This document provides an overview of redox reactions including:
- Redox reactions involve the transfer of electrons between chemical species, resulting in oxidation and reduction.
- Oxidizing agents gain electrons and are reduced, while reducing agents lose electrons and are oxidized.
- Latimer, Frost, and Pourbaix diagrams can be used to predict and understand redox reactions in aqueous solutions by showing the thermodynamic stability of different oxidation states.
- Key concepts like disproportionation, oxidizing/reducing abilities, and stable/unstable species can be determined from these types of diagrams.
1. Presented by :
Arvind Singh Heer
MSc-I
(Sem-I)
Inorganic Chemistry
MITHIBAI COLLEGE
REDOX REACTION
2. CONTENT
Introduction.
Oxidizing agent and Reducing agent.
Identifying OX, RD, SI Species.
Mechanisms and their characteristics.
Balancing Redox Reactions.
Redox reaction in aqueous media.
3. INTRODUCTION
What is redox reaction?
Redox reaction include all chemical reactions in
which atoms have their oxidation state changed; in
general, redox reactions involve the transfer of
electrons between species.
The term "redox" comes from two concepts involved
with electron transfer: reduction and oxidation. It
can be explained in simple terms:
• Oxidation is the loss of electrons or an increase in
oxidation state by a molecule, atom, or ion.
• Reduction is the gain of electrons or a decrease in
oxidation state by a molecule, atom, or ion.
4. OXIDIZING AND REDUCING AGENT
An oxidizing agent (also called an oxidant or oxidizer) is
a chemical compound that readily transfers oxygen atoms
or a substance that gains electrons in a redox chemical
reaction.
An oxidizing agent oxidizes other substances
and gains electrons; therefore, its oxidation state
will decrease.
An reducing agent (also called a reductant or reducer) is
the element or a compound in a redox reaction that
reduces another species. In doing so, it becomes oxidized,
and is therefore a electron donor in redox reaction.
A reducing agent reduces other substances
and lose electrons; therefore, its oxidation state
will increase.
5. In simple terms:
* The oxidizing agent is reduced.
* The reducing agent is oxidized.
* All atoms in a molecule can be assigned an
oxidation number. This number changes when an
oxidant acts on a substrate.
* Redox reactions occur when electrons are
exchanged.
A mnemonic for differentiating the reactions is "OIL
RIG": Oxidation Is Loss, Reduction Is Gain (of
electrons) or "LEO the lion says GER" (Lose Electrons:
Oxidation, Gain Electrons: Reduction).
6. EXAMPLE
Ca0 + 2 H+1Cl-1 Ca+2Cl-1
2 + H2
0
Since Ca0 is being oxidized and H+1 is being reduced,
the electrons must be going from the Ca0 to the H+1.
Since Ca0 would not lose electrons (be oxidized) if
H+1 weren’t there to gain them, H+1 is the cause, or
agent, of Ca0’s oxidation. H+1 is the oxidizing
agent.
Since H+1 would not gain electrons (be reduced) if
Ca0 weren’t there to lose them, Ca0 is the cause, or
agent, of H+1’s reduction. Ca0 is the reducing
agent.
7. IDENTIFYING OX, RD, SI SPECIES
Ca0 + 2 H+1Cl-1 Ca+2Cl-1
2 + H2
0
Oxidation = loss of electrons. The species becomes
more positive in charge. For example, Ca0 Ca+2,
so Ca0 is the species that is oxidized.
Reduction = gain of electrons. The species
becomes more negative in charge. For example,
H+1 H0, so the H+1 is the species that is reduced.
Spectator Ion = no change in charge. The species
does not gain or lose any electrons. For example,
Cl-1 Cl-1, so the Cl-1 is the spectator ion.
8. MECHANISMS AND THEIR CHARACTERISTICS
The Outer Sphere Electron Transfer Mechanism = electron
transfer with no change of coordination sphere
a)Example: Co(NH3)6
3+ + Cr(bipy)3
2+ Co(NH3)6
2+ + Cr(bipy)3
3+
oxidant reductant reduced oxidized
9. b)The rates of reaction depend on the ability of the electron to
“tunnel” through the ligands from one metal to the other
Tunneling = moving through an energy barrier (the ligands)
that is normally too high to allow the electron to pass
through. This is a quantum mechanical process having to do
with the wave nature of e-.
Ligands with p or p orbitals good for bonding more easily
allow tunnelling (CN-, F-) than those that don’t (NH3).
c)The ligands don’t change in Outer Sphere electron transfer, but
the M—L bond distances do
High Oxidation # = short bond distance
Low Oxidation # = longer bond distance
d)The stronger the ligand field, the less favored reduction is (or
more favored oxidation is), because more energy is gained by
losing high energy eg* electrons (NH3 > H2O)
Co(NH3)6
3+ + e- Co(NH3)6
2+ Eo = +0.108 V
Co(H2O)6
3+ + e- Co(H2O)6
2+ Eo = +1.108 V
10. The Inner Sphere Electron Transfer Mechanism = tunneling
of an electron through a bridging ligand.
Substitution links the reactants
e- transfer
Separation of products
This reaction could be followed by ion exchange
and UV-Vis
11. BALANCING REDOX REACTIONS
STEP 1. Split Reaction into 2 Half-Reactions
STEP 2. Balance Elements Other than H & O
STEP 3. Balance O by Inserting H2O into eqns. as
necessary
STEP 4. Balance H with H+ or H2O (see 4a, 4b)
STEP 5. Balance Charge by Inserting Electrons as
needed
STEP 6. Multiply Each 1/2 Reaction by Factor needed
to make no. of Electrons in each 1/2 Reaction Equal
STEP 7. Add Eqns. & Cancel Out Duplicate terms,
where possible
12. STEP 4a. In ACID: Balance H by Inserting H+, as
needed
STEP 4b. In BASE: Balance H by (i) inserting 1 H2O
for each missing H & (ii) inserting same no. of OH-
on OTHER SIDE OF REACTION as H2Os added in (i)
13. EXAMPLE
Complete and Balance Following Reaction:
CuS (s) + NO3
- (aq) Cu2+(aq) + SO4
2- (aq) + NO (g)
STEP1. Split into 2 Half-Reactions
a.1 CuS Cu2+ + SO4
2-
b.1 NO3
- NO
STEP 2. Balance Elements Other than H & O (It is already balanced).
STEP 3. Balance O by inserting H2O into equations as necessary.
a.3 CuS + 4H2O Cu2+ + SO4
2-
b.3 NO3
- NO + 2H2O
STEP 4. ACIDIC, so Balance H by inserting H+ as needed.
a4. CuS + 4H2O Cu2+ + SO4
2- + 8H+
b4. NO3
- + 4H+ NO + 2H2O
14. STEP 5. Balance Charge by inserting Electrons, where necessary
a5. CuS + 4H2O Cu2+ + SO4
2- + 8H+ + 8e-
b5. NO3
- + 4H+ + 3e- NO + 2H2O
STEP 6. Multiply each Eqn. by factor to make No. of Electrons in Each 1/2 Reaction the
Same
a6. Multiply by 3x
3CuS + 12H2O 3Cu2+ + 3SO4
2- + 24H+ + 24e-
b6.Multiply by 8x
8NO3
- + 32H+ + 24e- 8NO + 16H++ 24e-
STEP 7. Add Eqns. and Cancel Out Duplicated Terms
(a7 + b7)
8H+
3CuS + 12H2O + 8NO3
- + 32H+ + 24 e- 3Cu2+ + 3SO4
2- + 24H+ + 8NO +16H2O +24e-
4H2O
So, the final, balanced reaction is:
3CuS(s) + 8 NO3
-(aq) + 8H+ (aq) 3Cu2+(aq) + 3 SO4
2-(aq) + 8NO(g) + 4H2 O(l)
15. Redox Reactions in Aqueous Media
There are three main methods of predicting redox reactions
in aqueous solutions, summarising the thermodynamic
stabilities of oxidation states of elements in aqueous
solutions.
1. LATIMER DIAGRAM
2. VOLT EQUIVALENT [Also known as Frost (Free energy
oxidation state) diagram
These two are usually restricted to extremes of pH=0 or pH=14
solutions.
3. Pourbaix diagram
It expresses the variation in stabilities of oxidation states as a
function of pH between pH values 0 to 14.
16. Latimer Diagrams
It is a list of various oxidation states of an element arranged in
descending order from left to right.
The numerical value of standard potential is written over a
horizontal line connecting species of element in different
oxidation states.
Most oxidised form is on the left.
Species to the right are in successively lower oxidation states.
Available for all the elements exhibiting more than one oxidation
state.
17. Latimer diagram for Manganese
[acidic medium]
The Latimer diagram for Mn illustrates its standard
reduction potentials (in 1M HCl) in oxidation states from +7
to 0.
It compresses into shorthand notation all the standard
potentials for redox reactions of element Mn.
Values for multi-electron reactions can be also calculated by
first adding ∆Gº (nFEº) values and then dividing by the total
no of electrons.
18. Calculation of values for multi-electron reactions by
first adding ΔG°(=-nFE°) values and then dividing by
the total number of electrons
for the 5-electron reduction of MnO4
- to Mn2+, we write
for the three-electron reduction of MnO4
-(aq) to
MnO2(s),
19. Thermodynamically stable &
unstable oxidation states
An unstable species on a Latimer diagram will have a lower
standard potential to the left than to the right.
2 MnO4
-3 → MnO2 + MnO4
2- ; MnO4
-3 is unstable.
Eº = +4.27 − 0.274 = +3.997V ; (spontaneous
disproportionation)
Which Mn species are unstable with respect to
disproportionation?
MnO4
-3 ; 5+ → 6+, 4+
Mn3+ ; 3+→4+, 2+
So stable species are MnO4
-, MnO2, MnO4
2-, Mn2+, Mn0
Thermodynamically unstable ions can be quite stable
kinetically.
20. Disproportionation
In most redox reactions atoms of one element are oxidized
and atoms of a different element are reduced.
In some redox reactions a single substance can be both
oxidized and reduced. These are known as
disproportionation reactions.
Example : Decomposition reaction of H2O2
2 H2O2(aq) → 2 H2O(l) + O2(g)
The decomposition reaction of hydrogen peroxide produces
oxygen and water.
Oxygen is present in all parts of the chemical equation and
as a result it is both oxidized and reduced.
21. Frost Diagrams
In a Frost diagram, we plot ΔG°⁄F (=
nE°) vs. oxidation number.
The zero oxidation state is assigned a
nE° value of zero.
Contains the same information as in a
Latimer diagram, but graphically
shows stability and oxidizing power.
Stable and unstable oxidation
states can be easily identified in the
plot.
The standard potential for any
electrochemical reaction is given by
the slope of the line connecting the
two species on a Frost diagram.
22. What You Can Learn From a Frost
Diagram
Thermodynamic stability is found at the
bottom of the diagram. Thus, the lower
a species is positioned on the diagram,
the more thermodynamically stable it is
(from a oxidation-reduction perspective)
Mn (II) is the most stable species.
Any species located on the upper left
side of the diagram will be a strong
oxidizing agent.
MnO4
- is a strong oxidizer.
Any species located on the upper left
side of the diagram will be a strong
oxidizing agent.
MnO4
- is a strong oxidizer.
The information obtained from a Frost
diagram is for species under standard
conditions (pH=0 for acidic solution and
pH=14 for basic solution).
23. Pourbaix Diagrams
Plots of E versus pH for various couples in oxidation of an
element.
The Pourbaix diagram is a type of predominance diagram --
it shows the predominate form in an element will exist
under a given set of environmental conditions.
These diagrams give a visual representation of the oxidizing
and reducing abilities of the major stable compounds of an
element and are used frequently in geochemical,
environmental and corrosion applications.
Like Frost diagrams, Pourbaix diagrams display
thermodynamically preferred species.
24. How to Read a Pourbaix Diagram
Vertical lines separate species that
are in acid-base equilibrium.
Non vertical lines separate species
related by redox equilibria.
Horizontal lines separate species in
redox equilibria not involving
hydrogen or hydroxide ions.
Diagonal boundaries separate
species in redox equilibria in which
hydroxide or hydrogen ions are
involved.
Dashed lines enclose the practical
region of stability of the water
solvent to oxidation or reduction.
25. What You Can Learn From a
Pourbaix Diagram
Any point on the diagram will give the thermodynamically
most stable form of that element at a given potential and
pH condition.
Strong oxidizing agents and oxidizing conditions are found
only at the top of Pourbaix diagrams.
Strong oxidizing agents have lower boundaries that are also
high on the diagram.
Permanganate is an oxidizing agent over all pH ranges. It is
very strongly oxidizing at low pH.
A species that ranges from the top to the bottom of the
diagram at a given pH will have no oxidizing or reducing
properties at that pH.
26. Reducing agents and reducing conditions are found at
the bottom of a diagram and not elsewhere.
Strong reducing agents have low upper boundaries on
the diagram.
Manganese metal is a reducing agent over all pH
ranges and is strongest in basic conditions.
When the predominance area for a given oxidation
state disappears completely above or below a given
pH and the element is in an intermediate oxidation
state, the element will undergo disproportionation
MnO4
2- tends to disproportionate.
27. REFRENCES
1. Rayner- Canham, G. Descriptive Inorganic Chemistry;
Freeman:New York,1996; Chapter 9
2. Douglas, B; McDaniel, D.; Alexander, J. Concepts and Models of
Inorganic Chemistry, 3rd ed.; Wiley & Sons: New York, 1994;
Chapter 8.
3. J. Kotz, P. Treichel, J. Townsend, D. Treiche, Chemistry &
Chemical Reactivity, 9th ed. ; Cengage Learning.
4. J.E. Huheey, E.A. Keiter, R.L. Keiter, O.K. Medhi, Inorganic
Chemistry: Principles of Structure and Reactivity, 4th ed. ; Pearson
Education.
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