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Oxidation-Reduction

Oxidation-Reduction. Biology. Industry. Environment. Biology. Biology. 0. Industry. Extraction of elements. Synthesis of different compounds. Environment. Redox reactions - transfer of electrons between species. All the redox reactions have two parts:. Oxidation. Reduction.

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Oxidation-Reduction

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  1. Oxidation-Reduction Biology Industry Environment

  2. Biology

  3. Biology 0

  4. Industry Extraction of elements Synthesis of different compounds

  5. Environment

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

  7. Cu2+ Cu Mg Mg2+ • 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.

  8. Balancing Redox Equations Fe2+ + MnO4- + H+ Mn2+ + Fe3+ + H2O Assign oxidation numbers to each atom. Determine the elements that get oxidized and reduced. Split the equation into half-reactions. Balance all atoms in each half-reaction, except H and O. Balance O atoms using H2O. 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 thenumber of H+ ions shown in the overall equation. Note that H+ + OH- H2O

  9. 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 Example

  10. aOx1 +bRed2 a’Red1 + b’Ox2 [Red1]a’ [Ox2]b’ Q = [Ox1]a [Red2]b RT nF E = E0 - ln Q Nernst Equation E0 = Standard Potential R = Gas constant 8.314 J/K.mol F- Faraday constant = 94485 J/V.mol n- number of electrons

  11. 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 G0 = - n F  E0 Note: if G0 < 0, then  E0 must be >0 A reaction is favorable if  E0 > 0 (a) (b) (a-b) Reaction is favorable

  12. Hydrogen 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 reaction

  13. Galvanic Cell

  14. -

  15. Diagrammatic presentation of potential data Latimer Diagram Frost Diagram

  16. Latimer Diagram * Written with the most oxidized species on the left, and the most reduced species on the right. * Oxidation number decrease from left to right and the E0 values are written above the line joining the species involved in the couple. y z x w A+5 B+3 C+1 D0 E-2

  17. -0.44

  18. Fe3+ +0.036 +0.44 Fe2+ -0.036 +0.77 -0.44 What happens when Fe(s) react with H+? Iron +2 and +3 G = -nFE -2 x F x -0.44 = 0.88 V -0.440 -0.771 -1 x F x +0.771 = -0.771 V + 0.109 F = -3 x F x –0.036 Fe Fe2+ Fe Fe3+

  19. Fe2+ Fe Fe3+ [Fe(CN)6]3- [Fe(CN)6]4- -0.036 +0.77 -0.44 0.36 -1.16 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. E0 =? 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’’ ’+ ’’ How to extract E0 for nonadjacent oxidation state? +1 -1 0 1 Identify the two reodx couples 2 Find out the oxidation state of chlorine Write the balanced equation for the first couple Write the balanced equation for the second couple E = 1.5 V

  23. +0.37 +0.3 +0.68 +0.42 +1.36 ClO4- ClO3- ClO2- ClO- Cl2 Cl- +0.89 +0.42 ClO- Cl2 2ClO- (aq) + 2H2O(l) + 2e- Cl2(g) + 4OH-(aq) E0 = 0.42 V Latimer diagram for chlorine in basic solution +0.89 Balance the equation… Find out the E0 +0.89

  24. 2 M+(aq) M(s) + M2+(aq) E0 E0’ M(s) M2+(aq) 2M+(aq) +0.37 +0.3 +0.68 +0.42 +1.36 ClO4- ClO3- ClO2- ClO- Cl2 Cl- +0.42 +1.36 ClO- Cl2 Cl- Disproportionation Element is simultaneously oxidized and reduced. ‘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 asdisproportionation’.

  25. Cl2(g) + 2 e- 2Cl-(aq) +1.36 2ClO- (aq) 2H2O(l) +2e- Cl2(g) + 4OH-(aq) +0.42 Cl2 + 2OH- ClO- + Cl- + H2O +0.37 +0.3 +0.68 +0.42 +1.36 ClO4- ClO3- ClO2- ClO- Cl2 Cl- +0.42 +1.36 ClO- Cl2 Cl- E = E0 (Cl2/Cl-) –E0 (ClO-/Cl2) = 1.36 - +0.42 = 0.94 Reaction is spontaneous

  26. Latimer diagram for Oxygen 1.23 V

  27. Disproportionation 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 asdisproportionation.

  28. 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 Is it spontaneous? Yes the reaction is spontaneous

  29. Another example… 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

  30. Ag2+(aq) + Ag(s) 2Ag+(aq) E0 = + 1.18 V Comproportionation reaction Reverse of disproportionation …we will study this in detail under Frost diagram

  31. XN + Ne- X0 NE0 = -G0/F Frost Diagram Arthur A. Frost Graphically illustration of the stability of different oxidation states relative to its elemental form (ie, relative to oxidation state= 0)

  32. Look at the Latimer diagram of nitrogen in acidic solution d e a b c f g h

  33. c G = G’ + G’’ -nFE = -n’FE’ - n’’FE’’ E = n’E’+ n’’E’’ n’+n’’ a e b d g N2 f h

  34. 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

  35. Oxidation state: species NE0, N 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)

  36. Frost Diagram – N2

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

  38. Slope of the line joining any two points is equal to the std potential of the couple. N’E0’-N”E0” N’-N” Slope = E0= N’E0’ N’ N”E0’’ N’’

  39. N’E0’-N”E0” N’-N” Slope = E0=  1 V E0 of a redox couple HNO2/NO 3, 4.4 2, 3.4

  40. Oxidizing agent? Reducing agent? 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

  41. Identifying strong or weak agent?

  42. NO – Strong oxidant than HNO3

  43. 2 M+(aq) M(s) + M2+(aq) E0 E0’ M(s) M2+(aq) 2M+(aq) Disproportionation Element is simultaneously oxidized and reduced. ‘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 asdisproportionation’.

  44. Disproportionation What Frost diagram tells about this reaction?

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

  46. Disproportionation…. another example

  47. Comproportionation reaction

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