oxidation-reduction redox reactions - transfer of electrons between species. all the redox reactions...

Oxidation-ReductionOxidation-Reduction

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

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

OxidationOxidation ReductionReduction

• The The LLoss of oss of EElectrons is lectrons is OOxidationxidation– An element that loses electrons is said to be An element that loses electrons is said to be

oxidized. The species in which that element is oxidized. The species in which that element is present in a reaction is called the present in a reaction is called the reducing reducing agentagent..

• The The GGain of ain of EElectrons is lectrons is RReductioneduction– An element that gains electrons is said to be An element that gains electrons is said to be

reduced. The species in which that element is reduced. The species in which that element is present in a reaction is called the present in a reaction is called the oxidizing oxidizing agentagent..

What is the original definition for oxidation and reduction?What is the original definition for oxidation and reduction?

For metal oxide to be reduced to metal, one of the three following reactions must have a more negative Grx

o

(a) C(s) + ½ O2(g) CO(g) Grxo (C,CO)

(b) ½ C(s) + ½ O2(g) ½ CO2(g) grxo (C,CO2)

(c) CO(g) + ½ O2(g) CO2(g) Grxo(CO,CO2)

than a reaction of the form

(d) xM(s or l) + ½ O2(g) MxO(s) grxo (M,MxO)

For one of the following, one must have a more negative Grx

o

(a-d) MxO(s) + C(s) + xM(s or l) + CO(g)

Grxo(C,CO) - Grx

o(M,MxO)

and/or

(b-d) MxO(s) + ½ C(s) xM(s or l) + ½ CO2(g)

Grxo(C,CO2) - Grx

o(M,MxO)

and/or

(c.) MxO(s) + CO(g) xM(s or l) + CO2(g)

Grxo(CO,CO2) - Grx

o(M,MxO)

Ellingham Diagram

Grxo = Hrx

o - TSrxo

• As enthalpy and entropy are independent of temperature, slope of line should –Srx

o

• The entropies of gases are much larger than for solids

• Ellingham diagram should have a positive slope for reaction (d) when the net consumption of gas is negative

Ellingham Diagram

• For temperatures at which the C,CO line lies below the metal oxide, carbon can be used to reduce the metal oxide and itself is oxidized to carbon monoxide

• For temperatures at which the C,CO2 line lies below the metal oxide line, carbon can be used to achieve the reduction, but is oxidized to carbon dioxide

• For temperatures at which the CO,CO2 line lies below the metal oxide line, carbon monoxide can reduce the metal oxide to the metal and is oxidized to carbon dioxide

Balancing Redox EquationsBalancing Redox Equations

Assign oxidation numbers to each atom.Assign oxidation numbers to each atom. Determine the elements that get oxidized and reduced.Determine the elements that get oxidized and reduced. Split the equation into half-reactions.Split the equation into half-reactions. Balance all atoms in each half-reaction, except H and O.Balance all atoms in each half-reaction, except H and O. Balance O atoms using HBalance O atoms using H22O.O.

Balance H atoms using HBalance H atoms using H++..

7.7. Balance charge using electrons.Balance charge using electrons.

8.8. Sum together the two half-reactions, so that: eSum together the two half-reactions, so that: e-- lost = e lost = e-- gained gained

9.9. If the solution is basic, add a number of OHIf the solution is basic, add a number of OH-- ions to each side ions to each side of the equation equal to theof the equation equal to the number of Hnumber of H++ ions shown in the ions shown in the overall equation. Note that overall equation. Note that

HH++ + OH + OH-- H H22OO

ExampleExample

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

Nernst EquationNernst Equation

aOx1 +bRed2 a’Red1 + b’Ox2

Q =[Red1]a’ [Ox2]b’

[Ox1]a [Red2]b

E = E0 - ln Q RTnF

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

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

Cyclic VoltammetryCyclic Voltammetry

KK33[Fe(CN)[Fe(CN)66] + 0.1 M KCl] + 0.1 M KCl

Epa + Ep

c

E1/2 = 2

Diagrammatic presentation of potential dataDiagrammatic presentation of potential data

Latimer DiagramLatimer Diagram

Frost DiagramFrost Diagram

Pourbaix diagramPourbaix 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.

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

In basic solution…

ClO4- ClO3

- ClO2- ClO- Cl2 Cl-

+0.37 +0.3 +0.68 +0.42 +1.36

+0.89

ClO- Cl2

+0.42

ClO- (aq) + 2H2O(l) + 2e- Cl2(g) + OH-(aq) E0 = 0.42 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 known as disproportionation.

DisproportionationDisproportionation

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 detail under Frost diagram

XN + Ne- X0

NE0 = -G0/F

Frost DiagramFrost Diagram

Graphically illustration of the stability of different oxidation states relative to its elemental form (ie, relative to oxidationstate= 0)

The oxidizing agent in the couple with more positive slope - more positive E

The reducing agent of the couple with less positive slope

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.

Important information provided by Frost diagrams:Important information provided by Frost diagrams:

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

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

In acidic solution…

Mn and MnO2

Mn2+

Rate of the reaction hinderedinsolubility?

In basic solution…

MnO2 and Mn(OH)3

Mn2O3

The most useful aspect of Frost diagrams is that they allow us to predict whether a RedOx reaction will occur for a given pair of reagents and what the outcome of the reaction will be.

* 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

- is a strong oxidizer.

* Any species located on the upper left side of the diagram will be a reducing agent. Mn is a moderate reducing agent.

* 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.

Pourbaix Diagrams

Marcel Pourbaix 1904, Myshega, Russia

Marcel Pourbaix provided the brilliant means to utilize thermodynamics more effectively in corrosion science and electrochemistry in general.

 Graphical representations of thermodynamic and electrochemical equilibria between metal and water, indicating thermodynamically stable phases as a function of electrode potential and pH.

 - predicts the spontaneous direction of reactions.

- estimates the composition of corrosion products.- predicts environmental changes that will prevent or reduce

corrosion attack.

MMO

M+2

H+

H+

H2O

OH-

3.1 M/H3.1 M/H22O systemO system

 For metal M immersed in the pure water (25°C) in which the metal M is divalent and forms solvated M2+ ions and the simple oxide is MO.

M2+ +2e- = M

EM2+/M = E°M2+/M + 0.059/2 log aM2+

= E°M2+/M +0.0295 log10-6 (1)

MO + 2H++ 2e- = M + H2O

EMO/M = E°MO/M + 0.059 log aH+ = E°MO/M - 0.059pH (2)

Po

ten

tial

H2O is stable

H2 is stable

7 14

2.0

1.6

0.8

1.2

-0.4

0.4

0.0

-1.6

-0.8

-1.2

0

O2 is stable

- Water itself is sensitive to electrode potential: …decomposes to hydrogen below a potential EH2 given by :

2H+ + 2e- = H2

EH2 = 0.00 - 0.059/2 [log PH2 - log aH+2] = 0.00 - 0.059 pH at 1 atm

= -0.03 - 0.059 pH at 10 atm

…decomposes to oxygen above EO2 given by :

1/2 O2 + 2H+ + 2e- = H2O EO2 = 1.23V

EO2 = 1.23 + 0.059/2 [log PO21/2 aH+2]

= 1.23 - 0.059 pH at PO2= 1atm

= 1.23 + 0.015 - 0.059 pH at PO2 =10 atm

It is evident that the increase of pressure increases the range of stability of water

2H2O = O2 + 4H+ + 4e-

Equilibrium potential falls as pH increases

2H2O = O2 + 4H+ + 4e-

Equilibrium potential falls as pH increases

2H+ + 2e- = H2 Equilibrium

potential falls as pH increases

2H+ + 2e- = H2 Equilibrium

potential falls as pH increases

How to Read a Pourbaix Diagram How to Read a Pourbaix Diagram

•Vertical lines separate species that are in acid-base equilibrium. Vertical lines separate species that are in acid-base equilibrium.

•Non vertical lines separate species related by redox equilibria. Non vertical lines separate species related by redox equilibria.

•Horizontal lines separate species in redox equilibria not Horizontal lines separate species in redox equilibria not involving hydrogen or hydroxide ions. involving hydrogen or hydroxide ions.

•Diagonal boundaries separate species in redox equilibria in Diagonal boundaries separate species in redox equilibria in which hydroxide or hydrogen ions are involved. which hydroxide or hydrogen ions are involved.

•Dashed lines enclose the practical region of stability of the water Dashed lines enclose the practical region of stability of the water solvent to oxidation or reduction. solvent to oxidation or reduction.

What You Can Learn From a Pourbaix DiagramWhat You Can Learn From a Pourbaix Diagram

•Any point on the diagram will give the thermodynamically most stable (and theoretically most abundant) 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.

•Reducing agents and reducing conditions are found at the Reducing agents and reducing conditions are found at the bottom of a diagram and not elsewhere.bottom of a diagram and not elsewhere.Strong reducing agents have low upper boundaries on the Strong reducing agents have low upper boundaries on the diagram. diagram. Manganese metal is a reducing agent over all pH ranges Manganese metal is a reducing agent over all pH ranges and is strongest in basic conditions.and is strongest in basic conditions.

A species that ranges from the top to the bottom of the A species that ranges from the top to the bottom of the diagram at a given pH will have no oxidizing or reducing diagram at a given pH will have no oxidizing or reducing properties at that pH. properties at that pH.

•When the predominance area for a given oxidation state When the predominance area for a given oxidation state disappears completely above or below a given pH and the disappears completely above or below a given pH and the element is in an intermediate oxidation state, the element element is in an intermediate oxidation state, the element will undergo disproportionation. will undergo disproportionation.

•MnOMnO44

2-2- tends to disproportionate. tends to disproportionate.

Pourbaix diagram of Fe-HPourbaix diagram of Fe-H22O systemO system

FeFe2+2+, FeO, Fe, FeO, Fe22OO33, Fe, Fe33OO44,Fe(OH),Fe(OH)22, Fe(OH), Fe(OH)33. Fe. Fe3+3+

HH22, O, O22

HH++, OH, OH--

Equilibrium Reactions of iron in WaterEquilibrium Reactions of iron in Water• 1. 1. 2 e2 e-- + 2H + 2H++  = H  = H22

• 2. 2. 4 e4 e- - + O+ O22  + 4H  + 4H++  = 2H  = 2H22O O

• 3. 3. 2 e2 e- - + Fe(OH)+ Fe(OH)22  + 2H  + 2H++  = Fe + 2H  = Fe + 2H22O O

• 4. 4. 2 e2 e- - + Fe+ Fe2+2+  = Fe   = Fe

• 5. 5. 2 e2 e-- + Fe(OH) + Fe(OH)33--  + 3H  + 3H++  = Fe + 3H  = Fe + 3H22O O

• 6. 6.  ee-- + Fe(OH) + Fe(OH)33  + H  + H++  = Fe(OH)  = Fe(OH)22 + H + H22O O

• 7. 7.  ee-- + Fe(OH) + Fe(OH)33  + 3H  + 3H++  = Fe  = Fe2+2+ + 3H + 3H22O O

• 8. 8. Fe(OH)Fe(OH)3- 3-  + H + H++  = Fe(OH)2 + H  = Fe(OH)2 + H22O O

• 9.  9.   ee-- + Fe(OH) + Fe(OH)33  = Fe(OH)  = Fe(OH)33--

• 10. 10. FeFe3+3+  + 3H  + 3H22O  = Fe(OH)O  = Fe(OH)33 + 3H + 3H++

• 11. Fe11. Fe2+2+  + 2H  + 2H22O  = Fe(OH)O  = Fe(OH)2 2 + 2H+ 2H++

• 12.  e12.  e- - + Fe+ Fe3+3+  = Fe  = Fe2+2+

• 13. Fe13. Fe2+2+  + H  + H22O  = FeOHO  = FeOH++ + H + H++

• 14.14. FeOHFeOH++  + H  + H22O  = Fe(OH)O  = Fe(OH)22(sln) + H(sln) + H+ +

• 15.15. Fe(OH)Fe(OH)22(sln)  + H(sln)  + H22O  = Fe(OH)O  = Fe(OH)33-- + H + H++

• 16.16. FeFe3+3+  + H  + H22O  = FeOHO  = FeOH2+2+ + H + H++

• 17.17. FeOHFeOH2+2+  + H  + H22O  = Fe(OH)O  = Fe(OH)22++ + H + H++

• 18.18. Fe(OH)Fe(OH)22++  + H  + H22O  = Fe(OH)O  = Fe(OH)33(sln) + H(sln) + H++

• 19.19. FeOHFeOH2+2+  + H  + H++  = Fe  = Fe2+2+ + H + H22O O

• 20.20. e- + Fe(OH)e- + Fe(OH)22++  + 2H  + 2H++  = Fe  = Fe2+2+ + 2H + 2H22O O

• 21.21.  e- + Fe(OH) e- + Fe(OH)33(sln)  + H(sln)  + H++  = Fe(OH)  = Fe(OH)22(sln) + H(sln) + H22O O

• 22.22. e- + Fe(OH)e- + Fe(OH)33(sln)  + 2H(sln)  + 2H++  = FeOH+ + 2H  = FeOH+ + 2H22O O

• 23.23.  e- + Fe(OH) e- + Fe(OH)33(sln)  + 3H(sln)  + 3H++  = Fe  = Fe2+2+ + 3H + 3H22OO

Pourbaix Diagram for Iron

Pourbaix Diagram for Iron

Pot

enti

al

7 14

2.01.6

0.81.2

-0.4

0.40.0

-1.6

-0.8-1.2

0

Fe metal stable

Fe3+

Fe oxidesstable

Will iron corrode in

acid?

Will iron corrode in

acid?

Fe2+ stable

Yes - there is a reasonably wide

range of potentials where hydrogen

can be evolved and iron dissolved

Yes - there is a reasonably wide

range of potentials where hydrogen

can be evolved and iron dissolved

Will iron corrode in

neutral waters?

Will iron corrode in

neutral waters?Yes - although iron can form an oxide in neutral solution, it tends not to

form directly on the metal, as the potential

is too low, therefore it is not protective.

Yes - although iron can form an oxide in neutral solution, it tends not to

form directly on the metal, as the potential

is too low, therefore it is not protective.

Will iron corrode in alkaline solution?

Will iron corrode in alkaline solution?

No - iron forms a solid oxide at all potentials,

and will passivate

No - iron forms a solid oxide at all potentials,

and will passivate

Limitations of Pourbaix Diagrams

• Tell us what can happen, not necessarily what will happen.

• No information on rate of reaction.

• Can only be plotted for pure metals and simple solutions, not for alloys.