REDOX PROCESSES 91 OXIDATION AND REDUCTION ESSENTIAL IDEA Redox reductionoxidation reactions play a key role in many chemical and biochemical processes NATURE OF SCIENCE 19 How evidence is used changes in the definition of oxidation and reduction from one involving specific elemen ID: 552364
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Slide1
TOPIC 9REDOX PROCESSES
9.1
OXIDATION AND REDUCTIONSlide2
ESSENTIAL IDEA
Redox (reduction-oxidation) reactions play a key role in many chemical and biochemical processes.
NATURE OF SCIENCE (1.9)
How evidence is used – changes in the definition of oxidation and reduction from one involving specific elements (oxygen and hydrogen), to one involving electron transfer, to one invoking oxidation numbers is a good example of the way that scientists broaden similarities to general principles.Slide3
INTERNATIONAL-MINDEDNESS
Access to a supply of clean drinking water has been recognized by the United Nations as a fundamental human right, yet it is estimated that over one billion people lack this provision. Disinfection of water supplies commonly uses oxidizing agents such as chlorine or ozone to kill microbial pathogens.Slide4
THEORY OF KNOWLEDGE
Chemistry has developed a systematic language that has resulted in older names becoming obsolete. What has been lost and gained in this process?
Oxidation states are useful when explaining redox reactions. Are artificial conversions a useful or valid way of clarifying knowledge?Slide5
UNDERSTANDING/KEY IDEA 9.1.A
Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number.Slide6
Early definitions for oxidation and reduction were based upon observations of the gain and loss of oxygen and hydrogen during chemical change.
Oxidation: gain of oxygen or loss of hydrogen
Reduction: loss of oxygen or gain of hydrogen
It is now recognized that oxidation/reduction occur whenever there is a shift in electron density from one atom to another, whether complete or partial.Slide7
Oxidation is the loss of electrons.Atom gets more positive.
Reduction is the gain of electrons.
Atom gets more negative.
Reactions that involve oxidation and reduction are called
redox
reactions.
You can never have one without the other.Slide8
Lions roar http://www.youtube.com/watch?v=_22gJ5kB31k
Lion is also Know as LEOSlide9Slide10
Oxidation Reduction
Oxidation is the loss of electrons;
Reduction is the gain of electrons
Oxidation and reduction go together.
Whenever a substance loses electrons and another substance gains electronsSlide11
Oxidation Numbers
Oxidation Numbers are a system that we can use to keep track of electron transfersSlide12
Oxidation Numbers
Oxidation numbers always refer to single atoms
The oxidation number of an uncombined element is always 0
O
2
, H
2
, Ne
Zn
The oxidation number of Hydrogen is usually +1
Hydrides are an exception They are -1
HCl, H
2
SO
4
The oxidation number of Oxygen is usually -2
Peroxides are an exception They are –1 In OF
2
oxygen is a +2
H
2
O, NO
2
, et
Oxidation numbers of monatomic ions follow the charge of the ion
O
2-
, Zn
2+The sum of oxidation numbers is zero for a neutral compound. It is the charge on a polyatomic ionLiMnO4SO42-Slide13
Practice Assigning Oxidation Numbers
NO
2
N
2
O
5
HClO
3
HNO
3
Ca(NO
3
)
2
KMnO
4
Slide14
Practice Assigning Oxidation Numbers
NO
2
N= +4, O = -2
N
2
O
5
N = +5, O = -2
HClO
3
H=+1, Cl=+5, O = -2
HNO
3
H=+1, N = +5, O = -2
Ca(NO
3
)
2
Ca=+2, N =+5, O= -2
KMnO
4
K=+1, Mn=+7, O= -2Slide15
Practice Assigning Oxidation Numbers
Fe(OH)
3
K
2
Cr
2
O
7
CO
3
2-
CN
-
K
3
Fe(CN)
6
Slide16
Practice Assigning Oxidation Numbers
Fe(OH)
3
Fe =+3, O=-2, H=+1
K
2
Cr
2
O
7
K=+1, Cr=+6, O=-2
CO
3
2-
C=+4, O =-2
CN
-
C=+4, N=-5
K
3
Fe(CN)
6
K=+1, Fe=+3, C=+4, N=-5 Slide17
Naming compound using oxidation number
SO
2
Sulfur dioxide
SO
3
Sulfur trioxide
More Correctly now named as
Sulfur(IV) oxide
Sulfur(VI) oxide
Roman Numerals are used the oxidation numbers in the name of the compounds Slide18
Names of some compounds
H
2
SO
3
H
2
SO
4
PCl
3
PCl
5
N2O NO2 CuSO4 Cu2SO4 Sulfuric(IV) acidSulfuric(VI) acidPhosphorous (III) chloride
Phosphorous (V) chloride
Nitrogen (I) oxide
Nitrogen (IV) oxide
Copper(II) Sulfate
Copper(I) Sulfate Slide19
Names of some ions
SO
4
2-
Cr
2
O
7
2-
CrO
4
2-
PO
42- MnO42- Sulfate(VI) ionsDichromate (VI) ionChromate (VI) ionPhosphate(V) ions Manganate (VII) ionSlide20
Using Oxidation Numbers
Careful examination of the oxidation numbers of atoms in an equation allows us to determine what is oxidized and what is reduced in an oxidation-reduction reactionSlide21
Using Oxidation Numbers
An
increase
in the oxidation number indicates that an atom has
lost electrons
and therefore
oxidized.
A
decrease
in the oxidation number indicates that an atom has
gained electrons
and therefore
reduced
Example
Zn + CuSO
4
ZnSO
4
+ Cu
0
+2
+6-2
+2
+6-2
0
Zn: 0 + 2 Oxidized Cu: +2 0 ReducedSlide22Slide23
UNDERSTANDING/KEY IDEA 9.1.B
An oxidizing agent is reduced and a reducing agent is oxidized.Slide24
Redox reactions ALWAYS involve the simultaneous oxidation of one reactant with the reduction of another through the transfer of electrons.
The reactant causing the oxidation of the other reactant is called the oxidizing agent. (It is the one that was reduced.)
The reactant causing the reduction of the other reactant is called the reducing agent. (It is the one that is oxidized.)Slide25
APPLICATION/SKILLS
Be able to identify the species oxidized and reduced and the oxidizing and reducing agents, in redox reactions.Slide26
If you can determine which one was reduced (got more negative), then the other three are easy to find.Remember that if it got reduced, then it is the oxidizing agent and vice versa.Slide27
Oxidizing agent Vs Reducing agent
Oxidizing Agent
Reducing Agent
Also called as
Oxidant
An oxidizing agent takes electron away from other substance .
Also called as
reductants.
A reducing agent give electrons to other substance Slide28
Example
Identify the oxidizing and reducing agents in the following equation.
2Al + 3PbCl
2
→ 2AlCl
3
+ 3Pb
First write down the oxidation numbers.
0 +2 -1 +3 -1 0
2Al + 3PbCl
2
→ 2AlCl
3
+ 3PbAl went from 0 to +3 – oxidized (red agent)Pb went from +2 to 0 – reduced (ox agent)*Note the oxidizing agent is the whole compound (PbCl2), not just the element.*Slide29
Identify the substances being oxidized and reduced.
Cu + 2Ag
+
→
2Ag + Cu
2+
Copper goes from a zero to a +2.
It got more positive so it was oxidized.
Silver goes from +1 to zero.
It got more negative so it was reduced.
Slide30Slide31
Exercise
For each of the following reactions find the element oxidized and the element reduced
Cl
2
+ KBr
KCl + Br
2
Cu + HNO
3
Cu(NO
3
)
2
+ NO
2
+ H
2
O
HNO
3
+ I
2
HIO3 + NO2Slide32
Exercise
For each of the following reactions find the element oxidized and the element reduced
Cl
2
+ KBr
KCl + Br
2
0
+1
-1
+1
-1
0
Br increases from –1 to 0 -- oxidized
Cl decreases from 0 to –1 -- Reduced
K remains unchanged at +1 Slide33
Exercise
For each of the following reactions find the element oxidized and the element reduced
Cu + HNO
3
Cu(NO
3
)
2
+ NO
2
+ H
2
O
0
+1
+5
-2
+2
+5-2
+4
–2 +1-2
Cu increases from 0 to +2. It is oxidized
Only part of the N in nitric acid changes from +5 to +4. It is reduced
The nitrogen that ends up in copper nitrate remains unchangedSlide34
Exercise
For each of the following reactions find the element oxidized and the element reduced
HNO
3
+ I
2
HIO
3
+ NO
2
1
+5
-2 0
+1
+5
-2
+4
-2
N is reduced from +5 to +4. It is reduced
I is increased from 0 to +5 It is oxidized
The hydrogen and oxygen remain unchanged.Slide35
APPLICATION/SKILLS
Be able to deduce redox reactions using half-equations in acidic or neutral solutions.Slide36
½ EQUATIONSSince oxidation cannot occur without reduction, you can identify each type of reaction within a
redox
reaction.
These are called the ½ equations.
Include the electrons in your equations.Slide37
EXAMPLEDeduce the two ½ equations for the following reaction:
Zn + Cu
2+
→ Zn
2+
+ Cu
Zinc goes from 0 to 2+ (more pos – oxidized)
Cu goes from 2+ to 0 (more
neg
– red)
Oxidation: Zn → Zn
2+
+ 2e
-
Reduction: Cu
2+
+ 2e- → CuSlide38
Disproportionation Reactions
when the same element is both oxidized and reduced
.
3HNO
2
HNO
3
+ 2NO + H
2
O
3ClO
3-
2Cl
1-
+ ClO
31-Cu2O(s)+H2SO4(aq)→Cu(s)+CuSO4(aq)+H2O(l)Slide39
RULES FOR BALANCING ½ REACTIONS
WRITE DOWN THE ½ REACTIONS.
BALANCE ALL ELEMENTS BESIDES OXYGEN AND HYDROGEN.
BALANCE OXYGEN WITH WATER.
BALANCE HYDROGENS WITH H
+
DETERMINE OVERALL CHARGES ON BOTH SIDES.
BALANCE THE CHARGES ON EACH SIDE WITH e
-
MULTIPLY BY A FACTOR IF NEEDED SO THE e
- CANCEL. ADD EQUATIONS AND CANCEL OUT COMMON FACTORS ON BOTH SIDES. FOR BASIC SOLUTIONS, ADD OH
-
TO BOTH SIDES TO GET RID OF H
+
’S.Slide40
EXAMPLEBalance the following
redox
reaction:
NO
3
-
+ Cu
→ NO + Cu
2+
Step 1: Write the ½ reactions
Ox: Cu → Cu
2+
Red: NO
3
-
→ NOStep 2: Balance all elements besides H and O Cu and N are balanced already.Step 3: Balance oxygen with water Ox: Cu → Cu2+ Red: NO3
-
→ NO +
2H
2
OSlide41
Step 4: Balance
hydrogens
with H
+
Ox: Cu → Cu
2+
Red:
4H
+
+ NO
3
-
→ NO + 2H
2
OStep 5: Determine overall charges on both sides Ox: (zero) Cu → Cu2+ (+2)
Red:
(+3)
4H
+
+ NO
3
-
→ NO + 2H
2
O
(zero)
Step 6: Balance the charges on each side with e-
Ox: Cu → Cu
2+ + 2e- Red: 3e- + 4H+ + NO3- → NO + 2H2O Slide42
Step 7: Multiply by a factor if needed to cancel out the e-
3
(Cu → Cu
2+
+ 2e
-
)
2
(3e
-
+ 4H
+
+ NO
3
- → NO + 2H2O) 3Cu → 3Cu2+ + 6e-
6e
-
+ 8H
+
+ 2NO
3
-
→ 2NO + 4H
2
O
Step 8: Add both equations and cancel out anything common on both sides.
3Cu → 3Cu
2+
+ 6e- 6e- + 8H+ + 2NO3- → 2NO + 4H2O 8H+ + 3Cu + 2NO3- → 2NO + 3Cu2+ + 4H2OSlide43
Balancing Redox Equations 1
Cu + HNO
3
Cu(NO
3
)
2
+ NO
+ H
2
O
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms
Check your work. Make sure that both the atoms and charges balanceSlide44
Balancing Redox Equations 2
HNO
3
+ I
2
HIO
3
+ NO
2
+ H
2
O
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms
Check your work. Make sure that both the atoms and charges balanceSlide45
Balancing Ionic Redox Equations 3
Fe
2+
+MnO
4
-
Mn
2+
+Fe
3+ (acidic)
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms. You may need to add H+ or OH- and H
2
O in ionic equations
Check your work. Make sure that both the atoms and charges balanceSlide46
Balancing Ionic Redox Equations 5
VO
2
+
+ Zn
VO
2+
+ Zn
2+
(Acidic)
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms. You may need to add H
+
or OH
-
and H
2
O in ionic equations
Check your work. Make sure that both the atoms and charges balanceSlide47
UNDERSTANDING/KEY IDEA 9.1.C
Variable oxidation numbers exist for transition metals and for most main-group non-metals.Slide48
APPLICATION/SKILLS
Be able to deduce the oxidation states of an atom in an ion or a compound.Slide49
GUIDANCE
Oxidation states should be represented with the sign before the given number, not after like as in ions.Slide50
SIGN CONVENTIONOxidation numbers
are shown with the +or- in front of the number such as +7.
Ions
with their charges are shown with the number first followed by the charge such as 2+.Slide51
OXIDATION NUMBERSThe concept of oxidation numbers provides a way to keep track of electrons in
redox
reactions.
They are not really charges, but we will use them to assign numbers in covalent compounds to show that the overall charge in a compound is zero.
It is essentially a convention to assign which element has electron control.Slide52
OXIDATION NUMBER RULES1. Elements by themselves are zero.
2. In simple ions, the oxidation number is the same as its charge.
3. Oxidation numbers in a neutral compound must add up to zero.
4. Oxidation numbers in a polyatomic ion must add up to the charge of the ion.
You can predict many oxidation numbers from the periodic table.Slide53
GUIDANCE
Know that the oxidation state of hydrogen can be -1 in metal hydrides and oxygen can be -1 in peroxides.Slide54
COMMON EXAMPLESFluorine is always (-1).Oxygen is usually (-2) except:
Peroxides (-1)
OF
2
(+2)
Hydrogen is (+1) except:
Metal hydrides
NaH
(-1)
Chlorine is (-1) except:
When combined with O or F, then it is (+1).Slide55
EXAMPLES
Find the oxidation states for the elements in H
2
SO
4
and Na
2
C
2
O
4.
H
2
SO
4
: H +1 (2 of them for a total of +2)
O -2 (4 of them for a total of -8) S +6 (make the overall charge 0)Na2C2O4: Na +1 (2 of them for a total of +2) O -2 (4 of them for a total of -8) C +3 (2 of them to equal +6)Slide56
APPLICATION/SKILLS
Be able to deduce the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals.Slide57
GUIDANCE
Oxidation number and oxidation state are often used interchangeably, though IUPAC does formally distinguish between the two terms. Oxidation numbers are represented by Roman numerals according to IUPAC.Slide58
IONIC NAMING RULESYou know how to do this – Yeah!!!
Use Roman numerals when you have more than one choice of ion.
You may see covalent compounds using Roman numerals.
NO – nitrogen monoxide or
nitrogen II oxide
NO
2
– nitrogen dioxide or
nitrogen IV oxideSlide59
UNDERSTANDING/KEY IDEA 9.1.D
The activity series ranks metals according to the ease with which they undergo oxidation.Slide60
Not all oxidizing and reducing agents are the same strength.
Their strength depends upon how easily they lose or gain electrons.Slide61
Reducing AgentsMetals tend to give up electrons forming positive ions so they cause other elements to become more negative or to be reduced.
This is why metals are commonly reducing agents.
More reactive metals lose their electrons more readily so they are stronger reducing agents.Slide62
Reactivity Series
A sample reactivity series
Mg strongest reducing agent
Al (most readily oxidized)
Zn
Fe
Pb
Cu
Ag weakest reducing agent
(least readily oxidized)Slide63
Oxidizing AgentsNonmetals tend to gain electrons forming negative ions so they cause other elements to become more positive or to be oxidized.
This is why nonmetals are commonly oxidizing agents.
More reactive nonmetals gain their electrons more readily so they are stronger oxidizing agents.Slide64
Reactivity Series
A sample reactivity series
F
2
strongest oxidizing agent
Cl
2
(most readily reduced)
Br
2
I
2
weakest oxidizing agent
(least readily reduced)Slide65
You do not have to memorize the activity series, but you will have to interpret information from one.
Remember that any metal or nonmetal above another will cause a displacement reaction.
If it is not higher on the activity series, the reaction will not occur.
If you were given a series of
viable reactions,
you should be able to determine the activity series.Slide66
SAMPLE REDOX TITRATIONSRedox titrations are commonly used in the food and beverage industry.
They are very similar to acid-base titrations, but sometimes do not need an indicator as a color change can naturally occur at the equivalence point.Slide67
Analysis of Iron with Manganate VII
5Fe
2+
+ MnO
4
-
+ 8H
+
→ 5Fe
3+
+ Mn
2+
+ 4H
2
O
This reaction used potassium permanganate in an acidic solution as the oxidizing agent, which oxidizes Fe
2+ ions to Fe3+ ions. The manganese is reduced from Mn7+ to Mn2+.The reaction does not need an indicator as it goes from a deep purple to colorless at equivalence.http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/redoxNew/redox.htmlSlide68
Iodine-thiosulfate reaction
Several different redox titrations use an oxidizing agent to react with excess iodine ions to form iodine.
2I
-
+ oxidizing agent → I
2
+ reduced product
Examples of oxidizing agents are: KMnO
4
, KIO
3
, K
2
Cr
2
O
7 and NaOCl.The I2 is then titrated with sodium thiosulfate using starch as an indicator.The starch indicator is added during the titration (not at the start) and forms a deep blue color by forming a complex with the free I2. As the I2 is reduced to I- ions, the blue color disappears marking equivalence.Slide69
Redox equations:
oxidation: 2S
2
O
3
2-
→ S
4
O
6
2-
+ 2e
-
reduction: I
2
+ 2e- → 2I-Overall equation: 2S2O32 + I2 → 2I- + S4O62-Slide70
UNDERSTANDING/KEY IDEA 9.1.E
The Winkler Method can be used to measure biochemical oxygen demand (BOD), used as a measure of the degree of pollution in a water sample.Slide71
The dissolved oxygen content of water is one of the most important indicators of its quality.
As pollution increases, the dissolved oxygen content decreases as the oxygen is used by bacteria in decomposition reactions.
The BOD (biological oxygen demand) is used as a means of measuring the degree of pollution.
BOD is defined as the amount of oxygen used to decompose the organic matter in a sample of water over a specified time period, usually 5 days at a specified temperature.Slide72
APPLICATION/SKILLS
Be able to apply the Winkler Method to calculate BOD.Slide73
WINKLER METHODThe Winkler Method uses redox titrations to measure the dissolved oxygen in water to calculate the BOD.
The dissolved oxygen in the water is “fixed” by the addition of a manganese II salt such as MnSO
4
.
Reaction of this salt with oxygen in basic solution causes oxidation of
Mn
(II) to higher oxidation states such as
Mn
(IV).
2Mn
2+
+ O
2
+ 4OH
-
→ 2MnO2 + 2H2OSlide74
Acidified iodide ions are added to the solution
and are oxidized by the
Mn
(IV) to I2.
MnO
2
+ 2I
-
+ 4H
+
→ Mn
2+
+ I
2
+ 2H2O4. The iodine produced is then titrated with sodium thiosulfate as described earlier. 2S2O32 + I2 → 2I- + S4O6
2-
So we can see that for every 1 mole of O
2
in the water, 4 moles of S
2
O
3
2-
are used.
Work the sample problem page
384Slide75
Citations
International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford.
Higher Level Chemistry
. 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.
Most of the information found in this power point comes directly from this textbook.
The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.