1. Oxidation-Reduction Reactions (Redox)

2. What is the difference between acid/base reactions and redox reactions? Acid/base reactions – proton transfer (p + ) Redox reactions – electron transfer (e - )

3. Flow of electrons Electrons respond to differences in potential by moving from the region of high potential to the region of low potential. - + High E p Low E p e-e-

4. Flow of electrons Cl low electronegativity high electronegativity e-e- Li Lithium loses the e - tug-of-war with chloride.

6. Terminology Cations: – positively charged ions – generally metals – NH 4 + is the exception Anions: – negatively charged ions – non-metals – complex ions

7. Oxidation: – When a substances loses e -. Reduction: – When a substance gains e -.

8. oxidized reduced

9. Ca (s) + 2H + (aq)  Ca 2+ (aq) + H 2(g) Ca (s) has lost two e - to 2 H + (aq) to become Ca 2+ (aq). Ca (s) has been oxidized to Ca 2+ (aq) At the same time 2 electrons are gained by 2 H + (aq) to form H 2(g). We say H + (aq) is reduced to H 2(g).

10. Half-reactions Ca (s) → Ca 2+ (aq) + 2e - – Oxidation half reaction 2H + (aq) + 2e - → H 2(g) – reduction half reaction

11. Half-reactions add together Ca (s) → Ca 2+ (aq) + 2e - 2H + (aq) + 2e - → H 2(g) Ca (s) + 2H + + 2e -  Ca 2+ + 2e - + H 2(g) Ca (s) + 2H + (aq)  Ca 2+ (aq) + H 2(g) +

12. Half-reactions add together Cu (s) → Cu 2+ (aq) + 2e - Ag + (aq) + e - → Ag (s) Cu (s) + 2Ag + (aq) + 2e -  Cu 2+ (aq) + 2e - + 2Ag (s) Cu (s) + 2Ag + (aq)  Cu 2+ (aq) + 2Ag (s) + ( ) x 2

13. Electron Transfer and Terminology Lose Electrons: Oxidation Gain Electrons: Reduction. OIL - OXIDATION IS LOSS OF ELECTRONS RIG - REDUCTION IS GAIN OF ELECTRONS

14. Iron Iron comes from iron ore which is taken out of the ground by mining. The pure iron is obtained by heating the ore at very high temperatures in a furnace with limestone to remove impurities. The molten iron is taken out of the bottom of the furnace. It is further processed depending on how it is to be used.

19. Why is gaining electrons called reduction? Reduction originally meant the loss of oxygen from a compound. – 2Fe 2 O 3(s) + C (s) → 4Fe (s) + 3CO 2(g) Iron ore is reduced to metallic iron. The size of the pile gets smaller, hence the word reduction.

20. Why is losing electrons called oxidation? Oxidation originally meant the combination of an element with oxygen. 4Fe (s) + 3O 2(g) → 2Fe 2 O 3(g) C (s) + O 2(g) → CO 2(g)

21. It Takes Two: Oxidation-Reduction In all reduction-oxidation (redox) reactions, one species is reduced at the same time as another is oxidized.

22. It Takes Two: Oxidation-Reduction Oxidizing Agent: – the species which causes oxidation is called the oxidizing agent. – substances that gains electrons – the oxidizing agent is always reduced

23. It Takes Two: Oxidation-Reduction Reducing Agent: – the species which causes reduction is called the reducing agent. – the reducing agent is always oxidized. – substances that give up electrons

24. Example Cu (s) + 2 Ag + (aq) → Cu 2+ (aq) + Ag (s) oxidated reduced R.A. O.A.

25. Summary: Redox Theory 1) A redox reaction is a chemical reaction in which electrons are transferred. 2) Number of electrons lost by one species equals number of electrons gained by the other species. 3) Reduction is a process in which e - are gained. 4) Oxidation is a process in which e - are lost 5) A reducing agent donates e - and is oxidized. 6) A oxidizing agent gains e - and is reduced. WS 15-1

27. Only one of these two reactions is possible. Which one? Cu (s) + 2 Ag + (aq) → Cu 2+ (aq) + 2 Ag (s) Cu 2+ (aq) + 2 Ag (s) → Cu (s) + 2 Ag + (aq) Data table values E O, page 7 of your data books. 1) Cu 2+ (aq) + 2e - -- >> Cu (s) + 0.34 E O 2) Ag (s) -- >> Ag + (aq) + e - - 0.80 E O (R) 1) Cu (s) -- >> Cu 2+ (aq) + 2 e - -0.34 E O ( R) 2) Ag + (aq) + e - -- >> Ag (s) +0.80 E O

28. Electric potential (V), E o the electric potential under standard conditions of a half- reaction in which reduction is occurring. Standard conditions: – 25 o C with all ions at 1 M concentrations and all gases at 1 atm pressure

29. Standard Reduction Potentials We cannot measure the potential of an individual half-cell! We assign a particular cell as being our reference cell and then assign values to other electrodes on that basis.

30. [H + ] = 1.00 H 2 (g) e-e- Pt gauze The Standard Hydrogen electrode E o (H + /H 2 ) half-cell = 0.000 V p{H 2 (g)} = 1.00 atm

31. Electric potential (V), E o If the net potential is a positive number then the reaction is spontaneous. If the net potential is a negative number then the reaction is non-spontaneous. Half cell potentials are not doubled or tripled as per balancing. We are only comparing potentials.

32. Compare the two half reactions that make up the reaction. Cu 2+ (aq) + 2Ag (s) → Cu (s) + 2Ag + (aq) Cu 2+ + 2e - → Cu E o = 0.34 2Ag → 2Ag + + 2e - E o = -0.80 Cu 2+ (aq) + 2Ag (s) → Cu (s) + 2Ag + (aq) E o = -0.46 Negative potential, non-spontaneous +

33. Compare the two half reactions that make up the reaction. Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) Cu (s) → Cu 2+ + 2e - E o = -0.34 2Ag + + 2e - → 2Ag E o = 0.80 Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) E o = 0.46 Positive potential, spontaneous

34. Problem Write the oxidation/reduction half reactions and the net ionic equation when zinc is placed in Ni(NO 3 ) 2 solution. Identify the O.A. and R.A. and state if the reaction is spontaneous or non- spontaneous.

35. Problem Ni(NO 3 ) 2 → Ni 2+ (aq) + 2NO 3- (aq) Zn (s) + Ni 2+ (aq) → ? Oxidation: Zn (s) → Zn 2+ (aq) + 2e - +0.76 Reduction: Ni 2+ (aq) + 2e - → Ni (s) - 0.26 Spectator ion Add half reactions A piece of zinc is placed in a solution of nickel nitrate Ni(NO 3 ) 2

36. Problem Zn (s) + Ni 2+ (aq) → Zn 2+ (aq) + Ni (s) +0.50 R.A. O.A. Positive potential, spontaneous Zn is Oxidized Ni 2+ is Reduced

37. NOTE*** Spontaneous shortcut Locate the O.A. on the left and the R.A. on the right of the table. If the O.A. is higher up on the table than the R.A. then the reaction is spontaneous. O.A. R.A. SPONTANEOUS REACTION O.A. R.A. NON-SPONTANEOUS REACTION

38. highest attraction for electrons weak attraction for electrons

40. Problem Explain what happens when nickel is placed in a zinc nitrate solution. Ni (s) + Zn 2+ (aq) → ? + ? R.A. O.A. NICKEL  NiZINC NITRATE  Zn 2+ and NO 3 - REDUCING AGENT OXIDIZING AGENT ARE ON LEFT SIDE

41. On the table  Ni (s)   Zn 2+ (aq)  R.A. is above the O.A. NON SPONTANEOUS

42. Disproportionation Disproportionation redox reactions where the OA and the RA are the same species. ( p 577 – text) Example: Fe 2+ (aq) and Fe 2+ (aq) Fe 2+ (aq) + 2 e -  Fe (s) reduction of Fe 2+ 2[ Fe 2+ (aq)  Fe 3+ (aq) + e - ] oxidation of Fe 2+ 3 Fe 2+ (aq)  Fe (s) + 2 Fe 3+ (aq) net reaction NON – SPONTANEOUS REACTION

43. DISPROPORTIONATION TRY THE REACTION WHERE Cu 1+ ACTS AS THE OXIDIZING AND REDUCING AGENTS TRY THE REACTION WHERE Cr 2+ ACTS AS THE AS THE OXIDIZING AND REDUCING AGENTS

44. Predicting redox reactions 1) List all species present. 2) Choose the strongest oxidizing and reducing agent. 3) Write the reduction half reaction, as written in the data book. 4) Write the oxidation half reaction, reverse the equation in the data book. 5) Balance number of electrons. 6) Add the two half reactions together to form the net ionic equation. 7) Predict if reaction is spontaneous or not.

45. Problems A mixture of bromine gas and chlorine gas is added to a solution of copper (II) sulphate and a copper strip. (water) ( CuSO 4 ) (Br 2(g) ) (Cl 2(g) ) ( Cu (s) ) NOTE ( Go down S.O.A. / Go up S.R.A.) Br 2(g) Cl 2(g) H 2 0 (l) Cu 2+ (aq) Cu (s) SRA * SOA * Cl 2(g) + 2e - → 2 Cl - (aq) Cu (s) → Cu 2+ (aq) + 2e - Cl 2(g) + Cu (s) → 2 Cl - (aq) + Cu 2+ (aq) SPONTANEOUS

46. Problems Lead is placed in a zinc nitrate solution.(list species) NO 3 - (aq) H 2 0 (l) Zn 2+ (aq) Pb (s) SRA SOA Non-spontaneous OA is below RA Zn 2+ (aq) + 2e -  Zn (s) Pb (s)  Pb 2+ (aq) + 2e - Zn 2+ (aq) + Pb (s)  Zn (s) + Pb 2+ O.A. R.A.

47. Problems A few drops of Hg (l) are dropped into a solution which is 1.0 M in both sulphuric acid and potasium permanganate. MnO 4 - (aq) SO 4 2- (aq) H 2 0 (l) K + (aq) Hg (l) H + (aq) RA OA H + hydrogen ion (From acid) O.A. R.A YES

48. Problems A few drops of Hg (l) are droped into a solution which is 1.0 M in both sulphuric acid and potasium permanganate. MnO 4 - (aq) + 8 H + (aq) + 5e - → Mn 2+ (aq) + 4 H 2 O (l) Hg (l) → Hg 2+ (aq) + 2e - 2MnO 4 - (aq) + 16H + (aq) + 5Hg (l) → 2Mn 2+ (aq) + 8H 2 O (l) + 5Hg 2+ (aq) ( ) x2 ( ) x5 Oxidized (Balance electrons) LHS = RHS

50. General Rules Metal (+) ions are oxidizing agents. Nonmetal (-) ions are reducing agents. Metal elements are reducing agents. Nonmetal elements are oxidizing agents.

51. Building a redox table (method one) One can use experimental evidence to determine the relative strengths of oxidizing and reducing agents. The greater the number of spontaneous reactions, the stronger the oxidizing agent.

52. Building a redox table This means we can rank oxidizing agents according to the number of spontaneous reactions. By convention the strongest oxidizing agent is at the top left in a redox table and the strongest reducing agent is at the bottom right of the table.

53. Problem: Make a redox table Cu (s) Mg (s) Ag (s) Zn (s) Cu 2+ (aq) ________________ Mg 2+ (aq) ________________ Ag + (aq) ________________ Zn 2+ (aq) ________________ √ √ √ √ √√ Virtual Lab O.A. R.A.

54. Activity Series Ag + (aq) + 1e -  Ag (s) Cu 2+ (aq) + 2e -  Cu (s) Zn 2+ (aq) + 2e -  Zn (s) Mg 2+ (aq) + 2e -  Mg (s)

55. Example: Redox Reaction Based on the activity series, which reactions are spontaneous? a) Ag (s) + Mg(NO 3 ) 2 (aq)  ? ions b) Cu (s) + AgNO 3 (aq)  ? ions c) Zn (s) + Mg(NO 3 ) 2(aq)  ? ions d) Mg (s) + Mg(NO 3 ) 2 (aq) no reaction

56. Example: Redox Reaction a) Ag (s) vs. Mg 2+ (aq) chart (ions + metal) Ag (s) (RA) is above Mg 2+ (aq) (OA) non-spontaneous b) Cu (s) vs. Ag + (aq) Cu (s) (RA) is below Ag + (aq) (OA) spontaneous c) Zn (s) vs. Mg 2+ (aq) Zn (s) is above Mg 2+ (aq) non-spontaneous WS 15-28

57. Redox Table Building (method two) The spontaneity rule is used to order the oxidizing agents to produce a redox table. Consider the following redox equations which represent spontaneous reactions from an experiment. From this evidence construct a redox table.

58. Redox Table Building Co (s) + Pd 2+ (aq) → Co 2+ (aq) + Pd (s) Pd (s) + Pt 2+ (aq) → Pd 2+ (aq) + Pt (s) Mg (s) + Co 2+ (aq) → Mg 2+ (aq) + Co (s) Work with one equation at a time. ASSUME ALL REACTION ARE SPONTANEOUS

59. Redox Table Building Co (s) + Pd 2+ (aq) → Co 2+ (aq) + Pd (s) Pd 2+ (aq) + 2 e - → Pd (s) Co 2+ (aq) + 2 e - → Co (s) OA is above RA spontaneous reaction RAOA

60. Redox Table Building Pd (s) + Pt 2+ (aq) → Pd 2+ (aq) + Pt (s) Pt 2+ (aq) + 2 e - → Pt (s) Pd 2+ (aq) + 2 e - → Pd (s) OA is above RA spontaneous reaction

61. Redox Table Building Mg (s) + Co 2+ (aq) → Mg 2+ (aq) + Co (s) Co 2+ (aq) + 2 e - → Co (s) Mg 2+ (aq) + 2 e - → Mg (s) OA is above RA spontaneous reaction

62. Redox Table Building Pt 2+ (aq) + 2 e - → Pt (s) Pd 2+ (aq) + 2 e - → Pd (s) Co 2+ (aq) + 2 e - → Co (s) Mg 2+ (aq) + 2 e - → Mg (s)

64. Redox Stoichiometry Can be used to predict or analyze a chemical reaction. A method of reacting a solution with a known concentration with a solution of unknown concentration. Common oxidizing agents in redox reactions – MnO 4 - (aq) → Mn 2+ (aq) – purple colorless – Cr 2 O 7 2- (aq) → Cr 3+ (aq) – orange green

65. Redox Stoichiometry In a titration experiment all of the Br - (aq) ions in an acidic solution were oxidized to Br 2(l) by a 0.0200 M KMnO 4(aq) solution. The volume of Br - (aq) solution was 25.0 mL and the volume of KMnO 4(aq) was 15.0 mL. Calculate the concentration of Br - (aq) ions in solution.

66. We need a balanced chemical equation to do any stoichiometry. MnO 4 - (aq) + 8 H + (aq) + 5e - ↔ Mn 2+ (aq) + 4 H 2 O (l) 2 Br – (aq) ↔ Br 2(l) + 2e - ( )x2 ( )x5 2 MnO 4 - (aq) + 16 H + (aq) + 10 Br - (aq) ↔ 2 Mn 2+ (aq) + 8 H 2 O (l) +5 Br 2(l) 0.0200 M 0.015 L 0.0003 mol 0.025 L 0.0015 mol 2:10 c = 0.0600 M.0003 mol/ 2 x 10 =.0015 mol.0015mol /.025L

68. Oxidation States Some reactions are not adequately explained with redox theories. Chemists have developed a method of electron bookkeeping to describe the redox of molecules and complex ions.

69. Oxidation States Oxidation state: – apparent net charge that an atom would have if electron pairs belonged entirely to the more electronegative atom Oxidation number: – a positive or negative number assigned to a combined atom according to a set of arbitrary numbers.

70. Assigning Oxidation Numbers 1) Oxidation numbers for all uncombined elements (elemental/standard) = 0 K (s) = 0 N 2(g) = 0S 8(s) = 0 2) Oxidation number for all simple ions is equal to the charge of the ion. Br 1- (aq) = -1 Fe 3+ (aq) = +3 3) Oxidation for oxygen in a compound = -2 (except for peroxides = -1) H 2 O (l) H 2 O 2(l) -2

71. Assigning Oxidation Numbers 4) Hydrogen in compounds = +1 H 2 O (l) (except hydrides = -1) NaH (s) 5) Sum of oxidation numbers in a compound is = 0 H 2 O (l) → (2 x + 1) + (1 x - 2) = 0 6) Sum of oxidation numbers in a complex ion = charge of ion. NH 4 + (aq) → (4 x + 1) + (1 x - 3) = +1

72. Problems What is the oxidation number for Na (s) ? What is the oxidation number for H 2(g) ? What is the O# for hydrogen in HCl (g) ? What is the O# for Na + (aq) ? What is the O# for oxygen in H 2 O (l) ? 0 +1 -2

73. Assign oxidation numbers to chlorine in each of the following chemicals. HCl (aq) Cl 2(g) NaClO (s) Cl - (aq) HClO 3(aq) ClO 3(aq) 0 +1 +5 +6

74. Assign oxidation numbers to maganese in each of the following chemicals. Mn (s) MnO 2(s) MnO 4 -2 (aq) Mn 2+ (aq) Mn 2 O 7(aq) MnCl 2(s) 0 +4 +6 +2 +7 +2

75. Example What is the oxidation number for carbon in CO 3 2- (aq) ? C O# + 3 O O# = -2 ? + 3 (-2) = -2 ? + -6 = -2 ? = +4

76. Example What is the oxidation number for carbon in C 6 H 12 O 6 ? 6 C O# + 12 H O# + 6 O O# = 0 6 (?) + 12 (+1) + 6 (-2) = 0 6 (?) + 12 + - 12 = 0 ? = 0

77. Who cares about oxidation numbers? Determining oxidation numbers allows us to predict electron transfer. If there is an increase in oxidation number then oxidation occurs. If there is a decrease in oxidation number then reduction occurs.

78. Problem Determine the oxidation numbers for all atoms and ions in the following redox equation and indicate which substance is undergoing oxidation and reduction.

79. Problem CH 4(g) + 2 O 2(g) → CO 2(g) + 2 H 2 O (g) +1 -4 0-2 +4 -2 +1 C is oxidized O is reduced

81. Ion electron method Under Acidic conditions 1. Identify oxidized and reduced species Write the half reaction for each. 2. Balance the half rxn separately except H & O’s. Balance: Oxygen by adding H 2 O Balance: Hydrogen by adding H + Balance: Charge by adding e - 3. Multiply each half reaction by a coefficient. Must have the same # of e - in both half-rxn. 4. Add the half-rxn together, the e - should cancel.

82. Balancing Half Reactions MnO 4 ¯ → Mn 2+ MnO 4 ¯ → Mn 2+ + 4 H 2 O 8 H + + MnO 4 ¯ → Mn 2+ + 4 H 2 O 5 e¯ + 8 H + + MnO 4 ¯ → Mn 2+ + 4 H 2 O

83. Example: Acidic Conditions I - + S 2 O 8 -2  I 2 + S 2 O 4 2- Half Rxn (oxid): I -  I 2 Half Rxn (red): S 2 O 8 -2  S 2 O 4 2- Bal. chemical and e- : 2 I -  I 2 + 2 e - Bal. chemical O and H : 8e - + 8H + + S 2 O 8 -2  S 2 O 4 2 - + 4H 2 O Mult 1st rxn by 4: 8 I -  4 I 2 + 8e - Add rxn 1 & 2: 8I -  4 I 2 + 8e - 8e - + 8H + + S 2 O 8 -2  S 2 O 4 2 - + 4H 2 O 8I - + 8H + + S 2 O 8 -2  4 I 2 + S 2 O 4 2 - + 4H 2 O

84. Example: Acidic Conditions NO 3 - + Bi  NO 2 + Bi 3+ Half Rxn (oxid): Bi  Bi 3+ Half Rxn (red): NO 3 -  NO 2 Bal. chemical and e- : Bi  Bi 3+ + 3 e - Bal. chemical O and H : 1e - + 2H + + NO 3 -  NO 2 + H 2 O Mult 2nd rxn by 3: 3e - + 6H + + 3NO 3 -  3NO 2 + 3H 2 O Add rxn 1 & 2: Bi  Bi 3+ + 3 e - 3 e - + 6 H + + 3 NO 3 -  3NO 2 + 3H 2 O Bi + 6 H + + 3 NO 3 -  Bi 3+ + 3 NO 2 + 3 H 2 O

85. 1. Procedure identical to that under acidic conditions Balance the half rxn separately except H & O’s. Balance Oxygen by H 2 O Balance Hydrogen by H + Balance charge by e - 2. Mult each half rxn such that both half- rxn have same number of electrons 3. Add the half-rxn together, the e - should cancel. 4. Eliminate H+ by adding: OH - to both sides Redox Reactions - Ion electron method. Under Basic conditions

86. Example: Basic Conditions H 2 O 2 (aq) + Cr 2 O 7 -2 (aq )  Cr 3+ (aq) + O 2 (g) red: 6e - + 14H + + Cr 2 O 7 -2 (aq)  2Cr 3+ + 7 H 2 O oxid: (H 2 O 2 (aq)  O 2 + 2H + + 2e - ) x 3 8 H + + 3H 2 O 2 + Cr 2 O 7 2-  2Cr +3 + 3O 2 + 7H 2 O add: 8 OH -  8 OH - 3H 2 O 2 + Cr 2 O 7 2 - + 8H 2 O  2Cr +3 + 3O 2 + 7H 2 O + 8OH - 3H 2 O 2 + Cr 2 O 7 2 - + H 2 O  2Cr +3 + 3O 2 + 8OH - +3 +6 0

88. Breathalyzer The Breathalyzer device contains: – A system to sample the breath of the suspect – Two glass vials containing the chemical reaction mixture – A system of photocells connected to a meter to measure the color change associated with the chemical reaction

89. Breathalyzer To measure alcohol, a suspect breathes into the device. The breath sample is bubbled in one vial through a mixture of sulfuric acid, potassium dichromate, silver nitrate and water. The principle of the measurement is based on the following chemical reaction: 8H + + Cr 2 O 7 2- + 3C 2 H 5 OH → 2Cr 3+ + 3C 2 H 4 O + 7H 2 O yellow blue

90. The sulfuric acid removes the alcohol from the air into a liquid solution. The alcohol reacts with potassium dichromate to produce: – chromium sulfate – potassium sulfate – acetic acid – water The silver nitrate is a catalyst, a substance that makes a reaction go faster without participating in it. The sulfuric acid, in addition to removing the alcohol from the air, also might provide the acidic condition needed for this reaction.

91. During this reaction, the reddish-orange dichromate ion changes color to the green chromium ion when it reacts with the alcohol; the degree of the color change is directly related to the level of alcohol in the expelled air. To determine the amount of alcohol in that air, the reacted mixture is compared to a vial of unreacted mixture in the photocell system, which produces an electric current that causes the needle in the meter to move from its resting place. The operator then rotates a knob to bring the needle back to the resting place and reads the level of alcohol from the knob -- the more the operator must turn the knob to return it to rest, the greater the level of alcohol.

92. yellowblue

94. Bleaching Agents Bleaching agents are compounds which are used to remove color from substances such as textiles. In earlier times textiles were bleached by exposure to the sun and air. Today most commercial bleaches are oxidizing agents, such as sodium hypochlorite (NaOCl) or hydrogen peroxide (H 2 O 2 ) which are quite effective in "decolorizing" substances via oxidation.

95. Bleaching Agents The action of these bleaches can be illustrated in the following simplified way:

96. Bleaching Agents The decolorizing action of bleaches is due in part to their ability to remove electrons which are activated by visible light to produce the various colors. The hypochlorite ion (OCl - ), found in many commercial preparations, is reduced to chloride ions and hydroxide ions forming a basic solution as it accepts electrons from the colored material as shown below. OCl - + 2e - + HOH → Cl - + 2 OH -

97. Bleaching Agents Bleaches are often combined with "optical brighteners". These compounds are quite different from bleaches. They are capable of absorbing wavelengths of ultraviolet light invisible to the human eye, and converting these wavelengths to blue or blue-green light. The blue or blue-green light is then reflected by the substance making the fabric appear much "whiter and brighter" as more visible light is seen by the eye.

99. Photosynthesis An example of naturally-occuring biological oxidation-reduction reactions is the process of photosynthesis. It is a very complex process carried out by green plants, blue-green algae, and certain bacteria. These organisms are able to harness the energy contained in sunlight, and via a series of oxidation-reduction reactions, produce oxygen and sugar. The overall equation for the photosynthetic process may be expressed as: 6 CO 2(g) + 6 H 2 O (l) → C 6 H 12 O 6(aq) + 6 O 2(g)

100. Photosynthesis The equation is the net result of two processes. One process involves the splitting of water. This process is really an oxidative process that requires light, and is often referred to as the "light reaction". This reaction may be written as: 12 H 2 O (l) → 6 O 2(g) + 24 H + (aq) + 24e-

101. Photosynthesis Think of the light reaction, as a process by which organisms capture and store radiant energy as they produce oxygen gas. This energy is stored in the form of chemical bonds of compounds such as NADPH and ATP. The energy contained in both NADPH and ATP is then used to reduce carbon dioxide to glucose. This reaction, shown below, does not require light, and it is often referred to as the dark reaction. 6 CO 2 + 24 H + + 24 e - → C 6 H 12 O 6 + 6 H 2 O

102. Photosynthesis The chemical bonds present in glucose also contain a considerable amount of potential energy. This stored energy is released whenever glucose is broken down to drive cellular processes.

103. Photosynthesis In simplest terms, the process of photosynthessis can be viewed as one- half of the carbon cycle. In this half, energy from the sun is captured and transformed into nutrients which can be utilized by higher organisms in the food chain. The release of this energy during the metabolic re-conversion of glucose to water and carbon dioxide represents the second half of the carbon cycle and it may be referred to as "oxidative processes".

104. Cellular Respiration C 6 H 12 O 6(aq) + 2 O 2(g) → Energy + CO 2(g) + 2 H 2 O (g) +1 0 0 -2 +4 -2 +1 C is oxidized O is reduced -2