1. Oxidation-Reduction Redox reactions - transfer of electrons between species. All the redox reactions have two parts: OxidationReduction

2. The Loss of Electrons is OxidationThe 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 ReductionThe 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.

3. 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  G rx o (a) C (s) + ½ O 2 (g) CO (g)  G rx o (C,CO)‏ (b) ½ C (s) + ½ O 2 (g) ½ CO 2 (g)  g rx o (C,CO 2 )‏ (c) CO (g) + ½ O 2 (g) CO2(g)  G rx o (CO,CO 2 )‏ than a reaction of the form (d) xM (s or l) + ½ O 2 (g) M x O (s)  g rx o (M,M x O)

4. For one of the following, one must have a more negative  G rx o (a-d) M x O (s) + C (s) +  xM (s or l) + CO (g)  G rx o (C,CO) -  Gr x o (M,M x O)‏ and/or (b-d) M x O (s) + ½ C (s)  xM (s or l) + ½ CO 2(g)  G rx o (C,CO 2 ) -  Gr x o (M,M x O)‏ and/or (c.) M x O (s) + CO (g)  xM (s or l) + CO 2(g)  G rx o (CO,CO 2 ) -  Gr x o (M,M x O)‏

5. Ellingham Diagram

6.  G rx o =  H rx o - T  S rx o As enthalpy and entropy are independent of temperature, slope of line should –  S rx 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

7. Ellingham Diagram

8. 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,CO 2 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,CO 2 line lies below the metal oxide line, carbon monoxide can reduce the metal oxide to the metal and is oxidized to carbon dioxide

9. Balancing 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 H 2 O. Balance O atoms using H 2 O. Balance H atoms using H +. 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 -  H 2 O

10. Example Fe 2+ + MnO 4 - + H + Mn 2+ + Fe 3+ + H 2 O MnO 4 - Mn 2+ Reduction half reaction (+7)(+2)‏ Fe 2+ Fe 3+ Oxidation half reaction MnO 4 - + 8H + + 5e Mn 2+ + 4H 2 O 5Fe 2+ 5Fe 3+ +5e 5Fe 2+ + 8MnO 4 - + 8H + 8Mn 2+ + 5Fe 3+ + 4H 2 O

11. Nernst Equation aOx 1 +bRed 2 a’Red 1 + b’Ox 2 Q = [Red 1 ] a’ [Ox 2 ] b’ [Ox 1 ] a [Red 2 ] b E = E 0 - ln Q RT nF E 0 = Standard Potential R = Gas constant 8.314 J/K.mol F- Faraday constant = 94485 J/V.mol n- number of electrons

12.  G 0 = - n F  E 0 Note: if  G 0 0 A reaction is favorable if  E 0 > 0 Reaction is favorable 2H + (aq) + 2e H 2 (g) E 0 (H +, H 2 ) = 0 Zn 2+ (aq) + 2e Zn(s)E 0 (Zn 2+, Zn) = -0.76 V 2H + (aq) + Zn(s) Zn 2+ (aq) + H 2 (g)E 0 = +0.76 V

13. Hydrogen Electrode consists of a platinum electrode covered with a fine powder of platinum around which H 2(g) is bubbled. Its potential is defined as zero volts. Hydrogen Half-Cell H 2(g) = 2 H + (aq) + 2 e - reversible reaction

14. Galvanic Cell

16. Cyclic Voltammetry K 3 [Fe(CN) 6 ] + 0.1 M KCl E p a + E p c E 1/2 = 2

17. Diagrammatic presentation of potential data Latimer Diagram Frost Diagram Pourbaix diagram

18. Latimer Diagram * Oxidation number decrease from left to right and the E 0 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 reduced species on the right. reduced species on the right. Latimer diagram for chlorine in acidic solution

19. In basic solution… ClO 4 - ClO 3 - ClO 2 - ClO - Cl 2 Cl - +0.37+0.3+0.68+0.42+1.36 +0.89 ClO - Cl 2 +0.42 ClO - (aq) + 2H 2 O(l) + 2e - Cl 2 (g) + OH - (aq) E 0 = 0.42 V

20. 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. Disproportionation

22. 2 Cu + (aq) Cu 2+ (aq) + Cu(s)‏ Cu + (aq) + e - Cu(s) E 0 = + 0.52 V Cu 2+ (aq) + e - Cu + (aq) E 0 = =0.16 V Cu(I) undergo disproportionation in aqueous solution Another example…

23. Comproportionation reaction Ag 2+ (aq) + Ag(s) 2Ag + (aq) E 0 = + 1.18 V Reverse of disproportionation …we will study this detail under Frost diagram

24. X N + Ne - X 0 NE 0 = -  G 0 /F Frost Diagram Graphically illustration of the stability of different oxidation states relative to its elemental form (ie, relative to oxidation state= 0)‏

25. The oxidizing agent in the couple with more positive slope - more positive E The reducing agent of the couple with less positive slope

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

27. Important information provided by Frost diagrams:

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

29. Comproportionation is spontaneous if the intermediate species lies below the straight line joining the two reactant species. In acidic solution… Mn and MnO 2 Mn 2+ Rate of the reaction hindered insolubility? In basic solution… MnO 2 and Mn(OH) 3 Mn 2 O 3

32. 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.

33. * 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: MnO 4 3- MnO 2 and MnO 4 2- (in basic solution)‏ * Any species located on the upper right side of the diagram will be a strong oxidizing agent. MnO 4 - 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.

34. * 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 Mn 2+ 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.

36. 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.

37. M MO M +2 H+H+ H+H+ H2OH2O OH - 3.1 M/H 2 O system For metal M immersed in the pure water (25°C) in which the metal M is divalent and forms solvated M 2+ ions and the simple oxide is MO. M 2+ +2e - = M E M 2+ /M = E° M 2+ /M + 0.059/2 log a M 2+ = E° M 2+ /M +0.0295 log10 -6 (1) MO + 2H + + 2e - = M + H 2 O E MO/M = E° MO/M + 0.059 log a H + = E° MO/M - 0.059pH(2)

38. Potential H 2 O is stable H 2 is stable 714 2.0 1.6 0.8 1.2 -0.4 0.4 0.0 -1.6 -0.8 -1.2 0 O 2 is stable - Water itself is sensitive to electrode potential: …decomposes to hydrogen below a potential E H 2 given by : 2H + + 2e - = H 2 E H 2 = 0.00 - 0.059/2 [log P H 2 - log a H + 2 ] = 0.00 - 0.059 pH at 1 atm = -0.03 - 0.059 pH at 10 atm …decomposes to oxygen above E O 2 given by : 1/2 O 2 + 2H + + 2e - = H 2 O E O 2 = 1.23V E O 2 = 1.23 + 0.059/2 [log P O 2 1/2 a H + 2 ] = 1.23 - 0.059 pH at P O 2 = 1atm = 1.23 + 0.015 - 0.059 pH at P O 2 =10 atm It is evident that the increase of pressure increases the range of stability of water 2H 2 O = O 2 + 4H + + 4e - Equilibrium potential falls as pH increases 2H 2 O = O 2 + 4H + + 4e - Equilibrium potential falls as pH increases 2H + + 2e - = H 2 Equilibrium potential falls as pH increases

40. 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 involving hydrogen or hydroxide ions.Horizontal lines separate species in redox equilibria not involving hydrogen or hydroxide ions. Diagonal boundaries separate species in redox equilibria in which hydroxide or hydrogen ions are involved.Diagonal boundaries separate species in redox equilibria in which hydroxide or hydrogen ions are involved. Dashed lines enclose the practical region of stability of the water solvent to oxidation or reduction.Dashed lines enclose the practical region of stability of the water solvent to oxidation or reduction.

41. What 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.

42. Reducing agents and reducing conditions are found at the bottom of a diagram and not elsewhere. Strong reducing agents have low upper boundaries on the diagram. Manganese metal is a reducing agent over all pH ranges and is strongest in basic conditions.Reducing agents and reducing conditions are found at the bottom of a diagram and not elsewhere. Strong reducing agents have low upper boundaries on the diagram. Manganese metal is a reducing agent over all pH ranges and is strongest in basic conditions. A 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. 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.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. MnO 4 2- tends to disproportionate.MnO 4 2- tends to disproportionate.

43. Pourbaix diagram of Fe-H 2 O system Fe 2+, FeO, Fe 2 O 3, Fe 3 O 4,Fe(OH) 2, Fe(OH) 3. Fe 3+ H 2, O 2 H +, OH -

44. Equilibrium Reactions of iron in Water 1. 2 e - + 2H + = H 21. 2 e - + 2H + = H 2 2. 4 e - + O 2 + 4H + = 2H 2 O2. 4 e - + O 2 + 4H + = 2H 2 O 3. 2 e - + Fe(OH) 2 + 2H + = Fe + 2H 2 O3. 2 e - + Fe(OH) 2 + 2H + = Fe + 2H 2 O 4. 2 e - + Fe 2+ = Fe4. 2 e - + Fe 2+ = Fe 5. 2 e - + Fe(OH) 3 - + 3H + = Fe + 3H 2 O5. 2 e - + Fe(OH) 3 - + 3H + = Fe + 3H 2 O 6. e - + Fe(OH) 3 + H + = Fe(OH) 2 + H 2 O6. e - + Fe(OH) 3 + H + = Fe(OH) 2 + H 2 O 7. e - + Fe(OH) 3 + 3H + = Fe 2+ + 3H 2 O7. e - + Fe(OH) 3 + 3H + = Fe 2+ + 3H 2 O 8. Fe(OH) 3- + H + = Fe(OH)2 + H 2 O8. Fe(OH) 3- + H + = Fe(OH)2 + H 2 O 9. e - + Fe(OH) 3 = Fe(OH) 3 -9. e - + Fe(OH) 3 = Fe(OH) 3 - 10. Fe 3+ + 3H 2 O = Fe(OH) 3 + 3H +10. Fe 3+ + 3H 2 O = Fe(OH) 3 + 3H + 11. Fe 2+ + 2H 2 O = Fe(OH) 2 + 2H +11. Fe 2+ + 2H 2 O = Fe(OH) 2 + 2H + 12. e - + Fe 3+ = Fe 2+12. e - + Fe 3+ = Fe 2+ 13. Fe 2+ + H 2 O = FeOH + + H +13. Fe 2+ + H 2 O = FeOH + + H +

45. 14.FeOH + + H 2 O = Fe(OH) 2 (sln) + H +14.FeOH + + H 2 O = Fe(OH) 2 (sln) + H + 15.Fe(OH) 2 (sln) + H 2 O = Fe(OH) 3 - + H +15.Fe(OH) 2 (sln) + H 2 O = Fe(OH) 3 - + H + 16.Fe 3+ + H 2 O = FeOH 2+ + H +16.Fe 3+ + H 2 O = FeOH 2+ + H + 17.FeOH 2+ + H 2 O = Fe(OH) 2 + + H +17.FeOH 2+ + H 2 O = Fe(OH) 2 + + H + 18.Fe(OH) 2 + + H 2 O = Fe(OH) 3 (sln) + H +18.Fe(OH) 2 + + H 2 O = Fe(OH) 3 (sln) + H + 19.FeOH 2+ + H + = Fe 2+ + H 2 O19.FeOH 2+ + H + = Fe 2+ + H 2 O 20.e- + Fe(OH) 2 + + 2H + = Fe 2+ + 2H 2 O20.e- + Fe(OH) 2 + + 2H + = Fe 2+ + 2H 2 O 21. e- + Fe(OH) 3 (sln) + H + = Fe(OH) 2 (sln) + H 2 O21. e- + Fe(OH) 3 (sln) + H + = Fe(OH) 2 (sln) + H 2 O 22.e- + Fe(OH) 3 (sln) + 2H + = FeOH+ + 2H 2 O22.e- + Fe(OH) 3 (sln) + 2H + = FeOH+ + 2H 2 O 23. e- + Fe(OH) 3 (sln) + 3H + = Fe 2+ + 3H 2 O23. e- + Fe(OH) 3 (sln) + 3H + = Fe 2+ + 3H 2 O

46. Pourbaix Diagram for Iron

47. Potential 714 2.0 1.6 0.8 1.2 -0.4 0.4 0.0 -1.6 -0.8 -1.2 0 Fe metal stable Fe 3+ Fe oxides stable Will iron corrode in acid? Fe 2+ stable Yes - there is a reasonably wide range of potentials where hydrogen can be evolved and iron dissolved 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. Will iron corrode in alkaline solution? No - iron forms a solid oxide at all potentials, and will passivate

48. 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.