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Oxidation-Reduction. Biology. Industry. Environment. Biology. Biology. 0. Industry. Extraction of elements. Synthesis of different compounds. Environment. Redox reactions - transfer of electrons between species. All the redox reactions have two parts:. Oxidation. Reduction.

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slide1

Oxidation-Reduction

Biology

Industry

Environment

slide4

Industry

Extraction of elements

Synthesis of different compounds

slide6

Redox reactions - transfer of electrons between species.

All the redox reactions have two parts:

Oxidation

Reduction

slide7

Cu2+ Cu

Mg Mg2+

  • 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.
slide8

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 thenumber of H+ ions shown in the overall equation. Note that H+ + OH- H2O

slide9

Fe2+ + MnO4- + H+ Mn2+ + Fe3+ + H2O

MnO4- Mn2+ Reduction half reaction

(+7) (+2)

Fe2+ Fe3+ Oxidation half reaction

MnO4- + 8H+ + 5e Mn2+ + 4H2O

5Fe2+ 5Fe3+ +5e

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

Example

slide10

aOx1 +bRed2 a’Red1 + b’Ox2

[Red1]a’ [Ox2]b’

Q =

[Ox1]a [Red2]b

RT

nF

E = E0 - ln Q

Nernst Equation

E0 = Standard Potential

R = Gas constant 8.314 J/K.mol

F- Faraday constant = 94485 J/V.mol

n- number of electrons

slide11

2H+ (aq) + 2e H2(g) E0(H+, H2) = 0

Zn2+ (aq) + 2e Zn(s) E0(Zn2+, Zn) = -0.76 V

2H+ (aq) + Zn(s) Zn2+(aq) + H2(g) E0 = +0.76 V

G0 = - n F  E0

Note: if G0 < 0, then  E0 must be >0

A reaction is favorable if  E0 > 0

(a)

(b)

(a-b)

Reaction is favorable

slide12

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
slide16

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

A+5 B+3 C+1 D0 E-2

slide18

Fe3+

+0.036

+0.44

Fe2+

-0.036

+0.77

-0.44

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

-0.771

-1 x F x +0.771 = -0.771 V

+ 0.109 F

= -3 x F x –0.036

Fe

Fe2+

Fe

Fe3+

slide19

Fe2+

Fe

Fe3+

[Fe(CN)6]3-

[Fe(CN)6]4-

-0.036

+0.77

-0.44

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

slide21

Latimer diagram for chlorine in acidic solution

Can you balance the equation?

balance the equation

slide22

E0 =?

HClO(aq) + H+(aq) + e ½ Cl2(g) + H2O(l) +1.63 V

½ Cl2(g) + e Cl- (l) +1.36 V

G = G’ + G’’

-FE = - ’FE’ - ’’FE’’

E = ’E’+ ’’E’’

’+ ’’

How to extract E0 for nonadjacent oxidation state?

+1

-1

0

1

Identify the two reodx couples

2

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

slide23

+0.37

+0.3

+0.68

+0.42

+1.36

ClO4- ClO3- ClO2- ClO- Cl2 Cl-

+0.89

+0.42

ClO- Cl2

2ClO- (aq) + 2H2O(l) + 2e- Cl2(g) + 4OH-(aq) E0 = 0.42 V

Latimer diagram for chlorine in basic solution

+0.89

Balance the equation…

Find out the E0

+0.89

slide24

2 M+(aq) M(s) + M2+(aq)

E0

E0’

M(s)

M2+(aq)

2M+(aq)

+0.37

+0.3

+0.68

+0.42

+1.36

ClO4- ClO3- ClO2- ClO- Cl2 Cl-

+0.42

+1.36

ClO- Cl2 Cl-

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 asdisproportionation’.

slide25

Cl2(g) + 2 e- 2Cl-(aq) +1.36

2ClO- (aq) 2H2O(l) +2e- Cl2(g) + 4OH-(aq) +0.42

Cl2 + 2OH- ClO- + Cl- + H2O

+0.37

+0.3

+0.68

+0.42

+1.36

ClO4- ClO3- ClO2- ClO- Cl2 Cl-

+0.42

+1.36

ClO- Cl2 Cl-

E = E0 (Cl2/Cl-) –E0 (ClO-/Cl2) = 1.36 - +0.42 = 0.94

Reaction is spontaneous

slide27

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 asdisproportionation.

slide28

H2O2(aq) + 2H+ (aq) +2e- 2H2O(aq) +1.76 V

O2(g) + 2H+(aq) +2 e- H2O2(aq) +0.7 V

H2O2(aq) O2 (g) + H2O(l) +0.7 V

Is it spontaneous?

Yes the reaction is spontaneous

slide30

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

slide31

Ag2+(aq) + Ag(s) 2Ag+(aq) E0 = + 1.18 V

Comproportionation reaction

Reverse of disproportionation

…we will study this in detail under Frost diagram

slide32

XN + Ne- X0

NE0 = -G0/F

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)

slide34

c

G = G’ + G’’

-nFE = -n’FE’ - n’’FE’’

E = n’E’+ n’’E’’

n’+n’’

a

e

b

d

g

N2

f

h

slide35

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

slide36

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)

slide39

What do we really get from the Frost diagram?

the lowest lying species corresponds to the most stable oxidation state of the element

slide40

Slope of the line joining any two points is equal to the std potential of the couple.

N’E0’-N”E0”

N’-N”

Slope = E0=

N’E0’

N’

N”E0’’

N’’

slide41

N’E0’-N”E0”

N’-N”

Slope = E0=

 1 V

E0 of a redox couple

HNO2/NO

3, 4.4

2, 3.4

slide42

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

slide46

2 M+(aq) M(s) + M2+(aq)

E0

E0’

M(s)

M2+(aq)

2M+(aq)

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 asdisproportionation’.

slide47

Disproportionation

What Frost diagram tells about this reaction?

slide48

A species in a Frost diagram is unstable with respect to disproportionation

if its point lies above the line connecting two adjacent species.

slide51

Comproportionation is spontaneous if the intermediate species

lies below the straight line joining the two reactant species.

slide58

Comproportionation

In acidic solution…

Mn and MnO2

Mn2+

Rate of the reaction hindered

insolubility?

In basic solution…

MnO2 and Mn(OH)2

Mn2O3

slide59

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.
slide60

* 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.

slide61

*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