1. Oxidation-Reduction Biology Industry Environment

2. Biology

3. 0

4. Industry Synthesis of different compounds Extraction of elements

5. Environment

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

7. 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. Cu 2+ Cu MgMg 2+

8. 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 H 2 O. 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 -  H 2 O Fe 2+ + MnO 4 - + H + Mn 2+ + Fe 3+ + H 2 O

9. 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+ + MnO 4 - + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O

10. 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

11.  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 (a) (b) (a-b)

12. 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

13. Galvanic Cell

14. -

15. Diagrammatic presentation of potential data Latimer Diagram Frost Diagram

16. 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. A +5 B +3 C +1 D 0 E -2 x yz w

17. - 0.44

18. Iron+2 and +3 What 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 Fe 3+ +0.036 Fe 2+ +0.44 Fe 3+ Fe 2+ Fe -0.036 +0.77-0.44 -0.440 -0.771

19. Fe 3+ Fe 2+ -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

20. Oxidation of elemental copper

21. Latimer diagram for chlorine in acidic solution Can you balance the equation? balance the equation

22. Write the balanced equation for the first couple Write the balanced equation for the second couple 1 2 HClO(aq) + H + (aq) + e ½ Cl 2 (g) + H 2 O(l) +1.63 V ½ Cl 2 (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 E 0 for nonadjacent oxidation state? E 0 =? Identify the two reodx couples 0 +1

23. 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 2ClO - (aq) + 2H 2 O(l) + 2e - Cl 2 (g) + 4OH - (aq) E 0 = 0.42 V Latimer diagram for chlorine in basic solution +0.89 Balance the equation… Find out the E 0 +0.89

24. ClO 4 - ClO 3 - ClO 2 - ClO - Cl 2 Cl - +0.37+0.3+0.68+0.42+1.36Disproportionation Element is simultaneously oxidized and reduced. 2M + (aq) M 2+ (aq) M (s) E0E0E0E0 E 0’ 2 M + (aq) M(s) + M 2+ (aq) the species can oxidize and reduce itself, a process known as ‘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’. ClO - Cl 2 Cl - +0.42+1.36

25. ClO 4 - ClO 3 - ClO 2 - ClO - Cl 2 Cl - +0.37+0.3+0.68+0.42+1.36 Cl 2 + 2OH - ClO - + Cl - + H 2 O ClO - Cl 2 Cl - +0.42+1.36  E = E 0 (Cl 2 /Cl - ) –E 0 (ClO - /Cl 2 ) = 1.36 - +0.42 = 0.94 Reaction is spontaneous Cl 2 (g) + 2 e - 2Cl - (aq)+1.36 2ClO - (aq) 2H 2 O(l) +2e - Cl 2 (g) + 4OH - (aq)+0.42

26. Latimer diagram for Oxygen 1.23 V

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

28. H 2 O 2 (aq) + 2H + (aq) +2e - 2H 2 O(aq)+1.76 V O 2 (g) + 2H + (aq) +2 e - H 2 O 2 (aq) +0.7 V H 2 O 2 (aq) O 2 (g) + H 2 O(l)+0.7 VYes the reaction is spontaneous Is it spontaneous?

30. 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…

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

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

33. Look at the Latimer diagram of nitrogen in acidic solution a bc de fgh

34. N2N2 a b c d e f g h  G =  G ’ +  G ’’ -nFE = -n ’ FE ’ - n’’FE’’ E = n’E’+ n’’E’’ n’+n’’

35. NO 3 - + 6H + + 5e - ½ N 2 + 3H 2 O E 0 = 1.25V ½ N 2 O 4 + 4H + + 4e - ½ N 2 + 2H 2 O E 0 = 1.36V HNO 2 + 3H + + 3e - ½ N 2 + 2H 2 O E 0 = 1.45V NO + 2H + + 2e - ½ N 2 + H 2 O E 0 = 1.68V ½ N 2 O + H + + e - ½ N 2 + ½ H 2 O E 0 = 1.77V ½ N 2 + 2H + + H 2 O + e - NH 3 OH + E 0 = -1.87V ½ N 2 + 5/2 H + + 2e - ½ N 2 H 5 + E 0 = -0.23V ½ N 2 + 4H + + 3e - NH 4 + E 0 = 0.27V a b c d e f g h

36. N(V): NO 3 - (5 x 1.25, 5) N(IV): N 2 O 4 (4 x 1.36, 4) N(III): HNO 2 (3 x 1.35, 3) N(II): NO (2 x 1.68, 2) N(I): N 2 O (1 x 1.77, 1) N(-I): NH 3 OH + [-1 x (-1.87), -1] N(-II): N 2 H 5 + [-2 x (-0.23), -2] N(-III): NH 4 + (-3 x 0.27, -3) Oxidation state: speciesNE 0, N

37. Frost Diagram – N 2

39. the lowest lying species corresponds to the most stable oxidation state of the element What do we really get from the Frost diagram?

40. Slope of the line joining any two points is equal to the std potential of the couple. N’ N’E 0’ N”E 0’’ N’’ Slope = E 0 = N’E 0 ’-N”E 0” N’-N”

41. 3, 4.4 2, 3.4 Slope = E 0 = N’E 0 ’-N”E 0” N’-N”  1 V E 0 of a redox couple HNO 2 /NO

42. 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?

43. Identifying strong or weak agent?

44. NO – Strong oxidant than HNO 3

46. Disproportionation Element is simultaneously oxidized and reduced. 2M + (aq) M 2+ (aq) M (s) E0E0E0E0 E 0’ 2 M + (aq) M(s) + M 2+ (aq) the species can oxidize and reduce itself, a process known as ‘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’.

47. Disproportionation What Frost diagram tells about this reaction?

48. A 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.

49. Disproportionation…. another example

50. Comproportionation reaction

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

53. Favorable Favorable?

56. NE 0

57. Disproportionation

58. Comproportionation In acidic solution… Mn and MnO 2 Mn 2+ Rate of the reaction hindered insolubility? In basic solution… MnO 2 and Mn(OH) 2 Mn 2 O 3

59. * 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 - - 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….

60. * 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.

61. *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