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Oxidation-Reduction

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Oxidation-Reduction Biology Industry Environment N2 a b c d e f g h G = G + G -nFE = -n FE - n FE E = n E + n E n +n ... – PowerPoint PPT presentation

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Title: Oxidation-Reduction


1
Oxidation-Reduction
Biology
Industry
Environment
2
Biology
3
Biology
0
4
Industry
Synthesis of different compounds
Extraction of elements
5
Environment
6
Redox reactions - transfer of electrons between
species.
All the redox reactions have two parts
Oxidation
Reduction
7
  • The Loss of Electrons is Oxidation.
  • An element that loses electrons is said to be
    oxidized.
  • The species in which that element is present in a
    reaction is called the reducing agent.
  • The Gain of Electrons is Reduction.
  • An element that gains electrons is said to be
    reduced.
  • The species in which that element is present in a
    reaction is called the oxidizing agent.

8
Balancing Redox Equations
Fe2 MnO4- H Mn2
Fe3 H2O
  • Assign oxidation numbers to each atom.
  • Determine the elements that get oxidized and
    reduced.
  • Split the equation into half-reactions.
  • Balance all atoms in each half-reaction, except H
    and O.
  • Balance O atoms using H2O.
  • Balance H atoms using H.
  • 7. Balance charge using electrons.
  • 8. Sum together the two half-reactions, so that
    e- lost e- gained
  • 9. If the solution is basic, add a number of OH-
    ions to each side of the equation equal to the
    number of H ions shown in the overall equation.
    Note that H OH- ? H2O

9
Example
10
Nernst Equation
E0 Standard Potential R Gas constant 8.314
J/K.mol F- Faraday constant 94485 J/V.mol n-
number of electrons
11
?G0 - n F ? E0
Note if ?G0 lt 0, then ? E0 must be gt0
A reaction is favorable if ? E0 gt 0
(a)
(b)
(a-b)
Reaction is favorable
12
Hydrogen Electrode
  • consists of a platinum electrode covered with a
    fine powder of platinum around which H2(g) is
    bubbled. Its potential is defined as zero volts.
  • Hydrogen Half-Cell
  • H2(g) 2 H(aq) 2 e-
  • reversible reaction

13
Galvanic Cell
14
-
15
Diagrammatic presentation of potential data
Latimer Diagram
Frost Diagram
16
Latimer Diagram
Written with the most oxidized species on the
left, and the most reduced species on the
right.
Oxidation number decrease from left to right
and the E0 values are written above the line
joining the species involved in the couple.
y
z
x
w
A5 B3 C1 D0 E-2
17
-0.44
18
What happens when Fe(s) react with H?
Iron 2 and 3
?G -nFE
-2 x F x -0.44 0.88 V
-0.440
-1 x F x 0.771 -0.771 V
-0.771
0.109 F -3 x F x 0.036
Fe
Fe2
Fe
Fe3
19
0.36
-1.16
Oxidation of Fe(0) to Fe(II) is considerably more
favorable in the cyanide/acid mixture than in
aqueous acid.
(1) Concentration (2) Temperature (3) Other
reagents which are not inert
20
Oxidation of elemental copper
21
Latimer diagram for chlorine in acidic solution
Can you balance the equation?
balance the equation
22
How to extract E0 for nonadjacent oxidation state?
-1
1
0
1
2
Identify the two reodx couples
Find out the oxidation state of chlorine
Write the balanced equation for the first couple
Write the balanced equation for the second couple
E 1.5 V
23
Latimer diagram for chlorine in basic solution
0.89
Balance the equation
Find out the E0
0.89
24
Disproportionation
Element is simultaneously oxidized and reduced.
the potential on the left of a species is less
positive than that on the right- the species can
oxidize and reduce itself, a process known as
disproportionation.
25
?E E0 (Cl2/Cl-) E0 (ClO-/Cl2) 1.36 - 0.42
0.94
Reaction is spontaneous
26
Latimer diagram for Oxygen
1.23 V
27
Disproportionation
the potential on the left of a species is less
positive than that on the right- the species can
oxidize and reduce itself, a process known as
disproportionation.
28
Is it spontaneous?
Yes the reaction is spontaneous
29
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30
Another example
2 Cu(aq) Cu2(aq) Cu(s)
Cu(aq) e- Cu(s)
E0 0.52 V
Cu2(aq) e- Cu(aq)
E0 0.16 V
Cu(I) undergo disproportionation in aqueous
solution
31
Comproportionation reaction
Reverse of disproportionation
we will study this in detail under Frost diagram
32
Frost Diagram
Arthur A. Frost
Graphically illustration of the stability of
different oxidation states relative to its
elemental form (ie, relative to oxidation state
0)
33
Look at the Latimer diagram of nitrogen in acidic
solution
d
e
a
b
c
f
g
h
34
c
a
e
b
d
g
N2
f
h
35
NO3- 6H 5e-       ½ N2 3H2O E0
1.25V ½ N2O4 4H 4e-       ½ N2 2H2O
E0 1.36V HNO2 3H 3e-       ½ N2
2H2O E0 1.45V NO 2H 2e-       ½ N2
H2O E0 1.68V ½ N2O H e-       ½
N2 ½ H2O E0 1.77V ½ N2 2H H2O e-
      NH3OH E0 -1.87V ½ N2 5/2 H 2e-
      ½ N2H5 E0 -0.23V ½ N2 4H 3e-
      NH4 E0 0.27V
a
b
c
d
e
f
g
h
36
Oxidation state species NE0, N
N(V) NO3- (5 x 1.25, 5) N(IV) N2O4 (4 x
1.36, 4) N(III) HNO2 (3 x 1.35, 3) N(II)
NO (2 x 1.68, 2) N(I) N2O (1 x 1.77, 1)
N(-I) NH3OH -1 x (-1.87), -1 N(-II)
N2H5 -2 x (-0.23), -2 N(-III) NH4 (-3
x 0.27, -3)
37
Frost Diagram N2
38
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39
What do we really get from the Frost diagram?
the lowest lying species corresponds to the most
stable oxidation state of the element
40
Slope of the line joining any two points is equal
to the std potential of the couple.
41
E0 of a redox couple
HNO2/NO
3, 4.4
2, 3.4
42
Oxidizing agent? Reducing agent?
The oxidizing agent - couple with more positive
slope - more positive E
The reducing agent - couple with less positive
slope
If the line has ive slope- higher lying species
reducing agent If the line has ive slope
higher lying species oxidizing agent
43
Identifying strong or weak agent?
44
NO Strong oxidant than HNO3
45
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46
Disproportionation
Element is simultaneously oxidized and reduced.
the potential on the left of a species is less
positive than that on the right- the species can
oxidize and reduce itself, a process known as
disproportionation.
47
Disproportionation
What Frost diagram tells about this reaction?
48
A species in a Frost diagram is unstable with
respect to disproportionation if its point lies
above the line connecting two adjacent species.
49
Disproportionation. another example
50
Comproportionation reaction
51
Comproportionation is spontaneous if the
intermediate species lies below the straight
line joining the two reactant species.
52
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53
Favorable?
54
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56
NE0
57
Disproportionation
58
Comproportionation
In acidic solution
Mn and MnO2
Mn2
Rate of the reaction hindered insolubility?
In basic solution
MnO2 and Mn(OH)2
Mn2O3
59
From the Frost diagram for Mn.
Thermodynamic stability is found at the bottom
of the diagram. Mn (II) is the most stable
species.
A species located on a convex curve can undergo
disproportionation example MnO43-
MnO2 and MnO42- (in basic solution)
  • Any species located on the upper right side of
    the diagram will be
  • a strong oxidizing agent. MnO4- - strong
    oxidizing agent.
  • Any species located on the upper left side of the
    diagram will be
  • a reducing agent. Mn - moderate reducing agent.

60
Although it is thermodynamically favorable for
permanganate ion to be reduced to Mn(II) ion, the
reaction is slow except in the presence of a
catalyst. Thus, solutions of permanganate can be
stored and used in the laboratory.
Changes in pH may change the relative
stabilities of the species. The potential of
any process involving the hydrogen ion will
change with pH because the concentration of this
species is changing. Under basic
conditions aqueous Mn2 does not exist. Instead
Insoluble Mn(OH)2 forms.
61
All metals are good reducing agents Exception
Cu Reducing strength goes down smoothly from Ca
to Ni Ni- mild reducing agent Early transition
elements 3 state Latter 2 state Fe and Mn
many oxidation states High oxidation state
Strong oxidizing agents
62
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