1. Oxidation-Reduction Reactions

2. Oxidation and Reduction Oxidation-reduction reactions always occur simultaneoulsy. Redox Reactions Oxidation Loss of electrons Gain oxygen 2Fe 2 O 3 + 3C 2  4Fe + 3CO 2 Reduced Oxizided Reduction Gain Electrons Loss of Oxygen

3. Redox Reactions that Form Ions Between metal and nonmetals, electrons are transferred from the metal to the nonmetal. Increases stability Mg + S  Mg 2+ + S 2- Oxidation: Mg  Mg 2+ + 2e - (loss of electrons) Reduction: S + 2e -  S 2- (gain of electrons)

4. Oxidizing and Reducing Agents Reducing Agent: substance that loses electrons. Oxidizing Agent: substance that accepts the electrons is the oxidizing agent. Mg + S  Mg 2+ + S 2- Mg: reducing agent, oxidized S: oxidizing agent, reduced LEO the lion goes GER LEO: Losing Electrons is Oxidation GER: Gaining Electrons is Reduction

5. Redox with Covalent Compounds In covalent compounds, polar molecules involve unequal sharing of electrons The shift in electrons is redox for it is the partial gain and loss of electrons H 2 O Oxygen: electrons shift toward Reduced, oxidizing agent Hydrogen: electrons shift away Oxidized, reducing agent

6. Processes Leading to Redox Oxidation Complete loss of electrons (ionic reactions) Shift of electrons away from an atom in covalent bond Gain of Oxygen Loss of Hydrogen by a covalent compound Increase in oxidation number Reduction Complete gain of electrons (ionic reactions) Shift of electrons toward from an atom in covalent bond Loss of Oxygen Gain of Hydrogen by a covalent compound Decrease in oxidation number

7. Corrosion Iron, corrodes by being oxidized to ions of iorn by oxgyen 2Fe +O 2 + H 2 O  2Fe(OH) 2 To protect iron, a piece of magnesium is placed in electrical contact. When oxygen or water attack the iron object, iron lose electrons. Because Mg is more easily oxidized, the Mg immediately transfers electrons to the iron, preventing their oxidation to iron ions.

9. Oxidation Numbers

10. A positive or negative number assigned to an atom to indicate its degree of oxidation or reduction. Rule of Thumb: when bonded, the oxidation number is the same as its ionic charge. In a chemical reaction: Increase in oxidation number  oxidation Decrease in oxidation number  reduction

11. Rules for Assigning Oxidation Numbers Monatomic ions equal ionic charge; Br 1- : -1 H: compounds is +1; metal hydrides is -1 H 2 O: +1, NaH: -1 O: compounds is -2; peroxides is -1, or in compounds with F it’s + H 2 O: -2, H 2 O 2 : -1 Atoms in elemental form or diatomic is 0. S: 0, H 2 : 0 For compound, the sum of oxidation numbers must equal 0. H 2 O  H(+1), O(-2)  2(+1) + 1(-2) =0 For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion. NO 3 2- : N(4), O(-2)  1(+4) + 3(-2) = 2-

12. Oxidation Number Practice NaCl (+1,-1) H 2 O (+1,-2) SO 2 (+4, -2) CO 3 2- (+4,-2) Na 2 SO 4 (+1,+6,-2)

13. Oxidation-Number Changes in Chemical Reactions +1 +5 -2 0 +2 +5 -2 0 2AgNO 3 +Cu  Cu(NO 3 ) 2 + 2Ag Ag: reduced Cu: oxidized Let’s Try These: Cl 2 + 2HBr  2HCl +Br 2H 2 + O 2  2H 2 O 2KNO 3  2KNO 2 + O 2

14. Answers Let’s Try These: 0 +1 -1 +1 -1 0 Cl 2 + 2HBr  2HCl +Br 0 0 +1 -2 2H 2 + O 2  2H 2 O +1 +5 -2 +1 +3 -2 0 2KNO 3  2KNO 2 + O 2

15. Balancing Redox Reactions

16. How to tell if it’s a redox rxn If the oxidation number of an element in a reacting species changes 0 0 +2 -2 N 2 + O 2  2NO

17. Balancing by Oxidation No. 1) Assign oxidation numbers to all the atoms +3 -2 +2 -2 0 +4 -2 Fe 2 O 3 + CO  Fe + CO 2 2) Identify which atoms are oxidized and reduced. 3) Use brackets to connect that atoms undergoing oxidation, and other set for those reduced. +2 Oxidation +3 -2 +2 -2 0 +4 -2 Fe 2 O 3 + CO  Fe + CO 2 -3 reduction

18. Balancing by Oxidation No. Make the total increase in oxidation number equal the total decrease using coefficients (+2)x3=6 +3 -2 +2 -2 0 +4 -2 Fe 2 O 3 + 3CO  2Fe + 3CO 2 (-3)x2=6 Make sure the equation is balanced for both atoms and charge

19. Let’s Practice KClO 3  KCl + O 2 HNO 2 + HI  NO + I 2 + H 2 O Bi2S 3 + HNO 3  Bi(NO 3 ) 3 + NO + S + H 2 O SbCl5 + KI  SbCl 3 +KCl + I 2

20. Redox Reactions

21. Half-Reactions Equation showing just the oxidation or reduction portion of the redox reaction. S + HNO 3  SO 2 + NO +H 2 O 0 +4 -2 Oxidation Half: S  SO 2 +5 -2 +2 -2 Reduction Half: NO 3 -  NO

22. Balancing Half-Reactions To balance: Write the unbalanced ionic equation Write separate half reactions for oxidation & reduction Balance atoms in each half-reaction Add enough electrons to one side of each half-reaction to balance the charges Multiply each half-reaction by an appropriate number to make the numbers of electrons equal Add the half reaction to show the overall equation Add the spectator ions and balance the equation

23. Half-Reactions S + HNO 3  SO 2 + NO +H 2 O Ionic Form: S + H + + NO 3 -  SO 2 + NO +H 2 O 0 +4 -2 Oxidation Half: S  SO 2 +5 -2 +2 -2 Reduction Half: NO 3 -  NO Balancing Atoms in Half –Reactions 2H 2 O + S  SO 2 + 4H + 4H + + NO 3 -  NO + 2H 2 O

24. Half-Reactions Add e- to each side of half reactions to balance charges Oxidation: 2H 2 O + S  SO 2 + 4H + + 4e - Reduction: 4H + + NO 3 - + 3e -  NO + 2H 2 O Multiply each half reaction by an appropriate number to make the numbers of electrons equal Oxidation: 6H 2 O + 3S  3SO 2 + 12H + + 12e - Reduction: 16H + + 4NO 3 - + 12e -  4NO + 8H 2 O Subtract the terms that appear on both sides and add in the spectator ions 6H 2 O + 3S + 16H + + 4NO 3 - + 12e -  3SO 2 + 12H + + 12e -+ 4NO + 8H 2 O 3S + 4HNO 3  3SO 2 + 4NO + 2H 2 O

25. Half-Reactions Let’s Practice: KMnO 4 + HCl  MnCl 2 + Cl 2 + H 2 O + KCl