 +1 –2
+2
+1 –2
(a) 5 CH3OH(l) + 2 MnO4–(aq) + 6 H+(aq) o 5 CH2O(l) + 2 Mn2+(aq) + 8 H2O(l)
oxidation
(b) Carbon is oxidized from –2 to 0; CH3OH 
o CH2O
reduction
– 
o Mn2+
(c) Manganese is reduced from +7 to +2; MnO4
7. (a) 0
+1 +5 –2
0
+2 +5 –2
Cu(s) + 2 AgNO3(aq) o 2 Ag(s) + Cu(NO3)2(aq)
oxidation: Cu o Cu2+; reduction: Ag+ o Ag
+1 –1
(b) +2 +5 –2
Pb(NO3)2(aq) + 2 KI(aq) o
+2 –1
+1 +5 –2
reaction is redox
reaction is not redox
PbI2(s) + 2 KNO3(aq)
(c) 0
+1 –1
0
+1 –1
Cl2(aq) + 2 KI(aq) o I2(s) + 2 KCl(aq)
oxidation: I– o I in I2; reduction: Cl in Cl2 o Cl–
(d) +1 –1
0
0
2 NaCl(l) o 2 Na(l) + Cl2(g)
oxidation: Cl– o Cl in Cl2; reduction: Na+ o Na
reaction is redox
(e)
reaction is not redox
+1 –1
+1 –2 +1
+1 –2 +1
+1 –1
reaction is redox
HCl(aq) + NaOH(aq) o HOH(l) + NaCl(aq)
(f)
0
0
reaction is redox
+3 –1
2 Al(s) + 3 Cl2(g) o 2 AlCl3(s)
oxidation: Al o Al3+; reduction: Cl in Cl2 o Cl–
(g)
(h)
–5/2 +1
0
+4 –2
+1 –2
10 H2O(l)
2 C4H10(g) + 13 O2(g) o 8 CO2(g) +
oxidation: C in C4H10 o C in CO2;
reduction: O in O2 o O in CO2 and H2O
+1 –1
+1 –2
0
reaction is redox
reaction is redox
2 H2O2(l) o H2O(l) + O2(g)
oxidation: O in H2O2 o O in O2; reduction: O in H2O2 o O in H2O
8. (a) single replacement
(b) double replacement (precipitation)
(c) single replacement
(d) simple decomposition
(e) double replacement (neutralization)
(f) formation
(g) complete combustion
(h) other
Double replacement reactions do not appear to be redox reactions.
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9. The oxygen atoms in hydrogen peroxide, H2O(l), have an oxidation number of –1, and can be
either oxidized to O2(g), (oxidation number of 0), or reduced to an oxidation state of –2 (as in
H2O(l) or metallic oxides).
10. On Earth, carbon is easily oxidized to carbon dioxide by a combustion reaction involving
atmospheric oxygen as an oxidizing agent:
0
+4
C(s) + O2(g) o CO2(g)
On Saturn, carbon would likely be reduced in a reaction involving atmospheric hydrogen as a
reducing agent.
0
–4
C(s) + H2(g) o CH4(g)
Practice
(Page 593)
11. The oxidation number of an atom is calculated by counting electrons using a consistent rule:
shared pairs of electrons assigned to the more electronegative atom. Therefore, the gain or
loss of electrons by an atom is reflected by a change in the oxidation number that is equal to
the number of electrons transferred.
12. (a) +6 –2
–1
+3
0
2–
–
+
3+
Cr2O7 (aq) + 6 Cl (aq) + 14 H (aq) o 2 Cr (aq) + 3 Cl2(aq) + 7 H2O(l)
3 e–/Cr
1 e–/Cl
1 e–/Cl–
6 e–/Cr2O72–
u1
u6
(b) +5 –2
+1+4–2
+6–2
0
–
–
2–
2 IO3 (aq) + 5 HSO3 (aq) o 5 SO4 (aq) + I2(s) + 3 H+(aq) + H2O(l)
2 e–/S
5 e–/I
–
–
5 e /IO3
2 e–/HSO3–
u2
u5
+1 –1
+1 +6 –2
+4 –2
0
(c)
2 HBr(aq) + H2SO4(aq) o SO2(g) + Br2(l) + 2 H2O(l)
1 e–/Br
2 e–/S
–
1 e /HBr
2 e–/H2SO4
u2
u1
13. (a) +7 –2
+4 –2
+6 –2
+4 –2
2 MnO4–(aq) + 3 SO32–(aq) + H2O(l) o 3 SO42–(aq) + 2 MnO2(s) + 2 OH–(aq)
3 e–/Mn 2 e–/S
3 e–/MnO4–
2 e–/SO32–
u2
u3
(b) +5 –2
–2 +1
+2 –2
–1
4 ClO3–(aq) + 3 N2H4(aq) o 6 NO(g) + 4 Cl–(aq) + 6 H2O(l)
4 e–/N
6 e–/Cl
–
–
6 e /ClO3
8 e–/N2H4
u4
u3
14. MnO4– is the oxidizing agent and SO32– is the reducing agent.
ClO3– is the oxidizing agent and N2H4 is the reducing agent.
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15.
–3 +1
0
+4 –2
+1 –2
4 NH3(g) + 7 O2(g) o 4 NO2(g) + 6 H2O(g)
2 e–/O
7 e–/N
–
7 e /NH3 4 e–/O2
u4
u7
Case Study: Bleaching Wood Pulp
(Page 594)
1. The unintended consequences of using chlorine to bleach wood pulp were the production of
various harmful by-products, including dioxin (a carcinogen) and furans, which can
bioaccumulate in the environment.
2. To reduce the problems created by using elemental chlorine, the use of Cl2 was reduced
and/or replaced, thereby reducing dioxin and furan emissions; the recycling of paper
increased; and the consumption of water decreased. Overall, the effluent quality improved.
3. Perspectives that need to be considered include:
Ɣ Ecological: Will the replacement chemical be better for the environment?
Ɣ Economical: Will the replacement process and chemical be affordable?
Ɣ Technological: Does the replacement process work as well or better?
Ɣ Scientific: Is sufficient research available about the products of the reaction(s).
Ɣ Political: Has the government enacted, or will it enact legislation to mandate replacement
chemicals?
4. 0
–1
1 e–/Cl, 2 e–/mol Cl2
Cl2(g) o Cl–(aq)
+4
–1
ClO2(g) o Cl–(aq)
5 e–/Cl, 5 e–/mol ClO2
Chlorine dioxide transfers more than double the amount of electrons per mole of oxidizing
agent than does chlorine.
Extension
5. On its Web site, Domtar gives specifications on its many types of paper, including some that
are bleached using ozone, resulting in the production of non-detectable dioxins.
A brief report of a plan to improve environmental standards at a pulp and paper mill
in Henan Province, China, outlines some of the challenges, options, decisions, and effects of
reducing the amount of water, power, and toxic chemicals that are present during production.
The Swedish experience is different again, with huge changes having been made over
the past four decades to improve environmental protection. One of the main changes is the
changeover from chlorine bleaching to chlorine dioxide bleaching. As a result, the release of
chlorinated organic compounds has been reduced to very low levels.
6. (1) Cl2 | OCl– (aq) (Ca(OCl)2)
(2) ClO2 | OCl– (aq) (Ca(OCl)2)
(3) H2O2 | O2 (g) + C2H5OH(l)
In basic solutions, chlorine dioxide can disproportionate to form chlorite and chlorate ions:
+4
+3
+5
2 ClO2(aq) + 2 OH (aq) o ClO2 (aq) + ClO3(aq) + H2O(l)
The oxidation number of the chlorine in the chlorite ion, ClO2, is lower (reduction) and the
oxidation number of the chlorine in the chlorate ion, ClO3, is higher (oxidation) than in the
original chlorine dioxide molecule.
Chlorine dioxide is able to react with organic molecules (such as those that give
colour to pulp fibres) by removing one electron. By this process, chlorine dioxide is reduced
to chlorite, ClO2.
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Section 13.3 Questions
(Page 595)
1.
Electron transfer
loss of electrons
gain of electrons
Oxidation
Reduction
Oxidation states
increases (becomes more positive)
decreases (becomes more negative)
2. An oxidation number is a positive or negative number that corresponds to the apparent charge
that an atom in a molecule or ion would have if the electron pairs in the covalent bonds
belonged entirely to the more electronegative atom.
3. A redox reaction can be recognized using a chemical reaction equation by:
Ɣ looking for examples of formation, decomposition, single displacement, and combustion
reactions;
Ɣ looking for changes in the oxidation states of atoms/ions on the reactant side of the
chemical equation with those of atoms/ions of the same atoms/ions on the product side.
4. (a) carbon +2; oxygen –2
C +2; O –2
(b) oxygen 0
O0
(c) nitrogen –3; hydrogen +1; chlorine –1
N 3; H +1; Cl 1
(d) hydrogen +1; phosphorus +5; oxygen –2
H +1; P +5; O 2
(e) sodium +1; sulfur +2; oxygen –2
Na +1; S +2; O 2
(f) sodium +1; phosphorus +5; oxygen –2
Na +1; P +5; O 2
8
5. (a) The oxidation number of iron in Fe3O4 is calculated to be .
3
(b) The fractional value of the answer is unusual because it would involve a fractional
number of electrons, which is not possible. The formula, Fe3O4, might represent a
compound involving a combination of iron(II) oxide and iron(III) oxide, which could be
written FeO•Fe2O3.
+7 –2
+3 –2
+1
+2
+1 –2
+4–2
MnO4–(aq) + C2O42–(aq) + H+(aq) o Mn2+(aq) + H2O(l) + CO2(g)
Carbon is oxidized. Its oxidation number changes from +3 to +4.
Manganese is reduced. Its oxidation number changes from +7 to +2.
The oxidizing agent is MnO4– and the reducing agent is C2O42–.
CO2(g) + H2O(l) o H2CO3(aq)
This is not a redox reaction. The oxidation numbers of all of the atoms remain
unchanged.
8. (a)
Cu(s) oCu2+(aq) + 2 e–
2 [NO3–(aq) + 2 H+(aq) + e– oNO2(g) + H2O(l)]
Cu(s) + 2 NO3–(aq) + 4 H+(aq) oCu2+(aq) + 2 NO2(g) + 2 H2O(l)
6. (a)
(b)
(c)
(d)
7. (a)
(b)
(b)
Cr2O72(aq) + 14 H+(aq) + 6 e– o 2 Cr3+(aq) + 7 H2O(l)
3 [H2O2(aq) o O2(g) + 2 H+(aq) + 2 e–]
3 H2O2(aq) + Cr2O72(aq) + 8 H+(aq) o 2 Cr3+(aq) + 3 O2(g) + 7 H2O(l)
(c)
2 [Mn2+(aq) + 4 H2O(l) oҏMnO4–(aq) + 8 H+(aq) + 5 e–]
5 [HBiO3(aq) + 5 H+(aq) + 2 e– oBi3+(aq) + 3 H2O(l)]
2 Mn2+(aq) + 5 HBiO3(aq) + 9 H+(aq) o5 Bi3+(aq) + 2 MnO4(aq) + 7 H2O(l)
9. (a)
2 [Cr(OH)3(s) + 5 OH–(aq) oCrO42–(aq) + 4 H2O(l) + 3 e–]
IO3–(aq) + 3 H2O(l) + 6 e– oI–(aq) + 6 OH–(aq)
2 Cr(OH)3(s) + 4 OH–(aq) + IO3–(aq) o CrO42–(aq) + 5 H2O(l) + I–(aq)
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(b)
Ag2O(s) + H2O(l) + 2 e– o2 Ag(s) + 2 OH–(aq)
HCHO (aq) + 3 OH–(aq) o+CO2–(aq) + 2 H2O(l) + 2 e–
Ag2O(s) + OH–(aq) + HCHO(aq) oAg(s) + HCO2–(aq) + H2O(l)
2 [S2O42–(aq) + 8 OH–(aq) o 2 SO42–(aq) + 4 H2O(l) + 6 e–]
3 [O2(g) + 2 H2O(l) + 4 e– o 4 OH–(aq)]
2 S2O42–(aq) + 3 O2(g) + 4 OH–(aq) o 4 SO42–(aq) + 2 H2O(l)
10. oxidation
(c)
+1 –1
H2O2(l)
+1 –2
o
0
H2O(l)
+
O2(g)
reduction
Some oxygen atoms in H2O2 with a –1 oxidation number are oxidized to 0 and some are
reduced to –2.
11. Two general experimental designs that could help determine the balancing of the main
species in a redox reaction are gravimetric stoichiometry (mass measurement) and volumetric
stoichiometry (titration).
12. (a) +3 –2
+1–2
0
+2 –2
+1 +6 –2
2 Fe2S3(s) + 6 H2O(l) + 11 O2(g) o 4 FeO(s) + 6 H2SO4(aq)
(b) The S in Fe2S3 is oxidized from –2 to +6.
The Fe in Fe2S3 is reduced from +3 to +2 and the O in O2 is reduced from 0 to –2.
(c) The method presented to balance redox reactions involves only one entity that is oxidized
and one entity that is reduced. However, some redox reactions may involve two entities
gaining or losing electrons.
Extension
13. Pemmican is an example of technology providing a solution to the problems of storing and
transporting food. Pemmican provided a nutritious, high energy, protein snack that was light
and, therefore, easily transportable. As well, because it was dehydrated, it would be easily
preserved and would not need to be refrigerated. European settlers were seldom prepared for
their new environment. Many settlers were saved from starvation and disease by the food and
food preparation techniques of First Nations peoples.
14. The overall chemical reaction equation, showing the oxidation numbers, for cellular
respiration is:
0 +1 –2
0
+4 –2
+1 –2
C6H12O6(s) + 6 O2(g) o 6 CO2(g) + 6 H2O(l)
Oxygen is the oxidizing agent that removes electrons from carbon atoms (4 per carbon atom).
In doing so, oxygen, O2, gains 2 e– per oxygen atom and is reduced. Glucose is therefore the
reducing agent that causes the reduction of oxygen atoms. The electron transfer is from
C6H12O6 to O2.
The three main stages of aerobic respiration are glycolysis, the Krebs’ cycle, and the electron
transport chain.
15. (a) Some science terms and concepts mentioned include: energy, power, electric current,
electric charge, nano-particles, micro-sized titanium spheres, and carbonized titanium.
(b) Claims include reducing fatigue and pain, improving blood circulation, and promoting
relaxation and refreshment.
(c) The evidence presented is all anecdotal and provided by a selection of people who are
involved in sports. [Several people and organizations represented appear to be paid
representatives of the company.]
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(d) In a double-blind study, randomly selected control and experimental groups of 500
volunteers are studied for changes which affect their circulatory system. The control
group gets an ordinary bracelet (placebo) identical in appearance to the titanium bracelet
that the experimental group receives. Both groups undergo a thorough physical exam by a
team of doctors to determine their health and in particular, the characteristics of their
circulatory system. The physical exam is conducted before the test, at the mid-point of
the test period, and at the end of the test period.
13.4 REDOX STOICHIOMETRY
Lab Exercise 13.C: Analyzing for Tin
(Page 598)
Purpose
The purpose of this lab exercise is to use the stoichiometric method in a redox chemical analysis.
Problem
What is the amount concentration of tin(II) ions in a solution prepared for research on toothpaste?
Analysis
OA
SOA
K+(aq), MnO4–(aq),
OA
OA
OA
H+(aq), Sn2+(aq)
H2O(l)
SRA
RA
2 [MnO4–(aq) + 8 H+(aq) + 5 e– o Mn2+(aq) + 4 H2O(l)]
5 [Sn2+(aq) o Sn4+(aq) + 2 e–]
–
+
2 MnO4 (aq) + 16 H (aq) + 5 Sn2+(aq) o 2 Mn2+(aq) + 8 H2O(l) + 5 Sn4+(aq)
12.4 mL
10.00 mL
0.0832 mol/L
c
0.0832 mol
= 1.03 mmol
nMnO = 12.4 mL u
4
1L
5
= 2.58 mmol
nSn 2+ = 1.03 mmol u
2
2.58 mmol
[Sn 2+ (aq)]
0.258 mol/L
10.00 mL
0.0832 mol MnO 4 5 mol Sn 2+
1
or [Sn 2+ (aq)] = 12.4 m L MnO 4 u
u
u
1 L MnO 4
2 mol MnO 4
10.00 m L Sn 2+
= 0.258 mol/L
According to the evidence and the stoichiometric analysis, the amount concentration of tin(II)
ions in the solution is 0.258 mol/L.
Web Activity: Canadian Achievers—Imants Lauks
(Page 598)
1. The FlexCard™ technology functions as an entire blood diagnostic laboratory on a credit cardsized device. FlexCard is a card format similar to smart cards that are used in ATM, cheque,
and phone cards. It can hold large amounts of information on a small chip. The device on the
FlexCard contains a silicon chip with biosensor circuits and only a small sample needs to be
used, in order for the results to be read by a wireless card reader.2. This technology provides a
low-cost, accurate, and complete blood diagnostic system that can be used quickly and easily by
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