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OXIDATION AND REDUCTION REACTION

OXIDATION AND REDUCTION REACTION. Types of Reactions. Redox Reaction. Is the reaction that involve the transfer of electrons from a reducing agent to an oxidizing agent. OUT OF BODY. IN THE BODY. TYPE OF REDOX REACTION. Electrochemistry.

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OXIDATION AND REDUCTION REACTION

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  1. OXIDATION AND REDUCTION REACTION REDOX

  2. Types of Reactions REDOX

  3. Redox Reaction • Is the reaction that involve the transfer of electrons from a reducing agent to an oxidizing agent REDOX

  4. OUT OF BODY IN THE BODY TYPE OF REDOX REACTION REDOX

  5. Electrochemistry • Is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy • Electrochemical processes are redox reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to cause a nonspontaneous reaction to occur REDOX

  6. Oxidation: Is defined as the removal of electrons Reduction: is defined as the gain of electrons Definition of Oxidation and Reduction REDOX

  7. Common uses of the terms oxidation and reduction REDOX

  8. Oxidation of Ferrous to ferric ion REDOX

  9. OXIDATION NUMBER • Oxidation numbers are the charges atoms in a compound would have if the electrons of each bond belonged to the more electronegative atoms. REDOX

  10. Properties of oxidizing and reducing agents REDOX

  11. THE RULES FOR ASSIGNING OXIDATION NUMBER 1. Atoms of any element not combined with atoms from another element have oxidation numbers of zero. Examples: The oxidation number of the atom in N2, Br2, Cl2, P4 and S2 are zero continued REDOX

  12. continued 2. The oxidation numbers of monatomicions equal their ionic charge. Examples: The oxidation numbers of Na+ and K+ are +1, of Ca2+, Cu2+, and Mg2+ are +2, and Cl¯ and Br¯ are –1. REDOX

  13. 3. In their compounds, the oxidation number of any atoms of the - Group I A elements is +1 (e.g., Na+, K+) - Group II A elements is +2 (e.g., Cu2+, Mg2+) - Group III A is +3 continued REDOX

  14. 4.The oxidation number of any nonmetal in its binary compounds with metals equals the charge of the monatomic anion e.g., The oxidation number of Br in CrBr3 is –1 because the monatomic ion of Br is the bromide ion, which has a charge of 1-. continued REDOX

  15. 5. In compounds, the oxidation number of - O is almost always –2. (Exceptions occur only when the rules for H or F would be violated) - H is almost always +1. (the exceptions are binary compounds with metals, like NaH, in which the H has the oxidation number –1.) - F is always –1. (No exceptions. Fluorine is the most electronegative of all elements). continued REDOX

  16. 6. The sum of the oxidation numbers of all the atoms in the formula of the atom, ion, or molecule must equal the overall charge given for the formula – the sum rule continued REDOX

  17. Problem No. 1 Calomel, long use in medicine, has the formula Hg2Cl2. What are the oxidation numbers on the atoms in this compound? The rule for assigning oxidation number: No. 4 & 6 4. The oxidation number of any nonmetal in its binary compounds with metals equals the charge of the monatomic anion 6. The sum of the oxidation numbers of all the atoms in the formula of the atom, ion, or molecule must equal the overall charge given for the formula – the sum rule REDOX

  18. The oxidation number of Hg (x) Hg2Cl2 Rule 4 : Cl 2 atoms X (-1) = -2 Hg 2 atoms X (x) = 2x Rule 6 : sum = 0 The value of x comes from the sum, 2x + (-2) = 0 2x = +2 x = +1 The oxidation number of Hg in the Hg2Cl2 is +1 _ REDOX

  19. BALANCING REDOX REACTIONS REDOX

  20. BALANCING REDOX REACTIONS BY THE ION-ELECTRON METHOD Simply list the steps to balance a redox reaction in an acidic (or neutral) medium. Balancing redox reactions by the ion-electron method has 8 steps. REDOX

  21. Those are: 1. Write a skeletal equation that shows only the ions or molecules involved in the reaction. 2. Divide the skeletal equation into two half- reactions. 3. Balance all atoms that are not H or O. 4. Balance O by adding H2O. 5. Balance H by adding H+ (not H or H2 or H¯, but H+). 6. Balance the net charge by adding e¯. (remember its minus sign.) continued REDOX

  22. 7. Multiply Entire Half-reactions by simple whole numbers, as needed, to get the gain of e¯ in one half- reaction to match the loss of e¯ in the other. Then add the half-reactions. 8. Cancel whatever is the same on both sides of the arrow. REDOX

  23. The oxidation of methyl alcohol, CH3OH, to formic acid, HCHO2, using the dichromate ion, Cr2O72¯ in an acidic medium. As this reaction proceeds, the chromium in Cr2O72¯ change to Cr3+. Step 1. Write a skeletal equation showing reactants and products as given. CH3OH + Cr2O72¯ HCHO2 + Cr3+ Step 2. Divide the skeletal equation into two half-reactions. Except for H and O, the same elements must appear on both sides of each half-reaction. CH3OH HCHO2 Cr2O72¯ Cr3+ continued REDOX

  24. Step 3. Balance all atoms that are not H or O CH3OH HCHO2 (No change, yet.) Cr2O72¯ 2Cr3+ (Balances Cr atoms.) Step 4. Balance O by adding H2O CH3OH + H2O HCHO2 (Cs and Os balance.) Cr2O72¯ 2Cr3++ 7H2O (Crs and Os balance) Step 5. Balance H by adding H+ CH3OH + H2O HCHO2 + 4H+ (All atoms now balance.) Cr2O72¯ + 14H+ 2Cr3+ + 7H2O (All atoms now balance.) continued REDOX

  25. Step 5. Balance H by adding H+ (a)CH3OH + H2O HCHO2 + 4H+ (b) Cr2O72¯ + 14H+ 2Cr3+ + 7H2O Step 6. Balance the net charge by adding e¯. (a) CH3OH + H2O HCHO2 + 4H++ 4e¯ (b) Cr2O72¯ + 14H++ 6e¯ 2Cr3+ + 7H2O LOOK AT THE STEP 5, AND THEN MAKE THE NET CHARGE ON THE LEFT SIDE OF THE ARROW EQUAL TO THE NET CHARGE ON THE RIGHT SIDE. continued REDOX

  26. Step 7. Multiply half-reaction by whole numbers so that the electrons will cancel when the half-reactions are added. 3 X [CH3OH + H2O HCHO2 + 4H++ 4e¯] 2 X [Cr2O72¯ + 14H++ 6e¯ 2Cr3+ + 7H2O] Sum: 3CH3OH + 2Cr2O72 ¯+ 3H2O + 28H++ 12e¯ 3HCHO2 + 4Cr3++ 12H+ + 14H2O +12e¯ + continued REDOX

  27. Step 8. Cancel everything that can be canceled a). The 12 electrons on each side obviously cancel. b). Water molecule: there are 3 on the left and 14 on the right, so we can strike those on the left and change those on the right to 11. …. + 3H2O +…. …. + 14H2O +…. becomes: …. .… + 11H2O + …. c). And then also cancel some H+ …. + 28 H+ + …. …. + 12 H+ + …. becomes: …. + 16 H+ + …. ….. continued REDOX

  28. 3CH3OH(aq) + 2Cr2O72 ¯(aq) + 16H+(aq) 3HCHO2(aq) + 4Cr3+(aq) +11H2O Check to see that both material and electrical balance exist. REDOX

  29. Home work Problem No. 1 What is the oxidation number of carbon in ethane, C2H6? Problem No. 2 What are the oxidation numbers of the atoms in the nitrate ion, NO3¯? Problem No.3 Balance the following equation, which occurs in an acidic medium. Cu(s) + NO3¯(aq) Cu2+(aq) + NO2(g) REDOX

  30. MnO4¯(aq) + SO32¯(aq) MnO2(s) + SO42¯(aq) To Balance a Redox Equation When the Medium is Basic. Step 1. Through 8 for acidic solutions 2MnO4¯ + 3SO32¯ + 2H+ 2MnO2 + 3SO42¯ + H2O Step 9. Add as many OH as there are H+ to both sides of equation. There are two H+ on the left. So we add 2OH¯ to both sides. 2OH¯ + 2MnO4¯ + 3SO32¯ + 2H+ 2MnO2 + 3SO42¯ + H2O + 2OH¯ 1. First Balance it for an Acid Medium and 2. Then Neutralize the Acid REDOX

  31. Step 10. When they occur on the same side of the arrow, combine H+ and OH¯ into H2O. 2OH¯ + 2MnO4¯ + 3SO32¯ + 2H+ 2MnO2 + 3SO42¯ + H2O + 2OH¯ The left side has 2OH¯ and 2H+, so we combine them into 2H2O. The equation: 2H2O + 2MnO4¯ + 3SO32¯ 2MnO2 + 3SO42¯ + H2O + 2OH¯ continued REDOX

  32. Step 10. 2H2O + 2MnO4¯ + 3SO32¯ 2MnO2 + 3SO42¯ + H2O + 2OH¯ Step 11. Cancel H2O molecule as possible. The final equation: H2O + 2MnO4¯(aq) + 3SO32¯(aq) 2MnO2(s)+ 3SO42¯(aq) ……… + 2OH¯(aq) Step 11. Cancel H2O molecules as possible REDOX

  33. Reduction Potentials or Redox Potentials REDOX

  34. Rule for Combining Reduction Half-Reactions When two reduction half-reactions are combined into a full redox reaction the one with the more positive E° always runs as written, as a reduction, and it forces the other, with the less positive E°, to run in reverse, as an oxidation. REDOX

  35. Na(s) and Cl2(g).Will sodium react with chlorine? Na+(aq) + e¯ ⇌ Na(s) E°= -2,71 V Cl2(g) + 2 e¯ ⇌ 2Cl¯ E°= +1.36V When two reduction half-reactions are combined into a full redox reaction the one with the more positive E° always runs as written, as a reduction, and it forces the other, with the less positive E°, to run in reverse, as an oxidation. Na(s) → Na+(aq) + e¯ (oxidation) Cl2(g) + 2 e¯ → 2Cl¯ (aq) (reduction) Continued next slide REDOX

  36. to get the net reaction, multiply the coefficients of the first half-reaction by 2. 2X [Na(s) → Na+(aq) + e¯ (oxidation)] 2 Na(s) → 2Na+(aq) + 2e¯ (oxidation) Cl2(g) + 2e¯ → 2Cl¯ (aq) (reduction) Sum: 2Na(s) + Cl2(g)  2Na+(aq) + 2Cl¯ (aq) Thus sodium and chlorine spontaneously react. REDOX

  37. OUT OF BODY IN THE BODY TYPE OF REDOX REACTION REDOX

  38. Biological Oxidation • The processes of oxidation are essential for maintaining life because oxidation and the simultaneously occuring reduction supply the free energy for the vital work REDOX

  39. Organic Reaction Mechanisms • Organic reactions mechanisms has classified biochemical reactions into four categories: • 1. Group-transfer reactions • 2. Oxidations and reductions • 3. Eliminations, isomerizations, and rearrangements • 4. Reactions that make or break carbon- carbon bonds REDOX

  40. Covalent bonds • A covalent bond consists of an electron pair shared between two atoms. • In breaking such a bond, the electron pair can either remain with one of the atoms (heterolytic bond cleavage) or separate such that one electron accompanies each of the atoms (homolytic bond cleavage). REDOX

  41. Homolytic Bond Cleavage • Homolytic bond cleavage, which usually produces unstable radicals, occurs mostly in oxidation-reduction reactions REDOX

  42. Heterolytic Bond Cleavage • Heterolytic C-H bond cleavage involves either carbanion and proton (H+) formation or carbocation (carboniun ion) and hydride ion (H-) formation. REDOX

  43. Biologically Importan Nucleophillic Groups Nucleophiles are the conjugate bases of weak acids REDOX

  44. 1. Group-transfer reactions • Types of metabolic group-transfer: Acyl group transfer involves addition of a nucleophile (Y) to the electrophilic carbon atom of an acyl compound to form a tetrahedral intermediate. The original acyl carrier (X) is then expelled to form a new acyl compound. REDOX

  45. 2. Oxidations and reductions NAD+ IS AN ELECTRON ACEPTOR (OXIDATOR); TWO ELECTRON FROM GENERAL BASE ARE TRANSFERRRED TO AN ELECTRON ACCEPTOR SUCH AS NAD+. General Base Alcohol NAD+ General acid Ketone NADH REDOX

  46. In Living Systems The electron-transfer process connecting these half-reactions occurs through a multistep pathway that harnesses the liberated free energy to form ATP. REDOX

  47. For Example: Biologic Oxidation • The principal use of oxygen is in respiration. Which may be defined as the process by with cells derive energy in the form of ATP from the controlled reaction of hydrogen with oxygen to form water. REDOX

  48. For Example: oxidation of primary alcohols in the Body REDOX

  49. ACCUT TOXICITY Toxicity test SUBCHRONIC TOXICITY CHRONIC TOXICITY REDOX

  50. FREE ENERGY CHANGES CAN BE EXPRESSED IN TERMS OF REDOX POTENTIAL • In reaction involving oxidation and reduction, the free energy change is proportionate to the tendency of reactants to donate or accept electrons. Thus, in addition to expressing free energy change in terms of ΔG0. • It is possible, in an analogous manner, to express it numerically as an oxidation-reduction or redox potential (E0). REDOX

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