oxidation-reduction reactions (redox)
DESCRIPTION
Oxidation-Reduction Reactions (Redox). 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. - PowerPoint PPT PresentationTRANSCRIPT
Oxidation-Reduction Reactions(Redox)
What is the difference between acid/base What is the difference between acid/base reactions and redox reactions?reactions and redox reactions?
Acid/base reactions –proton transfer (p+)
Redox reactions– electron transfer (e-)
Flow of electronsFlow of electrons
Electrons respond to differences in potential by moving from the region of high potential to the region of low potential.
-+
High Ep
Low Ep
e-
Flow of electronsFlow of electrons
Cl
low electronegativity
high electronegativity
e-Li
Lithium loses the e- tug-of-war with chloride.
TerminologyTerminology
Cations:– positively charged ions– generally metals–NH4
+ is the exceptionAnions:
– negatively charged ions– non-metals–complex ions
Oxidation: –When a substances loses e-.
Reduction: –When a substance gains e-.
oxidizedreduced
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-reactionsHalf-reactions
Ca(s) → Ca2+(aq) + 2e-
–Oxidation half reaction2H+
(aq) + 2e- → H2(g)
– reduction half reaction
Half-reactions add togetherHalf-reactions add together
Ca(s) → Ca2+(aq) + 2e-
2H+(aq) + 2e- → H2(g)
Ca(s) + 2H+ + 2e- Ca2+ + 2e- + H2(g)
Ca(s) + 2H+(aq) Ca2+
(aq) + H2(g)
+
Half-reactions add togetherHalf-reactions add together
Cu(s) → Cu2+(aq) + 2e-
Ag+(aq) + e- → Ag(s)
Cu(s) + 2Ag+(aq) + 2e-
Cu2+(aq) + 2e- + 2Ag(s)
Cu(s) + 2Ag+(aq) Cu2+
(aq) + 2Ag(s)
+ ( ) x 2
Electron Transfer and TerminologyElectron Transfer and TerminologyLose Electrons:
OxidationGain Electrons:
Reduction.
OIL - OXIDATION IS LOSS OF ELECTRONS
RIG - REDUCTION IS GAIN OF ELECTRONS
IronIron
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.
The molten iron is taken out of the bottom of the furnace. It is further processed depending on how it is to be used.
Why is gaining electrons called reduction?Why is gaining electrons called reduction?
Reduction originally meant the loss of oxygen from a compound.
–2Fe2O3(s) + C(s) → 4Fe(s) + 3CO2(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?Why is losing electrons called oxidation?
Oxidation originally meant the combination of an element with oxygen.
4Fe(s) + 3O2(g) → 2Fe2O3(g)
C(s) + O2(g) → CO2(g)
It Takes Two: Oxidation-ReductionIt Takes Two: Oxidation-Reduction
In all reduction-oxidation (redox) reactions, one species is reduced at the same time as another is oxidized.
It Takes Two: Oxidation-ReductionIt Takes Two: Oxidation-Reduction
Oxidizing Agent:– the species which causes oxidation is called the oxidizing agent.
– substances that gains electrons– the oxidizing agent is always reduced
It Takes Two: Oxidation-ReductionIt Takes Two: Oxidation-Reduction
Reducing Agent:– the species which causes reduction is called the reducing agent.
– the reducing agent is always oxidized.
– substances that give up electrons
ExampleExample
Cu(s) + 2 Ag+(aq) → Cu2+
(aq) + Ag(s)
oxidated reduced
R.A. O.A.
Summary: Redox TheorySummary: Redox Theory1) A redox reaction is a chemical reaction in which
electrons are transferred.
2) Number of electrons lost by one species equals number of electrons gained by the other species.
3) Reduction is a process in which e- are gained.
4) Oxidation is a process in which e- are lost
5) A reducing agent donates e- and is oxidized.
6) A oxidizing agent gains e- and is reduced.
WS 15-1
Only one of these two reactions is possible. Only one of these two reactions is possible. Which one?Which one?
Cu(s) + 2 Ag+(aq) → Cu2+
(aq) + 2 Ag(s)
Cu2+(aq) + 2 Ag(s) → Cu(s) + 2 Ag+
(aq) Data table values EO, page 7 of your data books.
1) Cu 2+(aq) + 2e- -- >> Cu (s) + 0.34 EO
2) Ag (s) -- >> Ag +(aq) + e- - 0.80 EO (R)
1) Cu(s) -- >> Cu 2+ (aq) + 2 e- -0.34 EO ( R)
2) Ag +(aq) + e- -- >> Ag(s)
+0.80 EO
Electric potential (V), EElectric potential (V), Eoo
the electric potential under standard conditions of a half-reaction in which reduction is occurring.
Standard conditions: – 25oC with all ions at 1 M
concentrations and all gases at 1 atm pressure
Standard Reduction PotentialsStandard 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.
[H+] = 1.00
H2 (g)
e-
Pt gauze
The Standard Hydrogen electrode
• Eo (H+/H2) half-cell = 0.000 V
p{H2(g)} = 1.00 atm
Electric potential (V), EElectric potential (V), Eoo
If the net potential is a positive number then the reaction is spontaneous.
If the net potential is a negative number then the reaction is non-spontaneous.
Half cell potentials are not doubled or tripled as per balancing. We are only comparing potentials.
Compare the two half reactions that Compare the two half reactions that make up the reaction.make up the reaction.
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+
(aq)
Cu2+ + 2e- → Cu Eo = 0.34 2Ag → 2Ag+ + 2e- Eo = -0.80
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+
(aq)
Eo = -0.46Negative potential, non-spontaneous
+
Compare the two half reactions that Compare the two half reactions that make up the reaction.make up the reaction.
Cu(s) + 2Ag+(aq)
→ Cu2+(aq) + 2Ag(s)
Cu(s) → Cu2+ + 2e- Eo = -0.342Ag+ + 2e- → 2Ag Eo = 0.80
Cu(s) + 2Ag+(aq)
→ Cu2+(aq) + 2Ag(s)
Eo = 0.46Positive potential, spontaneous
ProblemProblem
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 non-spontaneous.
ProblemProblem
Ni(NO3)2 → Ni2+(aq) + 2NO3-
(aq)
Zn(s) + Ni2+(aq) → ?
Oxidation: Zn(s) → Zn2+(aq) + 2e- +0.76
Reduction: Ni2+(aq) + 2e- → Ni(s) - 0.26
Spectator ion
Add half reactions
A piece of zinc is placed in a solution of nickel nitrateNi(NO3)2
ProblemProblem
Zn(s) + Ni2+(aq) → Zn2+
(aq) + Ni(s) +0.50
R.A. O.A.Positive potential, spontaneous
Zn is Oxidized
Ni2+ is Reduced
NOTE*** Spontaneous shortcutNOTE*** Spontaneous shortcutLocate 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.
O.A.
R.A.
SPONTANEOUSREACTION
O.A.
R.A.
NON-SPONTANEOUSREACTION
highest attraction
for electrons
weak attraction
for electrons
ProblemProblem
Explain what happens when nickel is placed in a zinc nitrate solution.
Ni(s) + Zn2+(aq) → ? + ?
R.A. O.A.
NICKEL Ni ZINC NITRATE Zn2+ and NO3 -
REDUCING AGENT OXIDIZING AGENT ARE ON LEFT SIDE
On the tableOn the table
Ni(s)
Zn2+(aq)
R.A. is above the O.A.
NON SPONTANEOUS
DisproportionationDisproportionation
redox reactions where the OA and the RA are the same species. ( p 577 – text)
Example: Fe2+ (aq) and Fe 2+
(aq)
Fe2+ (aq)
+ 2 e - Fe (s) reduction of Fe2+
2[ Fe2+ (aq) Fe3+
(aq) + e - ] oxidation of Fe2+
3 Fe 2+(aq) Fe(s) + 2 Fe3+
(aq) net reaction
NON – SPONTANEOUS REACTION
DISPROPORTIONATIONDISPROPORTIONATION
TRY THE REACTION WHERE Cu 1+ ACTS
AS THE OXIDIZING AND REDUCING AGENTS
TRY THE REACTION WHERE Cr2+ ACTS AS THEAS THE OXIDIZING AND REDUCING AGENTS
Predicting redox reactionsPredicting redox reactions1) List all species present.2) Choose the strongest oxidizing and reducing
agent.3) Write the reduction half reaction, as written in
the data book.4) Write the oxidation half reaction, reverse the
equation in the data book.5) Balance number of electrons.6) Add the two half reactions together to form the
net ionic equation.7) Predict if reaction is spontaneous or not.
ProblemsProblemsA mixture of bromine gas and chlorine gas is added
to a solution of copper (II) sulphate and a copper strip. (water) ( CuSO4) (Br2(g)) (Cl2(g) ) ( Cu(s) )
NOTE ( Go down S.O.A. / Go up S.R.A.)Br2(g)
Cl2(g)
H20(l)
Cu2+(aq)
Cu(s)
SRA *
SOA *
Cl2(g) + 2e- → 2 Cl-(aq)
Cu(s) → Cu2+(aq) + 2e-
Cl2(g) + Cu(s) → 2 Cl-(aq) + Cu2+
(aq)
SPONTANEOUS
ProblemsProblems
Lead is placed in a zinc nitrate solution.(list species)
NO3-(aq)
H20(l)
Zn2+(aq)
Pb(s)SRA
SOA
Non-spontaneous OA is below RA
Zn2+(aq) + 2e- Zn (s)
Pb(s) Pb 2+(aq) + 2e-
Zn2+(aq) + Pb(s) Zn(s) + Pb2+
O.A.
R.A.
ProblemsProblemsA few drops of Hg(l) are dropped into a solution
which is 1.0 M in both sulphuric acid and potasium permanganate. MnO4
-(aq)
SO42-
(aq)
H20(l)
K+(aq)
Hg(l)
H+(aq)
RA
OA
H+ hydrogen ion(From acid)
O.A.
R.A
YES
ProblemsProblemsA 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)
Hg(l) → Hg2+(aq) + 2e-
2MnO4-(aq) + 16H+
(aq) + 5Hg(l) → 2Mn2+(aq) + 8H2O(l) + 5Hg2+
(aq)
( ) x2 ( ) x5
Oxidized
(Balance electrons) LHS = RHS
General RulesGeneral 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 Building a redox table (method one)(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.
Building a redox tableBuilding a redox tableThis 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.
Problem: Make a redox tableProblem: Make a redox table
Cu(s) Mg(s) Ag(s) Zn(s)
Cu2+(aq) ____ ____ ____ ____
Mg2+(aq) ____ ____ ____ ____
Ag+(aq) ____ ____ ____ ____
Zn2+(aq) ____ ____ ____ ____
√
√√
√
√√
Virtual Lab
O.A.R.A.
Activity SeriesActivity Series
Ag+(aq) + 1e- Ag(s)
Cu2+(aq) + 2e- Cu(s)
Zn2+(aq) + 2e- Zn(s)
Mg2+(aq) + 2e- Mg(s)
Example: Redox ReactionExample: Redox ReactionBased on the activity series, which reactions are spontaneous?a) Ag(s) + Mg(NO3)2 (aq) ? ions
b) Cu (s) + AgNO3 (aq) ? ions
c) Zn (s) + Mg(NO3)2(aq) ? ions
d) Mg(s) + Mg(NO3)2 (aq) no reaction
Example: Redox ReactionExample: Redox Reaction
a) Ag(s) vs. Mg2+(aq) chart (ions + metal)
Ag(s) (RA) is above Mg2+(aq) (OA) non-
spontaneous
b) Cu(s) vs. Ag+(aq)
Cu(s) (RA) is below Ag+(aq) (OA)
spontaneousc) Zn(s) vs. Mg2+
(aq)
Zn(s) is above Mg2+(aq)
non-spontaneousWS 15-28
Redox Table BuildingRedox Table Building (method two)(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 BuildingRedox Table Building
Co(s) + Pd2+(aq) → Co2+
(aq) + Pd(s)
Pd(s) + Pt2+(aq) → Pd2+
(aq) + Pt(s)
Mg(s) + Co2+(aq) → Mg2+
(aq) + Co(s)
Work with one equation at a time.
ASSUME ALL REACTION ARE SPONTANEOUS
Redox Table BuildingRedox Table Building
Co(s) + Pd2+(aq) → Co2+
(aq) + Pd(s)
Pd2+(aq) + 2 e- → Pd(s)
Co2+(aq) + 2 e- → Co(s)
OA is above RAspontaneous reaction
RA OA
Redox Table BuildingRedox Table Building
Pd(s) + Pt2+(aq) → Pd2+
(aq) + Pt(s)
Pt2+(aq) + 2 e- → Pt(s)
Pd2+(aq) + 2 e- → Pd(s)
OA is above RAspontaneous reaction
Redox Table BuildingRedox Table Building
Mg(s) + Co2+(aq) → Mg2+
(aq) + Co(s)
Co2+(aq) + 2 e- → Co(s)
Mg2+(aq) + 2 e-→ Mg(s)
OA is above RA
spontaneous reaction
Redox Table BuildingRedox Table Building
Pt2+(aq) + 2 e- → Pt(s)
Pd2+(aq) + 2 e- → Pd(s)
Co2+(aq) + 2 e- → Co(s)
Mg2+(aq) + 2 e-→ Mg(s)
Redox StoichiometryRedox StoichiometryCan 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– Cr2O7
2-(aq) → Cr3+
(aq)
– orange green
Redox StoichiometryRedox 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.
We need a balanced chemical equation to do any We need a balanced chemical equation to do any stoichiometry.stoichiometry.
MnO4-(aq) + 8 H+
(aq) + 5e- ↔ Mn2+(aq) + 4 H2O(l)
2Br–(aq) ↔ Br2(l) + 2e-
( )x2
( )x5
2 MnO4-(aq) + 16 H+
(aq) + 10 Br-(aq) ↔ 2 Mn2+
(aq) + 8 H2O(l) +5 Br 2(l)
0.0200 M
0.015 L
0.0003 mol
0.025 L
0.0015 mol2:10
c = 0.0600 M
.0003 mol/ 2 x 10 = .0015 mol
.0015mol / .025L
Oxidation StatesOxidation 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 StatesOxidation StatesOxidation 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 NumbersAssigning 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 NumbersAssigning Oxidation Numbers
4) Hydrogen in compounds = +1 H2O(l) (except hydrides = -1) 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
ProblemsProblems
What is the oxidation number for Na(s)?
What is the oxidation number for H2(g) ?
What is the O# for hydrogen in HCl(g) ?
What is the O# for Na+(aq) ?
What is the O# for oxygen in H2O(l) ?
0 0+1+1 -2
Assign oxidation numbers to chlorine in each of the Assign oxidation numbers to chlorine in each of the
following chemicals.following chemicals.
HCl(aq)
Cl2(g)
NaClO (s)
Cl-(aq)
HClO3(aq)
ClO3(aq)
-10+1-1+5+6
Assign oxidation numbers to maganese in each of the Assign oxidation numbers to maganese in each of the
following chemicals.following chemicals.
Mn(s)
MnO2(s)
MnO4-2
(aq)
Mn2+(aq)
Mn2O7(aq)
MnCl2(s)
0+4+6+2+7+2
ExampleExample
What is the oxidation number for carbon in CO3
2-(aq) ?
CO# + 3 OO# = -2 ? + 3 (-2) = -2 ? + -6 = -2 ? = +4
ExampleExample
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
Who cares about oxidation numbers?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.
ProblemProblem
Determine the oxidation numbers for all atoms and ions in the following redox equation and indicate which substance is undergoing oxidation and reduction.
ProblemProblem
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
+1
-4
0 -2
+4
-2
+1
C is oxidized
O is reduced
Ion electron methodIon electron methodUnder Acidic conditionsUnder 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 ReactionsBalancing Half Reactions
MnO4¯ → Mn2+
MnO4¯ → Mn2+ + 4 H2O
8 H+ + MnO4¯ → Mn2+ + 4 H2O
5 e¯ + 8 H+ + MnO4¯ → Mn2+ + 4 H2O
Example: Acidic ConditionsExample: Acidic ConditionsI- + S2O8
-2 I2 + S2O42-
Half Rxn (oxid): I- I2
Half Rxn (red): S2O8-2 S2O4
2-
Bal. chemical and e- : 2 I- I2 + 2 e-
Bal. chemical O and H : 8e- + 8H+ + S2O8-2 S2O4
2- + 4H2O
Mult 1st rxn by 4: 8I- 4 I2 + 8e-
Add rxn 1 & 2: 8I- 4 I2 + 8e-
8e- + 8H+ + S2O8-2 S2O4
2- + 4H2O
8I- + 8H+ + S2O8-2 4 I2 + S2O4
2- + 4H2O
Example: Acidic ConditionsExample: Acidic ConditionsNO3
- + Bi NO2 + Bi3+
Half Rxn (oxid): Bi Bi3+
Half Rxn (red): NO3- NO2
Bal. chemical and e- : Bi Bi3+ + 3 e-
Bal. chemical O and H : 1e- + 2H+ + NO3- NO2
+ H2OMult 2nd rxn by 3: 3e- + 6H+ + 3NO3
- 3NO2 + 3H2O
Add rxn 1 & 2: Bi Bi3+ + 3 e-
3 e- + 6 H+ + 3 NO3- 3NO2 + 3H2O
Bi + 6 H+ + 3 NO3- Bi3+
+ 3 NO2 + 3 H2O
1. Procedure identical to that under acidic conditions1. Procedure identical to that under acidic conditionsBalance the half rxn separately except H & O’s.
Balance Oxygen by H2O
Balance Hydrogen by H+
Balance charge by e-
2. Mult each half rxn such that both half- rxn have same 2. Mult each half rxn such that both half- rxn have same number of electronsnumber of electrons
3. Add the half-rxn together, the e3. Add the half-rxn together, the e-- should cancel. should cancel.
4. Eliminate H+ by adding: 4. Eliminate H+ by adding: OHOH-- to both sides to both sides
Redox Reactions - Ion electron method.Redox Reactions - Ion electron method.Under Basic conditionsUnder Basic conditions
Example: Basic ConditionsExample: Basic Conditions
H2O2 (aq) + Cr2O7-2
(aq ) Cr 3+ (aq) + O2 (g)
red: 6e- + 14H+ + Cr2O7-2
(aq) 2Cr3+ + 7 H2O
oxid: (H2O2 (aq) O2 + 2H+ + 2e-) x 3
8 H+ + 3H2O2 + Cr2O72- 2Cr+3 + 3O2
+ 7H2O
add: 8 OH- 8 OH-
3H2O2 + Cr2O72 - + 8H2O 2Cr+3 + 3O2 + 7H2O + 8OH-
3H2O2 + Cr2O72 - + H2O 2Cr+3 + 3O2 + 8OH-
+3+6 0-1
BreathalyzerBreathalyzer
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
BreathalyzerBreathalyzer
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 AgentsBleaching 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 AgentsBleaching Agents
The action of these bleaches can be illustrated in the following simplified way:
Bleaching AgentsBleaching AgentsThe 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 AgentsBleaching 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.
PhotosynthesisPhotosynthesis
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)
PhotosynthesisPhotosynthesis
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-
PhotosynthesisPhotosynthesis
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
PhotosynthesisPhotosynthesis
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.
PhotosynthesisPhotosynthesisIn simplest terms, the process of
photosynthessis can be viewed as one-half 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
C6H12O6(aq) + 2 O2(g) → Energy + CO2(g) + 2 H2O(g)
+1
0
0 -2
+4
-2
+1
C is oxidized
O is reduced
-2