Oxidation-Reduction Biology Industry Environment.

Oxidation-ReductionOxidation-Reduction

Biology

Industry

Environment

Biology

Biology

0

Industry Synthesis of different compoundsExtraction of elements

Environment

Redox reactions - transfer of electrons between species.Redox reactions - transfer of electrons between species.

All the redox reactions have two parts:All the redox reactions have two parts:

OxidationOxidation ReductionReduction

• 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.

Cu2+ Cu

Mg Mg2+

Balancing Redox Equations

1. Assign oxidation numbers to each atom.

2. Determine the elements that get oxidized and reduced.

3. Split the equation into half-reactions.

4. Balance all atoms in each half-reaction, except H and O.

5. Balance O atoms using H2O.

6. 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

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

Example

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

Nernst EquationNernst Equation

aOx1 +bRed2 a’Red1 + b’Ox2

Q =[Red1]a’ [Ox2]b’

[Ox1]a [Red2]bE = E0 - ln Q RT

nF

E0 = Standard PotentialR = Gas constant 8.314 J/K.molF- Faraday constant = 94485 J/V.moln- number of electrons

GG00 = - n F = - n F E E00

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

A reaction is favorable if E0 > 0

Reaction is favorable

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

(a)

(b)

(a-b)

Hydrogen ElectrodeHydrogen 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 reactionreversible reaction

Galvanic CellGalvanic Cell

-

Diagrammatic presentation of potential dataDiagrammatic presentation of potential data

Latimer DiagramLatimer Diagram

Frost DiagramFrost Diagram

Latimer DiagramLatimer Diagram

* Oxidation number decrease from left to right and the E0 values are written above the line joining the species involved in the couple.

* Written with the most oxidized species on the left, and the most * Written with the most oxidized species on the left, and the most reduced species on the right.reduced species on the right.

A+5 B+3 C+1 D0 E-2x y zw

-0.44

Fe3+ + 3e Fe

Iron +2 and +3What happens when Fe(s) react with H+?

G = -nFE

-2 x F x -0.44 = 0.88 V

-1 x F x +0.771 = -0.771 V

+ 0.109 F = -3 x F x –0.036

Fe

Fe3++0.036

Fe2++0.44

Fe3+ Fe2+ Fe

-0.036

+0.77 -0.44

-0.440

-0.771

Fe3+ Fe2+

-0.036

+0.77 -0.44 Fe

[Fe(CN)6]3- [Fe(CN)6]4- -1.16

0.36

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

Oxidation of elemental copper

Latimer diagram for chlorine in acidic solution

ClO4- ClO3

- HClO2 HClO Cl2 Cl-+1.2 +1.18 +1.65 +1.63 +1.36

+7 +5 +3 +1 0 -1

ClO4- ClO3

-+1.2

HClO Cl2+1.63

2 HClO(aq) + 2 H+(aq) + 2 e- Cl2(g) + 2 H2O(l) E0 = +1.63 V

Can you balance the equation?

balance the equation

ClO4- ClO3

- HClO2 HClO Cl2 Cl-+1.2 +1.18 +1.65 +1.63 +1.36

+7 +5 +3 +1 0 -1

Write the balanced equation for the first couple

Write the balanced equation for the second couple

1 2

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’’

’+ ’’

Find out the oxidation state of chlorine

E = 1.5 V

How to extract E0 for nonadjacent oxidation state?E0 =?

Identify the two reodx couples

0 -1+1

ClO4- ClO3

- ClO2- ClO- Cl2 Cl-

+0.37 +0.3 +0.68 +0.42 +1.36

+0.89

ClO- Cl2

+0.42

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

Latimer diagram for chlorine in basic solution

ClO4- ClO3

- HClO2 HClO Cl2 Cl-+1.2 +1.18 +1.65 +1.63 +1.36

+7 +5 +3 +1 0 -1

+0.89

Balance the equation…

Find out the E0

+0.89

ClO4- ClO3

- ClO2- ClO- Cl2 Cl-

+0.37 +0.3 +0.68 +0.42 +1.36

DisproportionationDisproportionationElement is simultaneously oxidized and reduced.

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

EE00 EE0’0’

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

‘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 the species can oxidize and reduce itself, a process known asas disproportionation’.

ClO- Cl2 Cl-+0.42 +1.36

ClO4- ClO3

- ClO2- ClO- Cl2 Cl-

+0.37 +0.3 +0.68 +0.42 +1.36

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

ClO- Cl2 Cl-+0.42 +1.36

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

Reaction is spontaneous

Cl2(g) + 2 e- 2Cl-(aq) +1.362ClO- (aq) 2H2O(l) +2e- Cl2(g) + 4OH-(aq) +0.42

Latimer diagram for Oxygen

1.23 V

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 the species can oxidize and reduce itself, a process known asknown as disproportionation.

DisproportionationDisproportionation

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

Yes the reaction is spontaneous Yes the reaction is spontaneous

Is it spontaneous?

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

Another example…

Comproportionation reactionComproportionation reaction

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

Reverse of disproportionation

…we will study this in detail under Frost diagram

Frost DiagramFrost DiagramGraphically illustration of the stability of different oxidation states relative to its elemental form (ie, relative to oxidation state= 0)

Arthur A. FrostArthur A. Frost

XN + Ne- X0

NE0 = -G0/F

Look at the Latimer diagram of nitrogen in acidic solution

a b c d e

f g h

N2

a

b

c

d

e

f

g

h

G = G’ + G’’

-nFE = -n’FE’ - n’’FE’’E = n’E’+ n’’E’’

n’+n’’

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

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)

Oxidation state: species NE0, N

Frost Diagram – NFrost Diagram – N22

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

What do we really get from the Frost diagram?What do we really get from the Frost diagram?

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

N’

N’E0’

N”E0’’

N’’ Slope = E0= N’E0’-N”E0”

N’-N”

3, 4.4

2, 3.4

Slope = E0= N’E0’-N”E0”

N’-N”

1 V

EE00 of a redox couple of a redox couple

HNOHNO22/NO/NO

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

Oxidizing agent? Reducing agent?Oxidizing agent? Reducing agent?

Identifying strong or weak agent?Identifying strong or weak agent?

NO – Strong oxidant than HNONO – Strong oxidant than HNO33

DisproportionationDisproportionationElement is simultaneously oxidized and reduced.

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

EE00 EE0’0’

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

‘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 the species can oxidize and reduce itself, a process known asas disproportionation’.

DisproportionationDisproportionationWhat Frost diagram tells about this reaction?

A species in a Frost diagram is unstable with respect to disproportionationA species in a Frost diagram is unstable with respect to disproportionation if its point lies above the line connecting two adjacent species.if its point lies above the line connecting two adjacent species.

Disproportionation…. another example

Comproportionation reaction

Comproportionation is spontaneous if the intermediate species lies below the straight line joining the two reactant species.

FavorableFavorable?

NE0

Disproportionation

Comproportionation

In acidic solution…

Mn and MnO2

Mn2+

Rate of the reaction hinderedinsolubility?

In basic solution…

MnO2 and Mn(OH)2

Mn2O3

* 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 MnO4

2- (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.

From the Frost diagram for Mn….

* 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.

* 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.

*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 stateLatter +2 state

*Fe and Mn – many oxidation states *High oxidation state:

Strong oxidizing agents