Oxidation-Reduction Reactions
(Redox)
Copyright© by Houghton Mifflin
Company. All rights reserved.
What is the difference between acid/base
reactions and redox reactions?

Acid/base reactions
– proton transfer (p+)

Redox reactions
– electron transfer (e-)
Flow of electrons
 Electrons
respond to differences in
potential by moving from the
region of high potential to the
region of low potential.
Low
Ep
+
High
e Ep
-
Flow of electrons
high
electronegativity
Cl
-
low
electronegativity
e
+
Li
Lithium loses the e- tug-of-war with chloride.
Terminology
 Cations:
– positively charged ions
– generally metals
– NH4+ is the exception
 Anions:
– negatively charged ions
– non-metals
– complex ions
Oxidation:
–When a substances loses e
Reduction:
–When a substance gains e
Electron Transfer and Terminology
 Lose
electrons:
Oxidation
 Gain electrons:
Reduction.
oxidized
reduced
Half-reactions
Ca(s) →
(aq) +
– oxidation half reaction
2+
Ca
2e
2H+(aq) + 2e- → H2(g)
– reduction half reaction
Half-reactions add together
Ca(s) →
(aq) +
2H+(aq) + 2e- → H2(g)
2+
Ca
+
2e
Ca(s) + 2H+ + 2e-  Ca2+ + 2e- + H2(g)
+
2+
Ca(s) + 2H (aq)  Ca (aq) + H2(g)
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g)
Ca(s) has lost two e- to 2 H+(aq) to become Ca2+(aq). Ca(s)
has been oxidized to Ca2+(aq)
 At the same time 2 electrons are gained by 2 H+(aq) to
form H2(g) . We say H+(aq) is reduced to H2(g) .

Half-reactions add together
Cu(s) →
+
( Ag+(aq) + e- → Ag(s) ) x 2
2+
Cu (aq)
+
2e
Cu(s) + 2Ag+(aq) + 2e-  Cu2+(aq) + 2e- + 2Ag(s)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)

Iron comes from iron ore which is
taken out of the ground by mining.

The pure iron is obtained by heating the
ore at very high temperatures in a furnace
with limestone to remove impurities.
This heap of iron ore pellets will be used in steel production.
1. Hot blast from Cowper
stoves
2. Melting zone
3. Reduction zone of
ferrous oxide
4. Reduction zone of ferric
oxide
5. Pre-heating zone
6. Feed of ore, limestone
and coke
7. Exhaust gases
8. Column of ore, coke and
limestone
9. Removal of slag
10. Tapping of molten pig
iron
11. Collection of waste
gases
Why is gaining electrons called reduction?
 Reduction
originally meant the loss of
oxygen from a compound.
– 2 Fe2O3(s) + C(s) → 4 Fe(s) + 3 CO2(g)
 Iron
ore is reduced to metallic iron. The
size of the pile gets smaller, hence the
word reduction.
Why is losing electrons called oxidation?
 Oxidation
originally meant the
combination of an element with oxygen.
– 4 Fe(s) + 3 O2(g) → 2 Fe2O3(g)
– C(s) + O2(g) → CO2(g)
It Takes Two: Oxidation-Reduction
In
all reduction-oxidation (redox)
reactions, one species is reduced at
the same time as another is
oxidized.
Oxidizing Agent:
– the species which causes
oxidation is called the
oxidizing agent.
– substances that gains
electrons
– the oxidizing agent is
always reduced
Reducing
Agent:
– the species which causes
reduction is called the
reducing agent.
– the reducing agent is always
oxidized.
– substances that give up
electrons
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + Ag(s)
oxidated
R.A.
reduced
O.A.
1)
2)
3)
4)
5)
6)
A redox reaction is a chemical reaction in which
electrons are transferred.
Number of electrons lost by one species equals
number of electrons gained by the other species.
Reduction is a process in which e- are gained.
Oxidation is a process in which e- are lost
A reducing agent donates e- and is oxidized.
A oxidizing agent gains e- and is reduced.
WS 15-1
Electric potential (V),
o
E

Work that must be done to move an electric
charge between specified points.
 Electric potential differences are measured in
volts .
 Standard conditions:
– At 25oC with all ions at 1 mol/L
concentrations and all gases at 1.00 atm
pressure
Standard Reduction Potentials
 We
cannot measure the potential
of an individual half-cell!
 We assign a particular cell as
being our reference cell and then
assign values to other electrodes
on that basis. ( H2 – half cell )
The Standard Hydrogen
electrode
• Eo (H+(aq)/H2(g)) half-cell = 0.000 V
ep{H2(g)} = 1.00 atm
H2 (g)
[H+] = 1.00 mol/L
Pt gauze
Electric potential (V),
o
E

If the net potential is a positive number then the
reaction is spontaneous. Products are favoured.

If the net potential is a negative number then the
reaction is non-spontaneous. Reactants are favoured.

Half cell potentials are not doubled or tripled as per
balancing. We are only comparing potentials.
Only one of these two reactions is possible.
Which one?
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
Cu2+(aq) + 2 Ag(s) → Cu(s) + 2 Ag+(aq)
Use data table values, electrical
potential, on page 7 of your data
books. (2009)
Compare the two half reactions that
make up the reaction.
+
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq
Cu2+(aq) + 2e- → Cu (s)
Eo = +0.34
2Ag(s) → 2Ag+(aq) + 2e- Eo = -0.80
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
Eo = -0.46
Negative potential, non-spontaneous
Compare the two half reactions that
make up the reaction.
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Cu(s) → Cu2+(aq) + 2eEo = -0.34
2Ag+(aq) + 2e- → 2Ag(s)
Eo = +0.80
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
Eo = +0.46
Positive potential, spontaneous
Problem
 Write
the oxidation/reduction
half reactions and the net ionic
equation when zinc is placed in
Ni(NO3)2 solution. Identify the
O.A. and R.A. and state if the
reaction is spontaneous or nonspontaneous.
Problem
 Ni(NO3)2
Spectator ion
→ Ni2+(aq) + 2NO3- (aq)
Zn(s) + Ni2+(aq) → ?
Zn(s) → Zn2+(aq) + 2e- +0.76
 Reduction: Ni2+(aq) + 2e- → Ni(s) - 0.26
 Oxidation:
Add half reactions
Problem
Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
R.A.
O.A.
+0.50
Positive
potential,
spontaneous
SOA
high attraction
for electrons
decreasing
strength
low attraction
for electrons
decreasing
strength
SRA
Spontaneous shortcut
 Locate
the O.A. on the left and
the R.A. on the right of the table.
If the O.A. is higher up on the
table than the R.A. then the
reaction is spontaneous.
Problem
 Explain
what happens when nickel is
placed in a zinc nitrate solution.
Ni(s) + Zn2+(aq) →
R.A.
O.A.
Zn +2 (aq) is the strongest Oxidizing agent
and therefore is Reduced
 Zn+2 (aq) + 2e  Zn (s)
 Ni(s) is the strongest Reducing agent and
 Therefore is Oxidized
 Ni(s)  Ni +2 (aq) + 2e (must reverse)

NET REDOX REACTION

ADD REACTIONS/REMOVE ELECTRONS
 Zn2+ (aq) + 2e  Zn(s)
- 0.76 V
 Ni (s)  Ni 2+ (aq) + 2e
+ 0.26 V (r)
Zn2+ (aq) Ni(s)  Zn(s) + Ni2+ (aq) - 0.50V
 NON SPONTANEOUS REACTION

On the table
 Ni(s)





 R.A.
is above the
O.A.
 Non-spontaneous
Zn2+(aq) 
WS 15-23
DISPROPORTIONATION
redox reactions in which the
oxidizing agent and the
reducing agent are the same
species
2 H2O2 (l) ------> 2 H2O (l) + 2 O2 (g)

O ( -1)
O ( -2)
O ( 0)
 Oxygen
-------- > reduced / OA
 Oxygen ---------------------> oxidized / RA
Predicting redox reactions
1)
2)
3)
4)
5)
6)
7)
List all species present.
Choose the strongest oxidizing and reducing
agent. Watch for acids ( H +) Also water H2O
Write the reduction half reaction, as written in
the data book.
Write the oxidation half reaction, reverse the
equation in the data book.
Balance number of electrons.
Add the two half reactions together to form the
net ionic equation.
Predict if reaction is spontaneous or not.
Problems

A mixture of bromine gas and chlorine gas is added
to a solution of copper (II) sulphate and a copper
strip.
Br2(g)
Cl2(g) OA
H20(l)
Cu2+(aq)
Cu(s)
RA
SO42-(aq)
Cl2(g) + 2e- → 2 Cl-(aq)
Cu(s) → Cu2+(aq) + 2eCl2(g) + Cu(s) → 2 Cl-(aq) + Cu2+(aq)
Problems

Lead is placed in a zinc nitrate solution.
NO3-(aq)
H20(l)
Zn2+(aq) OA
Pb(s) RA
Non-spontaneous
OA is below RA
Problems

A few drops of Hg(l) are dropped into a solution
which is 1.0 M in both sulphuric acid and potasium
permanganate.
MnO4-(aq) OA
SO42-(aq)
H20(l)
K+(aq)
Hg(l) RA
H+(aq)
Problems

A few drops of Hg(l) are droped into a solution
which is 1.0 M in both sulphuric acid and potasium
permanganate.
( MnO4-(aq) + 8 H+(aq) + 5e- → Mn2+(aq) + 4 H2O(l)) x2
( Hg(l) → Hg2+(aq) + 2e- ) x5
2MnO4-(aq) + 16H+(aq) + 5Hg(l) → 2Mn2+(aq) + 8H2O(l) + 5Hg2+(aq)
General Rules
 Metal
(+) ions are oxidizing agents.
 Nonmetal (-) ions are reducing
agents.
 Metal elements are reducing agents.
 Nonmetal elements are oxidizing
agents.
Building a redox table (method one)

One can use experimental evidence to determine the
relative strengths of oxidizing and reducing agents.
 The greater the number of spontaneous reactions, the
stronger the oxidizing agent.
 This means we can rank oxidizing agents according to the
number of spontaneous reactions.
 By convention the strongest oxidizing agent is at the top
left in a redox table and the strongest reducing agent is at
the bottom right of the table.
Reduction Table
SOA
SRA
Problem: Make a redox table
Cu2+(aq)
Mg2+(aq)
Ag+(aq)
Zn2+(aq)
Cu(s)
Mg(s)
Ag(s)
Zn(s)
____
____
____
√
____
√
____
____
____
√
____
√
____
____
____
____
____
√
____
____
√
____
Virtual Lab
REDOX TABLE

Mg(s) is most reactive - 3 times
 Zn(s) is second
- 2 times
 Cu(s) is third
- 1 times
 Ag(s) is most unreactive
0 times
 THEREFORE Mg oxidizes easiest and

is the strongest reducing agent
 Mg is placed in the lower right hand side
 then Zn(s) …Cu(s) and Ag(s) is last.
 Now you fill in the reduction reactions
Activity Series - Redox Table
Strongest
oxidizing
agents
Weaker
Ag+(aq) + 1e-  Ag(s)
Cu2+(aq) + 2e-  Cu(s)
Zn2+(aq) + 2e-  Zn(s)
Mg2+(aq) + 2e-  Mg(s)
Weaker
Strongest
reducing
agents
Mg is the strongest reducing agent as it oxidizes the most
and is on the lower right side of the table.
Redox Table Building (method two)
 The
spontaneity rule is used to order
the oxidizing agents to produce a redox
table.
 Consider the following redox equations
which represent spontaneous reactions
from an experiment. From this
evidence construct a redox table.
Redox Table Building
3 Equations given
all are spontaneous reactions
2+
Co(s) + Pd (aq)
Pd(s) + Pt2+(aq)
2+
Mg(s) + Co (aq)
→
→
→
2+
Co (aq) + Pd(s)
Pd2+(aq) + Pt(s)
2+
Mg (aq) + Co(s)
Work with one equation at a time.
Redox Table Building
RA
OA
2+
2+
Co(s) + Pd (aq) → Co (aq) + Pd(s)
2+
Pd (aq) + 2 e → Pd(s) reduced/stays as is
2+
Co (aq) + 2 e → Co(s) oxidized/reverse
OA is above RA
Reverse the oxidation reaction to
compare Pd(s) with Co(s)
NOTE POSITION OF THE REDUCING
AGENTS

Pt(s) is above Pd(s) in position

Pt(s)

Pd(s) stronger reducing agent
Redox Table Building
RA
OA
2+
2+
Pd(s) + Pt (aq) → Pd (aq) + Pt(s)
Pt2+(aq) + 2 e- → Pt(s)
Pd2+(aq) + 2 e- → Pd(s)
OA is above RA
spontaneous reaction
NOTE POSITION OF THE REDUCING
AGENTS


Pt(s) is above Pd(s) in position
Pt (s)



Pd (s)
Stronger reducing agent
Redox Table Building
RA
OA
2+
2+
Mg(s) + Co (aq) → Mg (aq) + Co(s)
2+
Co (aq) + 2 e → Co(s) ( stays as is)
2+
Mg (aq) + 2 e → Mg(s) (reverse)
OA is above RA
spontaneous reaction
MAKING THE TABLE

USING THE 4 REDUCING REAGENTS
 Mg(s) Pt(s)
Co(s) Pd(s)
 Place into the correct order using the
previous information you collecte
 Note the position with respect to each
other….
 Make the complete ½ reactions
 MAKE THE REDOX TABLE
Redox Table Building
+ 2 → Pt(s)
Pd2+(aq) + 2 e- → Pd(s)
2+
Co (aq) + 2 e → Co(s)
2+
Mg (aq) + 2 e → Mg(s)
2+
Pt
(aq)
e
Oxidation States
 Some
reactions are not adequately
explained with redox theories.
 Chemists have developed a method
of electron bookkeeping to
describe the redox of molecules
and complex ions.
Oxidation States
 Oxidation
state:
– apparent net charge that an atom
would have if electron pairs
belonged entirely to the more
electronegative atom
 Oxidation
number:
– a positive or negative number
assigned to a combined atom
according to a set of arbitrary
numbers.
Assigning Oxidation Numbers
1) Oxidation numbers for all uncombined
elements (elemental/standard) = 0
•
K(s) = 0
N2(g) = 0
S8(s) = 0
2) Oxidation number for all simple ions is
equal to the charge of the ion.
•
Br1-(aq) = -1
Fe3+(aq) = +3
3) Oxidation for oxygen in a compound = -2
(except for peroxides = -1)
H2O(l)
H2O2(l)
-2
-1
Assigning Oxidation Numbers
4) Hydrogen in compounds = +1
(except hydrides = -1)
H2O(l)
NaH(s)
5) Sum of oxidation numbers in a compound
is = 0
•
H2O(l) → (2 x +1) + (1 x -2) = 0
6) Sum of oxidation numbers in a complex
ion = charge of ion.
•
NH4+(aq) → (4 x +1) + (1 x -3) = +1
Example
What is the oxidation number for
carbon in CO32-(aq) ?
CO# + 3 OO# = -2
? + 3 (-2) = -2
? + -6 = -2
? = +4
Example
What is the oxidation number for
carbon in C6H12O6 ?
6 CO# + 12 HO# + 6 OO# = 0
6 (?) + 12 (+1) + 6 (-2) = 0
6 (?) + 12 + -12 = 0
? =0
If you have 2 unknowns?
First ionize the substance in water. Then work out
the two resulting ions separately.
+2 +6 -2
Example:
CuSO4(aq)
CuSO4(s) → Cu2+(aq) + SO42-(aq)
Simple ion
Solve as a
complex ion
4(-2) + 1(x) = -2
x = +6
Assign oxidation numbers to chlorine in each of the
following chemicals.
 HCl(aq)
 -1
 Cl2(g)
0
 NaClO (s)
 Cl-(aq)
 +1
 HClO3(aq)
 +5
 ClO3(aq)
 +6
 -1
Who cares about oxidation numbers?
 Determining
oxidation numbers
allows us to predict electron
transfer.
 If there is an increase in oxidation
number then oxidation occurs.
 If there is a decrease in oxidation
number then reduction occurs.
Problem
 Determine
the oxidation
numbers for all atoms and ions
in the following redox equation
and indicate which substance is
undergoing oxidation and
reduction.
Problem
O is reduced
+1
0
-2
-2
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
+4
-4
C is oxidized
+1
Identifying Redox Reactions….Which of
these are Redox reactions?
+4 -2
+4 -2
1) N2O4(g) → 2NO2(g) NO
0
+1 -1
+1 -1
0
2) Cl2(g) + 2NaBr(aq) → 2 NaCl(aq) + Br2(l) YES
+2 -1
+1 +6 -2
+1 -1
+2 +6 -2
3) PbCl2(aq)+ K2SO4(aq) → 2KCl(aq) + PbSO4(aq) NO
+1 -2 +1
+1 +6 -2
+1 +6 -2
+1 -2
4) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) NO
0
+1 -2
+1 -2+1
0
5) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) YES
Ion electron method
Under Acidic conditions
1. Identify oxidized and reduced species
Write the half reaction for each.
2. Balance the half rxn separately except H & O’s.
Balance: Oxygen by adding H2O
Balance: Hydrogen by adding H+
Balance: Charge by adding e 3. Multiply each half reaction by a coefficient.
Must have the same # of e- in both half-rxn.
4. Add the half-rxn together, the e - should cancel.
Balancing Half Reactions
MnO4¯ → Mn2+
MnO4¯ → Mn2+ + 4 H2O
8 H+ + MnO4¯ → Mn2+ + 4 H2O
5 e¯ + 8 H+ + MnO4¯ → Mn2+ + 4 H2O
Note All elements balance
Note The charge is balanced / LHS 2 + = RHS 2 +
Balancing Half Reactions
NO3¯ → HNO2
NO3¯ → HNO2 + H2O
3 H+ + NO3¯ → HNO2 + H2O
2 e¯ + 3 H+ + NO3¯ → HNO2 + H2O
NOTE : CHARGE IS BALANCED / LHS = RHS ( 0)
Example: Acidic Conditions
I- + S2O8-2  I2 + S2O42 Oxidation:
 Reduction:
I-  I2
S2O8-2  S2O422 I-  I2 + 2 e-
 Bal.
chemical and e- :
 Bal.
chemical O and H : 8e- + 8H+ + S2O8-2  S2O42- + 4H2O
 Mult
1st rxn by 4:
 Add
rxn 1 & 2:
8I-  4 I2 + 8e8I-  4 I2 + 8e-
8e- + 8H+ + S2O8-2  S2O42- + 4H2O
 8I-
+ 8H+ + S2O8-2  4 I2 +
S2O42- + 4H2O
Example: Acidic Conditions
NO3- + Bi  NO2 + Bi3+
 Half Rxn (oxid):
Bi  Bi3+
 Half Rxn (red):
NO3-  NO2
chemical and e- :
Bi  Bi3+ + 3 e Bal. chemical O and H : 1e- + 2H+ + NO3-  NO2 + H2O
 Mult 2nd rxn by 3:
3e- + 6H+ + 3NO3-  3NO2 + 3H2O
 Bal.
 Add
Bi  Bi3+ + 3 e3 e- + 6 H+ + 3 NO3-  3NO2 + 3H2O
rxn 1 & 2:
Bi + 6 H+ + 3 NO3-  Bi3+ +
3 NO2 + 3 H2O
Redox Reactions - Ion electron method.
Under Basic conditions/ NOT PART OF PROGRAM
1. Procedure identical to that under acidic conditions
Balance the half rxn separately except H & O’s.
Balance Oxygen by adding H2O
Balance Hydrogen by H+ (ADD OH - TO BOTH SIDES )
Balance charge by e( NEUTRALIZE H + ) result is
OH - AND H + => H2O (water)
2. Mult each half rxn such that both half- rxn have same
number of electrons – MUST CANCEL OUT e 3. Add the half-rxn together, the e- should cancel.
Example: Basic Conditions
-1
+6
+3
0
H2O2 (aq) + Cr2O7-2(aq )  Cr 3+ (aq) + O2 (g)
6e- + 14H+ + Cr2O7-2 (aq)  2Cr3+ + 7 H2O
 oxid:
(H2O2 (aq)  O2 + 2H+ + 2e-) x 3
 red:
8 H+ + 3H2O2 + Cr2O72-  2Cr+3 + 3O2 + 7H2O
add:
8 OH-  8 OH3H2O2 + Cr2O72 - + 8H2O  2Cr+3 + 3O2 + 7H2O + 8OH-
3H2O2 + Cr2O72 - + H2O  2Cr+3 + 3O2 + 8OH-
Redox Stoichiometry

Can be used to predict or analyze a chemical
reaction.
 A method of reacting a solution with a known
concentration with a solution of unknown
concentration.
 Common oxidizing agents in redox reactions
–
–
–
–
MnO4-(aq) → Mn2+(aq)
purple
colorless
Cr2O72-(aq) → Cr3+(aq)
orange
green
Redox Stoichiometry
 In
a titration experiment all of the
Br-(aq) ions in an acidic solution were
oxidized to Br2(l) by a 0.0200 M
KMnO4(aq) solution. The volume of
Br-(aq) solution was 25.0 mL and the
volume of KMnO4(aq) was 15.0 mL.
Calculate the concentration of Br-(aq)
ions in solution.
MAKE HALF REACTIONS
 K+ MnO4- H+ Br- ions
 FIND SOA AND SRA / MAKE ½ RXN
 MnO4- + H+ is the SOA
 Br- is the SRA
Make half reactions / data book P. 11
continue next slide -- >
 KMnO4
We need a balanced chemical equation to do any
stoichiometry.
(
MnO4-(aq) + 8 H+(aq) + 5e- ↔ Mn2+(aq) + 4 H2O(l)) x2
( 2Br–(aq)
↔ Br2(l) + 2e- ) x 5
2 MnO4-(aq) + 16 H+(aq) + 10 Br-(aq) ↔ 2 Mn2+(aq) + 8 H2O(l) +5 Br 2(l)
0.0200 mol/L
0.025 L
0.015 L
0.0015 mol
0.0003 mol
C=n/V
c = 0.0600 M
U
K
10 Br2 MnO4 -
=
x
0.0003 mole
Br- = 0.0015 mole
Breathalyzer
 The
Breathalyzer device contains:
– A system to sample the breath of the suspect
– Two glass vials containing the chemical
reaction mixture
– A system of photocells connected to a meter to
measure the color change associated with the
chemical reaction
Breathalyzer
To measure alcohol, a suspect breathes into the
device. The breath sample is bubbled in one vial
through a mixture of sulfuric acid, potassium
dichromate, silver nitrate and water. The
principle of the measurement is based on the
following chemical reaction:
8H+ + Cr2O72- + 3C2H5OH → 2Cr3+ + 3C2H4O + 7H2O
yellow
blue

The sulfuric acid removes the alcohol from the
air into a liquid solution.
 The alcohol reacts with potassium dichromate to
produce:
– chromium sulfate
– potassium sulfate
– acetic acid
– water

The silver nitrate is a catalyst, a substance that
makes a reaction go faster without participating in it.
The sulfuric acid, in addition to removing the
alcohol from the air, also might provide the acidic
condition needed for this reaction.

During this reaction, the reddish-orange dichromate
ion changes color to the green chromium ion when
it reacts with the alcohol; the degree of the color
change is directly related to the level of alcohol in
the expelled air. To determine the amount of alcohol
in that air, the reacted mixture is compared to a vial
of unreacted mixture in the photocell system, which
produces an electric current that causes the needle
in the meter to move from its resting place. The
operator then rotates a knob to bring the needle back
to the resting place and reads the level of alcohol
from the knob -- the more the operator must turn the
knob to return it to rest, the greater the level of
alcohol.
yellow
blue
Bleaching Agents
 Bleaching
agents are compounds which are
used to remove color from substances such
as textiles. In earlier times textiles were
bleached by exposure to the sun and air.
Today most commercial bleaches are
oxidizing agents, such as sodium
hypochlorite (NaOCl) or hydrogen peroxide
(H2O2) which are quite effective in
"decolorizing" substances via oxidation.
Bleaching Agents
 The
action of
these bleaches
can be
illustrated in the
following
simplified way:
Bleaching Agents
 The
decolorizing action of bleaches is due in
part to their ability to remove electrons
which are activated by visible light to
produce the various colors. The
hypochlorite ion (OCl-), found in many
commercial preparations, is reduced to
chloride ions and hydroxide ions forming a
basic solution as it accepts electrons from
the colored material as shown below.
OCl- + 2e- + HOH → Cl- + 2 OH-
Bleaching Agents

Bleaches are often combined with "optical
brighteners". These compounds are quite
different from bleaches. They are capable of
absorbing wavelengths of ultraviolet light
invisible to the human eye, and converting these
wavelengths to blue or blue-green light. The blue
or blue-green light is then reflected by the
substance making the fabric appear much
"whiter and brighter" as more visible light is
seen by the eye.
Photosynthesis

An example of naturally-occuring biological
oxidation-reduction reactions is the process of
photosynthesis. It is a very complex process
carried out by green plants, blue-green algae, and
certain bacteria. These organisms are able to
harness the energy contained in sunlight, and via
a series of oxidation-reduction reactions, produce
oxygen and sugar. The overall equation for the
photosynthetic process may be expressed as:
6 CO2(g) + 6 H2O(l) → C6H12O 6(aq) + 6 O2(g)
Photosynthesis
 The
equation is the net result of two
processes. One process involves the splitting
of water. This process is really an oxidative
process that requires light, and is often
referred to as the "light reaction". This
reaction may be written as:
12 H2O(l)
→ 6 O2(g) + 24 H+(aq) + 24e-
Photosynthesis

Think of the light reaction, as a process by which
organisms capture and store radiant energy as
they produce oxygen gas. This energy is stored in
the form of chemical bonds of compounds such as
NADPH and ATP.
 The energy contained in both NADPH and ATP
is then used to reduce carbon dioxide to glucose.
This reaction, shown below, does not require
light, and it is often referred to as the dark
reaction.
6 CO2 + 24 H+ + 24 e- → C6H12O6 + 6 H2O
Photosynthesis
 The
chemical bonds present in glucose
also contain a considerable amount of
potential energy. This stored energy is
released whenever glucose is broken
down to drive cellular processes.
Photosynthesis
 In simplest terms, the process of
photosynthessis can be viewed as onehalf of the carbon cycle. In this half,
energy from the sun is captured and
transformed into nutrients which can be
utilized by higher organisms in the food
chain. The release of this energy during
the metabolic re-conversion of glucose to
water and carbon dioxide represents the
second half of the carbon cycle and it may
be referred to as "oxidative processes".
Cellular Respiration
O is reduced
+1 -2
0
-2
-2
C6H12O6(aq) + 2 O2(g) → Energy + CO2(g) + 2 H2O(g)
0
+4
C is oxidized
+1