6-1

Oxidation and Reduction• Transfer of electrons between species

Thermodynamic Kinetics

• Conditions• Procedures for evaluating redox• Trends in oxidation

• Reduction Electron gain

• Oxidation Electron loss

• Reducing agent supplies electrons• Oxidizing agent removes electrons

For an element, redox results in change of oxidation state

6-2

Periodic Variations of Oxidation State1 18

2 13-17

6-123 4 5

Steps of 1constant Steps of 2

Mainly 3+

6-3

Oxidizing and Reducing AgentsOxidizing Agents Reducing AgentsF2 F-

Cl2 Cl-

Br2 Br-

Ag+ AgI2 I-

Cu2+ CuH+ H2Fe2+ FeZn2+ ZnAl3+ AlNa+ Na

weak

Strong

6-4

Extraction of the elements

• Origin of the term Oxidation From the reaction of an element with O2

Reduction from oxide to metal• Separation by reduction

Metal oxides reduced at higher temperature in the presence of a reducing agent Carbon used in early smelting

Electrolysis Electric current used for reduction

6-5

Thermodynamics• Gibbs free energy

∆G=-RTlnK= ∆H-T ∆S Negative value

spontaneous Phase changes effect

free energy due to entropy

* Exploited in metal reduction

• Temperature dependence on free energy Metal reduced by C or

CO Eappl≥ ∆G/nF

6-6

Pyrometallurgy

6-7

Extraction by oxidation

• Separations of halogens 2 Cl- +2 H2O 2 OH-+H2+Cl2 ∆G=422 kJ/mol

• Apply overpotential Overpotential η always reduces theoretical cell potential when

current is flowing η = Ecurrent - Eequilibrium

Overpotential due to electrode polarization: concentration polarization - mass transport limited adsorption/desorption polarization - rate of surface

attach/detachment charge-transfer polarization - rate of redox reaction reaction polarization - rate of redox reaction of

intermediate in redox reaction Overpotential means one must apply greater potential before

redox chemistry occurs

6-8

Redox Reactions

• Zn + Cu2+ <--> Zn2+ + Cu Zn is oxidized, Cu is reduced Transfer of electrons from one metal to another

• May not involved charge species C + O2 <--> CO2

• Oxidation agent oxidizes another species and is reduced• Reduction agent reduces another species and is

oxidized

6-9

Balancing Redox Equations• Balancing can be accomplished through examining ion-

electron half reactions H+ + NO3

- + Cu2O <--> Cu2+ + NO + H2O• Identify reduced and oxidized species

Cu2O to Cu2+ (1+ to 2+): oxidized NO3

- to NO (5+ to 2+): reduced• Balance oxidized/reduced atoms

Cu2O <--> 2Cu2+

• Add electrons to balance redox of element Cu2O <--> 2Cu2+ + 2e-

NO3- + 3e- <--> NO

6-10

Balancing Redox Equations

• Add H+ (or OH-) to balance charge of reaction 2H+ + Cu2O <--> 2Cu2+ + 2e-

4 H+ + NO3- + 3e- <--> NO

• Add water to balance O and H, then balance other atoms if needed 2H+ + Cu2O <--> 2Cu2+ + 2e- + H2O 4 H+ + NO3

- + 3e- <--> NO + 2 H2O• Multiple equations to normalize electrons

3(2H+ + Cu2O <--> 2Cu2+ + 2e- + H2O) 2(4 H+ + NO3

- + 3e- <--> NO + 2 H2O)

6-11

Balancing Equations

• Add the reactions together 14H+ + 2NO3

- + 3Cu2O <--> 6Cu2++2NO +7 H2O• Important for reactions involving metal with multiple

oxidation statesDisproportionation

• Some elements with intermediate states can react to form species with different oxidation states

• Species acts as both oxidation and reduction agent 2 Pu4+ <--> Pu3+ + Pu5+

6-12

Electrochemistry• Chemical transformations produced by

electricity Corrosion Refining

• Electrical Units Coulomb (C)

Charge on 6.25 x 1018 electrons Amperes (A)

Electric current A=1C/sec

6-13

Electrochemistry• Volt (V)

Potential driving current flow V= 1 J/C

• Ohm’s law ε = IR

ε = potential, I =current, and R=resistancesymbol unit relationships

Charge q Coulomb (C)Current I Ampere (A) I=q/t (t in s)Potential ε Volt (V) ε=IRPower P Watt (W) P= εIEnergy E Joule (J) Pt= εIt= εqResistance R Ohm (Ω) R= ε/I

6-14

Electrolysis

• Production of a chemical reaction by means of an electric current 2 H2O <--> 2H2 + O2

• Cathode Electrode at which reduction occurs Cations migrate to cathode

Cu2+ + 2e- <--> Cu• Anode

Electrode at which oxidation occurs Anions migrate to anode

2Cl- <-->Cl2 + 2e-

6-15

Electrolysis

• Redox depends upon tendencies of elements or compounds to gain or lose electrons electrochemical series

Lists of elements or compounds Half cell potentials

• Related to periodic tendencies

6-16

Electrolysis of CuCl2

C electrode

Cl2

Anode: 2Cl-->Cl2+2e-

Cathode: Cu2++2e-->Cu

C electrode

Cu Plating on C electrode

6-17

NaCl Solutions• Dilute NaCl solution

anode: 2 H2O <--> O2 + 4H+ + 4e-

cathode: 2 H2O + 2e- <--> H2 + 2OH-

• Concentrated NaCl (Brines) anode: 2Cl- <--> Cl2 + 2e-

cathode: 2 H2O + 2e- <--> H2 + 2OH-

• Molten Salt anode: 2Cl- <--> Cl2 + 2e-

cathode: Na+ + e- <-->Na Na metal produced by electrolysis of NaCl and

Na2CO3 Lower melting point than NaCl

6-18

Faraday Laws

• In 1834 Faraday demonstrated that the quantities of chemicals which react at electrodes are directly proportional to the quantity of charge passed through the cell

• 96487 C is the charge on 1 mole of electrons = 1F (faraday)

6-19

Faraday Laws

• Cu(II) is electrolyzed by a current of 10A for 1 hr between Cu electrode anode: Cu <--> Cu2+ + 2e-

cathode: Cu2+ + 2e- <--> Cu Number of electrons

(10A)(3600 sec)/(96487 C/mol) = 0.373 F 0.373 mole e- (1 mole Cu/2 mole e-) =

0.186 mole Cu

6-20

Electrochemical cell

6-21

Conduction in a cell

• Charge is conducted Electrodes Ions in solution Electrode surfaces

Oxidation and reduction Oxidation at anode Reduction at cathode

• Reaction can be written as half-cell potentials

6-22

Half-cell potentials• Standard potential

Defined as ε°=0.00V H2(atm) <--> 2 H+ (1.000M) + 2e-

• Cell reaction for Zn and Fe3+/2+ at 1.0 M Write as reduction potentials

Fe3+ + e- <--> Fe2+ ε°=0.77 V Zn2+ + 2e- <-->Zn ε°=-0.76 V

Fe3+ is reduced, Zn is oxidized

6-23

Half-Cell Potentials• Overall

2Fe3+ +Zn <--> 2Fe2+ + Zn2+ ε°=0.77+0.76=1.53 V• Half cell potential values are not multiplied

Application of Gibbs• If work is done by a system

∆G = -ε°nF (n= e-)• Find ∆G for Zn/Cu cell at 1.0 M

Cu2+ + Zn <--> Cu + Zn2+ ε°=1.10 V

2 moles of electrons (n=2) ∆G =-2(96487C/mole e-)(1.10V) ∆G = -212 kJ/mol

6-24

Reduction PotentialsElectrode Couple "E0, V"Na+ + e- --> Na -2.7144Mg2+ + 2e- --> Mg -2.3568Al3+ + 3e- --> Al -1.676Zn2+ + 2e- --> Zn -0.7621Fe2+ + 2e- --> Fe -0.4089Cd2+ + 2e- --> Cd -0.4022Tl+ + e- --> Tl -0.3358Sn2+ + 2e- --> Sn -0.141Pb2+ + 2e- --> Pb -0.12662H+ + 2e- --> H2(SHE) 0S4O62- + 2e- --> 2S2O32- 0.0238Sn4+ + 2e- --> Sn2+ 0.1539SO42- + 4H+ + 2e- --> H2O + H2SO3(aq) 0.1576Cu2+ + e- --> Cu+ 0.1607S + 2H+ + 2e- --> H2S 0.1739AgCl + e- --> Ag + Cl- 0.2221Saturated Calomel (SCE) 0.2412UO22+ + 4H+ + 2e- --> U4+ + 4H2O 0.2682

6-25

Reduction PotentialsHg2Cl2 + 2e- --> 2Cl- + 2Hg 0.268Bi3+ + 3e- --> Bi 0.286Cu2+ + 2e- --> Cu 0.3394Fe(CN)63- + e- --> Fe(CN)64- 0.3557Cu+ + e- --> Cu 0.518I2 + 2e- --> 2I- 0.5345I3- + 2e- --> 3I- 0.5354H3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O 0.57482HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2Cl- 0.6011Hg2SO4 + 2e- --> 2Hg + SO42- 0.6152I2(aq) + 2e- --> 2I- 0.6195O2 + 2H+ + 2e- --> H2O2(l) 0.6237O2 + 2H+ + 2e- --> H2O2(aq) 0.6945Fe3+ + e- --> Fe2+ 0.769Hg22+ + 2e- --> Hg 0.7955Ag+ + e- --> Ag 0.7991Hg2+ + 2e- --> Hg 0.85192Hg2+ + 2e- --> Hg22+ 0.9083NO3- + 3H+ + 2e- -->HNO2(aq) + H2O 0.9275

6-26

Reduction PotentialsVO2+ + 2H+ + e- --> VO2+ + H2O 1.0004HNO2(aq) + H+ + e- --> NO + H2O 1.0362Br2(l) + 2e- --> 2Br- 1.0775Br2(aq) + 2e- --> 2Br- 1.09782IO3- + 12H+ + 10e- -->6H2O + I2 1.2093O2 + 4H+ + 4e- --> 2H2O 1.2288MnO2 + 4H+ + 2e- -->Mn2+ + 2H2O 1.1406Cl2 + 2e- --> 2Cl- 1.3601MnO4- + 8H+ + 5e- -->4H2O + Mn2+ 1.5119

2BrO3- + 12H+ + 10e- -->6H2O + Br2 1.5131

6-27

Nernst Equation• Compensated for non unit activity (not 1 M)• Relationship between cell potential and activities• aA + bB +ne- <--> cC + dD

• At 298K 2.3RT/F = 0.0592• What is potential of an electrode of Zn(s) and 0.01 M

Zn2+

• Zn2+ +2e- <--> Zn ε°= -0.763 V• activity of metal is 1

ε = ε° −2.30RT

nFlog[C]c[D]d

[A]a[B]b

ε = −0.763 −0.0592

2log 1

0.01= −0.822V

6-28

Kinetic Factors• Overpotential of 0.6 V for

metal reactions Oxidation of water Reduction of hydrogen

ions Added potential

beyond equilibrium conditions

• Electron transfer Outer sphere transfer

Minimal change in complex* No change in

coordination sphere

* Minimal bond distance

Rapid kinetics dN/dtαexp(∆Eº)

6-29

Kinetic Factors• Inner-sphere electron transfer

Change in coordination sphere Ligand change after

redox• Non-complementary redox

reaction Changes in oxidation

numbers of agents are unequal

Slow reactions since multiple steps Oxidation of p-

block oxoanions by d-block ions* Steps of 2 by

steps of 1• High oxidation state, smaller

radius, slower reduction

6-30

Cyclic Voltammetry

• Oxidation and reduction• Variation of rates• Peak potentials

Anode (bottom peak) Cathode (top peak)

Difference 0.0592/n• Peak currents

Cathode (line to peak) Anode (slope to bottom)

Peak currents equal and opposite sign

• Mechanisms and rates of redox

6-31

Redox stability in water

• Water redox Reduced to H2

2 H2O + 2e- H2 + 2OH- E=-0.059V*pH

Oxidized to O2 O2+4H++4e- 2

H2O E=1.23V-0.059V*pH

• Oxidation by water M + H+ M++0.5 H2

s block except Be, Group 4 to Group 7

May need to occur in acid rather than water

• Reduction by water Ion reduced by water with

O2 evolution

6-32

Oxidation by O2

• Oxidation of metals by O2 in air Fe3+/Fe2+ = 0.77 V

Consider O2 half reaction* O2+4H++4e- 2 H2O E=1.23V-

0.059V*pH➋ At pH=0, E=0.46 V➋ At pH 7, E=0.047 V

Consider Cu metal oxidation Cu2+/Cu=0.34 V

6-33

Potential Diagrams

• Latimer Diagrams Quantitative data for each element

• Frost Diagrams Qualitative description of relative oxidation

state stabilities

• Latimer Diagrams Shows changes in oxidation state and

species for a given element

6-34

6-35

Potential Diagrams• ClO4

- and ClO3- represents the half-reaction

ClO4- +2H+ +2e- ClO3

- + H2O, E=1.20 V Reaction is acidic, so proton is added

For basic conditions the diagram is

Cl2+ 2 e- 2 Cl- has no protons Need to include OH- for basic reaction for

balance• Non-adjacent species redox

Find based on ∆G = -ε°nF (n= e-) and ∆G = ∆G’+ ∆G”

-ε°nF= -ε°’n’F + -ε°”n”F , n=n’+n” ε°= (ε°’n’ + ε°”n”)/(n)

6-36

Disproportionation

• 2 M+ M + M2+

Positive E, spontaneous Evaluate half reactions to determine thermodynamics Tends to disproportionate if right potential is higher than

left potential• 2H2O2 2H2O + O2

H2O2+2H++2e- 2H2O E= 1.76 V O2+2H++2e- 2H2O2 E= 0.69 V E=1.07 V

6-37

Frost Diagrams• Plot of nE versus oxidation number

nE= -∆G/F Most stable oxidation state is

lowest nE value Slope related to potential

Can construct Frost diagrams from Latimer diagram

Need to consider electrons transferred in reactions

• Consider Tl Tl to Tl+ nE=-0.336 Tl to Tl3+ nE=2.16

Construct Frost diagram

-0.5

0

0.5

1

1.5

2

2.5

-0.5 0 0.5 1 1.5 2 2.5 3 3.5

nE (V

)

Oxidation number

6-38

Utilization of Frost Diagrams• Thermodynamic stability at the bottom of the diagram

Mn (II) is the most stable species. • species located on a convex curve can undergo

disproportionation MnO4

2- and Mn (III) tends to disproportionate. • species on a concave curve do not typically

disproportionate MnO2 does not disproportionate

• species on the upper left side is a oxidizing agent MnO4- is a strong oxidizer.

• species located on the upper rignt side of the diagram is reducing agent

manganese metal is a moderate reducing agent • information obtained from a Frost diagram is for species

under standard conditions (pH=0 for acidic solution and pH=14 for basic

solution)• Changes in pH may change the relative stabilities of the

species• potential of any process involving the hydrogen ion will

change with pH because the concentration of this species is changing

Need to consider in speciation Under basic conditions aqueous Mn2+ does not

exist, insoluble Mn(OH)2 forms

6-39

Pourbaix Diagram

• Eh-pH diagrams shows thermodynamically foavored form of

an element as a function of potential and pH

provide a visual representation of the oxidizing and reducing abilities of the major stable compounds of an element

used frequently in geochemical, environmental and corrosion applications

6-40

Pourbaix Diagram • Strong oxidizing agents and

oxidizing conditions are found at the top

• Reducing agents and reducing conditions are found at the bottom

• 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

• 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

• Horizontal lines indicate no proton involved in reaction

Mn2++2e- Mn

6-41

Complexation and potentials

• Metal complexes have different potentials than aquocoordinated ions Can stabilize oxidation state

Against both oxidation and reduction Fe(CN)6 3+/2+ = 0.36 V

* Related to stability➋ More stable complexes are more

resistant to reduction Consider Fe3+ to Fe

Which complex is more stable, hydroxide or S?