Transcript 12.1 Redox Intro

Redox
Reactions
Oxidation
In the 1700s, chemists determined that when metals corrode
and get a “rust” layer, this rust is a combination of the metal
with oxygen. Atoms of the metal join chemically with atoms
of oxygen in the air.
4 Fe
+ 3 O2

2 Fe2O3
Different Colours of Iron Rust
Iron rust, Fe2O3.X H2O, forms various hydrated compounds
(with different numbers [X] of water molecules arranged
around a central Fe2O3, making many different-coloured rust
compounds.
Reduction
When iron III oxide (rust) is heated to a high temperature with
carbon, the products are pure iron metal and carbon dioxide.
In the 1800s this reaction was named a reduction because the
iron oxide was reduced to pure iron.
(coke)
2Fe2O3 + 3C  4Fe + 3CO2
(iron III oxide – “iron ore”)
The First Meaning of Oxidation and Reduction
The first understanding of oxidation was that it involved the
joining of oxygen to an element or compound. Reduction
was seen as the removal of oxygen from a compound.
4 Fe
+ 3 O2

2 Fe2O3
In the above reaction, iron is oxidized because oxygen is
added to it.
2Fe2O3 + 3C  4Fe + 3CO2
In the above reaction, iron is reduced because oxygen is
removed from it.
The Second Meaning for Oxidation and Reduction
In the late 1800s, chemists studying oxidation and
reduction reactions concluded that oxidation and
reduction involved the giving or taking of electrons.
Oxidation of iron:
0
4 Fe

+ 3 O2
+3
-2
2 Fe2O3
In the above reaction, each iron atom in the compound gives
up or loses three electrons.
Reduction of iron:
+3
-2
0
2Fe2O3 + 3C  4Fe + 3CO2
In the above reaction, each iron atom in the compound iron is
gaining three electrons.
Oxidizing and Reducing Agents
An oxidizing agent is the substance helping to oxidize
another substance. An oxidizing agent is reduced itself as it
helps oxidize another substance. A reducing agent is
oxidized as it helps to reduce another substance.
Redox Reactions
When oxidation and reduction were understood as the loss
and gain of electrons, it was realized that when any
substance is oxidized, another substance must be reduced.
For this reason oxidation and reduction reactions are called
Redox reactions.
Half Reactions
In a Redox reaction, half reactions are written to show just
the oxidation or reduction reactions of the overall reaction.
Example :
Redox reaction: Cu (s) + 2AgNO3 (aq)  2Ag (s) + Cu(NO3)2
Reduction half reaction: 2Ag+ + 2e -  2Ag (s)
Oxidation half reaction: Cu (s)  Cu2+ + 2e -
Copper and Silver Nitrate Reaction
Cu + 2Ag + + 2NO3-  2Ag + Cu2+ + 2NO3 Reduction half reaction: 2Ag+ + 2e -  2Ag (s)
Oxidation half reaction: Cu (s)  Cu2+ + 2e -
LEO says GER
The mnemonic LEO says GER is used to remember that a
Loss of Electrons is an Oxidation while a Gain of Electrons is
a Reduction. Since
Zn  Zn2+ + 2 eCl2 + 2 e-  2 Cl-
(Half reaction showing zinc oxidized)
(Half reaction showing chlorine reduced)
Electronegativity
The total attraction an atom has for an electron is called its
electronegativity. Fluorine, F, has the largest
electronegativity, followed by oxygen.
Oxidation Numbers
An oxidation number is the charge that an atom would have if the
species containing the atom were made up of ions. In reality, this
is not true for species which have covalent bonds. However the
concept is a useful one, so atoms are assigned oxidation numbers
based on electronegativity differences. The greater an atom’s
elecronegativity, the stronger its attraction for shared electrons.
Shared electrons are assigned to the atom with greater
electronegativity which then has a negative sign.
Ex: In the compound, Cl2O, the two shared electrons are assigned
to oxygen, giving it a 2- oxidation number. Cl has a +1 oxidation
number. Oxidation numbers are nearly identical to valence
numbers.
+1 -2
Cl2O
Note F has the largest electroneg.
Rules for Oxidation Numbers
1.
2.
3.
The sum of all positive and negative charges must equal
the overall charge of the species.
F always has a -1 oxidation number.
Oxygen always has a 2- Ox.# except with F where it has a
2+ and except in peroxides where it has 1-.
2+ 1-
OF2
1+
4.
5.
6.
7.
8.
2-
1+
1-
Na2O sodium oxide Na2O2 sodium peroxide
H always has Ox# = 1 except in metal hydrides (-1). NaH
The atoms in an element have an Ox# of zero.
Alkali metals have an Ox# = +1, alkaline earths Ox# = +2
Halogens have an Ox# = -1 except when with a more
electronegative element (oxygen) where their Ox# = +1
The charge on a single ion is its Ox#
Assigning Oxidation Numbers
In a chemical species, the sum of all the oxidation number
charges for each individual atom in the species must equal
the overall charge of the species.
Example: Determine the oxidation number of P in H4P2O7.
Solution: H has a 1+ and O a 2-. The overall species charge
equals zero. Let x equal Ps Ox#.
This means 4(1+) + 2(x) + 7(2-) = 0
Simplified, 4 + 2x -14 = 0 or, 2x -10 = 0
2x = 10
Thus x = +5
What is the Oxidation Number of S in SO42- ?
Solution:
Let x = sulfur’s Ox#
1(x) + 4(2-) = 2X - 8 = -2
X = +6
Oxidation Number Changes in Redox Reactions
Consider:
0
4 Fe
0
+ 3 O2
3+

2-
2 Fe2O3
Note that as iron has been oxidized its Ox# has risen.
Note that as oxygen has been reduced its Ox# has been
reduced.
In redox reactions, elements with rising oxidation numbers
are being oxidized while elements with falling oxidation
numbers are being reduced.
The Standard Reduction Potentials Chart
An excerpt from the Standard Reduction Potentials is
included below. Note the following patterns:
1.
2.
3.
4.
5.
6.
In general, metals are found on the bottom right of the chart.
Halogens and oxyions are at the top right of the chart.
Metals with more than one Ox.# have more than one ½ reaction
HO is found at the top left and bottom right.
In general, top reactions favour the forward reaction.
In general, bottom reactions favour the reverse reaction.
Voltages for ½ Reactions in the SRP Chart
In the Standard Reduction Potentials (SRP) chart, the voltages
given are for forward reduction ½ reactions. If the reactions are
written in reverse, the voltage sign must be reversed also. To
obtain the theoretical voltage from an oxidation ½ reaction and a
reduction ½ reaction, add their voltages. For example, consider
Cu2+ + 2e-  Cu (+0.34) and Pb  Pb2+ + 2e- (+0.13)
The overall theoretical voltage expected from this reaction is:
+0.34 + 0.13 = +0.47 . A + voltage like this reaction would be
spontaneous, producing energy.
The Spontaneity of Redox Reactions
Redox Reactions will be spontaneous if they involve a
forward ½ reaction to the right that is above a reverse ½
reaction to the left.
Consider these ½ r. Ag+ + e-  Ag and Zn  Zn2+ + 2e2Ag+ + 2e-  2Ag
Zn  Zn2+ + 2e- These two ½ reactions are spontaneous together
Net overall reaction:
2Ag+ + Zn  Zn2+ + 2Ag
Full chemicals: 2Ag+ + 2NO3- + Zn  Zn2+ + 2Ag + 2NO3-
Nonspontaneous Reactions
Consider Cu2+ and Zn2+ as possible reactants in a redox
reaction. These reactants are both on the left side of the
SRP chart. In this case, there is no possible oxidation
reaction for the two possible reduction reactions. No
reaction will occur.
Nonspontaneous Reactions : Second Case
Consider Ag and Zn as possible reactants in a redox reaction.
These reactants are both on the right side of the SRP chart.
In this case, there is no possible reduction reaction for the
two possible oxidation reactions. No reaction will occur.
Nonspontaneous Reactions : Third Case
Consider Cu and Zn2+ as possible reactants in a redox reaction. In
this case the forward reduction reaction is below the reverse
oxidation reaction. No reaction will occur. Note that the overall
voltages are Cu  Cu2+ + 2e- (- 0.34) and Zn2+ + 2e-  Zn (0.76) Total theoretical V = - 1.1 V which means energy must be
added. This is not spontaneous.
The Rule for Spontaneous Reactions
A reaction will be spontaneous only if there is a reactant to
be reduced (on left side) which is above a reactant to be
oxidized (right side).
1. From the SRP table below, find a reactant that will
spontaneously react with Ni2+.
2. From the SRP table, find a reactant that will
spontaneously react with Pb.
Hydrogen Ion Reductions
A number of reactions require H+ to occur. Without the H+,
these reactions do NOT occur.
There is just ONE reaction in which H+ is the only substance
being reduced:
The reduction of H has been arbitrarily set to zero as the
standard against which all other reductions are measured.
(More on this in a later slide)
Balancing Half Reactions
The steps in balancing half reactions are as follows:
1. Balance all the MAJOR atoms by inspection (Not O and H)
2. Balance the Oxygen atoms by adding H2O molecules.
3. Balance the Hydrogen atoms by adding H+.
4. Balance the overall charge by adding electrons.
The mneumonic, Major OH- is used to remember the steps.
OH-
Sample Balancing Problem 1
Balance the half-reaction, RuO2  Ru
1. The Major Ru is balanced on both sides (no O and H balancing yet)
2. Balance O by adding H2O : RuO2  Ru + 2H2O
3. Balance H by adding H+ : RuO2 + 4H+  Ru + 2H2O
4. Balance the charge by adding electrons to one side (-) .
The left side has 4+ while right side has 0. So adding 4- to
left side brings it to zero also, balancing the charge. The
rule for charge balancing is that the electrons are always
added to the side which has the greater charge.
RuO2 + 4H+ + 4e-  Ru + 2H2O
The above reaction occurs in an acid solution
What kind of reaction is this half-reaction?
Sample Balancing Problem 2
Balance the half-reaction Cr2O72-  Cr3+
1. Balance the Major Cr: Cr2O72-  2Cr3+
2. Balance O by adding H2O : Cr2O72-  2Cr3+ + 7H2O
3. Balance H by adding H+ : Cr2O72- + 14H+  2Cr3+ + 7H2O
4. Balance the charge by adding electrons to one side (-) .
The left side has 12+ total while right side has 6+ total. So
adding 6- to left side brings it to 6+ like the right side.
The rule for charge balancing is that the electrons are
always added to the side which has the greater charge.
Cr2O72- + 14H+ + 6e-  2Cr3+ + 7H2O
The above reaction occurs in an acid solution
What kind of reaction is this half-reaction?
Sample Balancing Problem 3
Balance the half-reaction, Pb  HPbO2- , in a basic solution
1. Balance the Major Pb: Pb  HPbO2- Pb is balanced
2. Balance O by adding H2O: Pb + 2H2O  HPbO23. Balance H by adding H+: Pb + 2H2O  HPbO2- + 3H+
4. Balance the charge by adding electrons to one side (-) . The
left side has 0 total while right side has 2+ total. So adding 2to the right side brings it to 0 like the left side. Electrons are
added to the side with the greater charge.
Pb + 2H2O 
HPbO2- + 3H+ + 2e5. Add the same number of OH- to both sides so that the H+ are
neutralized to form water.
6. Pb + 2H2O + 3OH-  HPbO2- + 3H+ + 3OH- + 2ePb + 2H2O + 3OH-  HPbO2- + 3H2O + 2e7. Cross-cancel H2O : Pb + 3OH-  HPbO2- + H2O + 2eThe above reaction is an oxidation happening in an alkaline solution.
Sample Balancing Problem 4
Balance the half-reaction, H2  , in an acidic solution
1. Balance the Major: No major to balance
2. Balance O by adding H2O: No O to balance
3. Balance H by adding H+: H2  2H+
4. Balance the charge by adding electrons to one side (-) . The
left side has 0 total while right side has 2+ total. So adding 2to the right side brings it to 0 like the left side. Electrons are
added to the side with the greater charge.
H2  2H+ + 2eThe above reaction is an oxidation happening in an acidic solution.
Balancing Redox Equations : The ½ Reaction Method
There are two methods for balancing redox equations. You
need only use one, but it is advantageous to know both
methods.
Balance Os + IO3-  OsO4 + I2 (acidic solution)
1. Split the reaction into its half reactions:
Os  OsO4
IO3-  I2
2. Balance the half reactions using the Major OH- method.
Os + 4H2O  OsO4 + 8H+ + 8e2IO3- + 12H+ + 10e-  I2 + 6H2O
3. Multiply half reactions to ensure # e lost = # e gained
5(Os + 4H2O  OsO4 + 8H+ + 8e-)
4(2IO3- + 12H+ + 10e-  I2 + 6H2O)
Balancing Redox Eq. : The ½ Reaction Method (Cont.)
From step 3:
5(Os + 4H2O  OsO4 + 8H+ + 8e-)
4(2IO3- + 12H+ + 10e-  I2 + 6H2O)
4. Add the two half reactions and simplify:
5Os + 20H2O  5OsO4 + 40H+ + 40e8IO3- + 48H+ + 40e-  4I2 + 24H2O
Overall Balanced Equation:
5Os + 8IO3- + 8H+  5OsO4 + 4I2 + 4H2O
A Second Example of Redox Equation Balancing
Balance MnO4- + C2O42-  MnO2 + CO2 (in basic solution)
1. Make ½ reactions
MnO4-  MnO2 & C2O42-  CO2
2. Balance ½ reac.
3e- + 4H+ + MnO4-  MnO2 + 2H2O
C2O42-  2CO2 + 2e3. Mult for = #e- 2(3e- + 4H+ + MnO4-  MnO2 + 2H2O)
3(C2O42-  2CO2 + 2e-)
4. Add ½ reac.
6e- + 8H+ + 2MnO4-  2MnO2 + 4H2O
3C2O42-  6CO2 + 6eSimp. 8H+ + 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2
5. Add OH- to both sides for a basic solution
8OH- + 8H+ + 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2 + 8OH8H2O + 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2 + 8OH-
4H2O + 2MnO4- + 3C2O42-  2MnO2 + 6CO2 + 8OH-
A Third Example of Redox Balancing: Disproportionation
Balance ClO2-  ClO3- + Cl- (in basic solution)
1. Make ½ reactions
ClO2-  ClO3- & ClO2-  Cl2. Balance ½ reac.
H2O + ClO2-  ClO3- + 2H+ + 2e4e- + 4H+ + ClO2-  Cl- + 2H2O
3. Mult for = #e2(H2O + ClO2-  ClO3- + 2H+ + 2e-)
1(4e- + 4H+ + ClO2-  Cl- + 2H2O)
4. Add ½ reac.
2H2O + 2ClO2-  2ClO3- + 4H+ + 4e4e- + 4H+ + ClO2-  Cl- + 2H2O
Simp.
3ClO2-  2ClO3- + Cl5. No OH- to add because no H+ present. (Reaction happens
in basic solution but equation doesn’t show this)
3ClO2-  2ClO3- + ClThis reaction is called a disproportionation reaction because
one species is both reduced and oxidized.
A Second, Faster Method For Balancing Redox Eq.
Balance
U4+ + MnO4-  Mn2+ + UO22+
1. Balance majors (done already)
2. Assign ox.#s
+4
+7
+2
+6
U4+ +
MnO4-  Mn2+ +
UO22+
3. Note total changes in ox#s
1U goes up 2 (oxidation) while 1 Mn goes down 5
(reduction)
ΔU = +2 and ΔMn = -5
4. Find cross-mult. #s so that the
product Δ up = product Δ down (Because #e lost = #e gained)
2(-5Δ) = -10 = (Mn Δ)
5(+2Δ) = +10 = (U Δ)
5. Apply cross-mult #s to reactants and then products
5U4+ +
2MnO4-  2Mn2+ +
5UO22+
A Second Method For Balancing Redox (Cont)
6.
Balance the O atoms by adding H2O and the H atoms by
adding H+
5U4+ +
2MnO4-  2Mn2+ +
5UO22+
2 H2O + 5U4+ + 2MnO4-  2Mn2+ + 5UO22+
2 H2O + 5U4+ + 2MnO4-  2Mn2+ + 5UO22+ + 4H+
Check that the charge is balanced (It should be because the
oxidation # rise was set equal to the oxidation # drop).
A final charge balance check helps to detect any balancing
errors made.
Ex 2 of Redox Equation Balancing by Ox. #s
Balance
Zn + As2O3  AsH3 + Zn2+ (in basic solution)
1. Balance the majors
Zn + As2O3  2AsH3 + Zn2+
2. Assign ox.#s
0
+3
-3
+2
Zn + As2O3  2AsH3 + Zn2+
3. Note total changes in ox#s
1 Zn goes up 2 while 2As go down 6
total Δ up (1)(2) = +2Δ
total Δ down (2)(-6) = -12Δ
4. Find cross-mult. #s so that the total pr. Δ up = total Δ
down
6(2+Δ) = +12
and 1(-12) = -12
5. Apply cross-mult #s to reactants and then products
6Zn + 1As2O3  1.2AsH3 + 6Zn2+
6Zn + 1As2O3  2AsH3 + 6Zn2+
Ex 2 of Redox Equation Balancing by Ox. #s (Cont.)
6. At this point O atoms could be balanced by adding H2O and
H atoms could be balanced by adding H+ and finally OHcould be added to produce a basic solution.
6. Alternate shortcut: a) Use OH- to charge balance and then
b) add H2O to balance O and H atoms at the same time.
a) 6Zn + As2O3  2AsH3 + 6Zn2+ + 12OHb) 6Zn + As2O3 + 9H2O  2AsH3 + 6Zn2+ + 12OH-
6Zn + As2O3 + 9H2O  2AsH3 + 6Zn2+ + 12OH-
Ex 3 of Redox Equation Balancing by Ox. #s
Balance this disproportionation P4  H2PO2- + PH3 (acid)
1. Write equation so same reactant can undergo ox. and red.
P4 + P4  H2PO2- + PH3
2. Balance majors
P4 + P4  4H2PO2- + 4PH3
3. Assign ox.#s
0
0
+1
-3
P4 + P4  4H2PO2- + 4PH3
4. Note total changes in ox#s
4 P go up 1
while
4 P go down 3
total Δ up = (4)(+1 Δ) = +4 Δ , total Δ down = (4)(-3 Δ) = -12 Δ
5. Find cross-mult. #s so that the total Δ up = total Δ down
3(4+Δ) = +12
and 1(-12) = -12
6. Apply cross-mult #s to reactants and then products
3P4 + 1P4  12H2PO2- + 4PH3
Ex 3 of Redox Equation Balancing by Ox. #s (Cont.)
Apply cross-mult #s to reactants and then products
3P4 + 1P4  12H2PO2- + 4PH3
7. Simplify and reduce to lowest terms
4P4  12H2PO2- + 4PH3
P4  3H2PO2- + PH3
8. Balance O atoms by adding H2O
P4 + 6H2O  3H2PO2- + PH3
9. Balance H atoms by adding H+
P4 + 6H2O  3H2PO2- + PH3 + 3H+
9.
Check charge balance to detect mistakes
P4 + 6H2O 
3H2PO2-
+ PH3 + 3H+
Redox Titrations
Redox reactions can be used to accurately determine
unknown concentrations of substances. The end point of the
redox reaction is determined by a colour change in the agent
as it is oxidized or reduced.
Oxidizing Agents for Redox Titrations
In an acidic solution of KMnO4, the ion, MnO4- is a strong
oxidizing agent (has a strong tendency to reduce) as shown
by this reaction: MnO4- + 8H+ + 5e-  Mn2+ + 4H2O ;Eo = 1.49V
MnO4- is purple in solution while Mn2+ is colourless. As the
MnO4- is added to a solution, it is oxidized and becomes clear
Mn2+ until the end point where all the reducing agent is used
up and a single added drop of MnO4- remains purple.
Example of Redox Titration Using MnO4When 25.00 mL of a solution of Fe2+ is titrated to an endpoint
with acidic KMnO4 , the titration requires 17.52 mL of
acidified 0.1000 M KMnO. What is the Fe2+ in the solution?
MnO4- + 8H+ + 5Fe2+  Mn2+ + 4H2O + 5Fe3+
1. The mols MnO4- = (0.1000 mmol/mL)(17.52 mL)
= 1.752 mmol MnO42. The mols Fe2+ = (1.752 mmol MnO4-)(5mmol Fe2+/1 mmolMnO4-)
= 8.760 mmol Fe2+
3. [Fe2+] = 8.760 mmol Fe2+ / 25.00 ml = 0.3504 M (mol/L)
Reducing Agents for Redox Titrations
NaI or KI are commonly used as reducing agents in redox
titrations. The I- ion (reducing agent) is oxidized as
follows:
2I-  I2 + 2eTwo steps are generally used:
(1) Titrations using I- to reduce a given species produces I2,
and (2) this I2 is then generally reduced back to I- by a
second reducing agent like S2O32- (thiosulfate ion).
For example, to determine the concentration of the
hypochlorite ion (OCl-) in bleach (NaOCl), an excess of KI
solution is added to acidified bleach to consume all the
moles of OCl-. The I2 turns the solution brownish.
2 H+ + OCl- + 2I-  Cl- + H2O + I2
+

Reducing Agents for Redox Titrations (Cont.)
The brownish I2 solution is then reacted with a known
concentration of Na2S2O3 until a pale yellow colour
remains which indicates just a small amount of I2 left.
Only the I2 reacts with the S2O32-. The excess I- does not
react with the S2O32-.
2S2O32- + I2  S4O62- + 2IStarch is added to the yellow solution (solution becomes
dark blue) and then more S2O32- is added until a single
drop turns the mixture from blue to clear (the endpoint).
Sample Redox Titration Problem With I- and S2O32A 25.00 mL sample of bleach is reacted with excess KI. The
resulting brown solution takes 46.84 mL of 0.7500 M Na2S2O3
to reach the endpoint. What was the [OCl-] in the bleach?
2 H+ + OCl- + 2I-  Cl- + H2O + I2
2S2O32- + I2  S4O62- + 2I-
End of Presentation