1. Chapter 16 Oxidation-Reduction Reactions

2. Objectives 16.1 Analyze the characteristics of an oxidation reduction reaction 16.1 Distinguish between oxidation reactions and reduction reactions by definition 16.1 Identify the substances that are oxidized and those that are reduced in a redox reaction 16.1 Distinguish oxidizing and reducing agents

3. Objectives 16.2 Analyze common redox processes to identifying the oxidizing and reducing agents 16.2 Identify some redox reactions that take place in living cells Balance Oxidation-Reduction Reactions

4. Review and New Intro * Bonds are formed by the sharing of electrons (in covalent compounds) or they can also be formed by the transfer of electrons (in ionic compounds).

5. Review and New Consider the two reactions below. 2 Na(s) + Cl 2 (g)  2 NaCl (s) – Electrons completely transferred 2 C (s) + O 2 (g)  2 CO (g) – Electrons shared, but unevenly

6. Loss of Gain – In the two examples previously, electrons were transferred from one atom to another. The sodium completely lost its electron to the chlorine, and the carbon partially lost its electrons. The loss or gain of electrons is referred to oxidation and reduction

7. Redox Any time an atom loses, either completely or partially, one or more electrons, it is called oxidation. Any time an atom gains, either completely or partially, one or more electrons, it is called reduction.

8. Redox – Oxidation and reduction go hand in hand. You can not have oxidation without reduction. A single reaction in which oxidation and reduction take place is called an oxidation- reduction reaction, or redox reaction.

9. Memorize OIL RIG: Oxidation is lose, Reduction is gain LEO the lion goes GER: Lose Electrons Oxidation, Gain Electrons Reduction

10. Examples 2 Na(s) + Cl 2 (g)  2 NaCl (s) – Na (s) is Oxidized (loses electrons) – Cl 2 (g) is Reduced (gains electrons) 2 C (s) + O 2 (g)  2 CO (g) – C (s) is Oxidized (loses electrons) – O 2 (g) is Reduced (gains electrons)

11. Oxidizing Agent A substance that causes other substances to lose electrons. Oxidizing agents are easily able to accept electrons, since they accept electrons, they themselves are reduced. – Oxidizing Agent = Molecule (Atom) Reduced

12. Reducing Agent A substance that causes other substances to gain electrons. Reducing agents are easily able to give up electrons, since they lose electrons, they themselves are oxidized. – Reducing Agent = Molecule (Atom) Oxidized

13. Summarized Oxidizing agents (Oxidizers)  Want electrons, make others give them electrons, and they themselves are reduced Reducing agents (Reducer)  Don’t want electrons, make others take their electrons, they themselves are oxidized

14. Assigning Oxidation Numbers To identify whether atoms are reduced or oxidized, oxidation numbers are used. The oxidation number is: – The number of electrons that must be added to or removed from an atom to convert the atom into the elemental form. What number (charge) do we assign to atoms in different situations?

15. Elemental Form How many electrons it should normally have in its valence shell. – Columns below – H = 1; Be = 2; B = 3; C = 4; N = 5; O = 6; F = 7; He = 8 (except He )

16. Assigning Oxidation Numbers Oxidation number of an atom (that is not combined with another type of atom) in atomic or molecular form is zero. Na (s) = 0, O 2 (g) = 0, F 2 (g) = 0, Mn (s) = 0

17. Assigning Oxidation Numbers The oxidation number of a monatomic ion is equal to the charge on the ion Fe +2 = +2, Fe +3 = +3, F - = -1, O -2 = -2

18. Assigning Oxidation Numbers Fluorine is always -1 when with any atom other than itself (because it is the most electronegative) Example: OF 2 (g) = F is -1 oxidation #

19. Assigning Oxidation Numbers For binary compounds, the more electronegative element is assigned the number equal to the charge it would have if it were an anion. Example: CO 2 (g) = O is -2, C is +4 Na 3 N (s) = N is -3

20. Assigning Oxidation Numbers In compounds, any atom from Group 1, 2, or 3 (H, Be, B) have oxidation numbers of +1, +2, and +3 (Metals like to give up their electrons) Na 3 N (s) = Na is +1 CaO (s) = Ca is +2 AlN (s) = Al is +3

21. Assigning Oxidation Numbers Hydrogen with a metal is -1, Hydrogen with a non-metal is +1 Example: LiH = H is -1 H 2 O (l) = H is +1

22. Assigning Oxidation Numbers Oxygen is always -2 with the two exceptions: 1) When attached to Flourine (+2) 2) When in a peroxide (H 2 O 2 ) (-1) MnO 4 - (permanganate) O is -2 OF 2 (g) = O is +2

23. Assigning Oxidation Numbers In covalent bonds, the negative oxidation number is given to the more electronegative Using these you can now assign Unknown Oxidation Numbers

24. Assigning Oxidation Numbers The sum of the oxidation numbers for all the atoms in a molecule is zero HgO (l), We know O is -2, therefore, Hg must be +2

25. Redox Reactions Is the following a redox reaction? – 2 H 3 O + (aq) + Zn (s)  H 2 (g) + 2 H 2 O (l) + Zn +2 (aq) To figure out whether or not a reaction is a redox one, is to see if any of the atoms were oxidized or reduced. – So, we assign oxidation numbers to each atom to see if any atom was oxidized or reduced

26. Assigning +1 -2 0 0 +1 -2 H 3 O + (aq) + Zn (s)  H 2 (g) + 2 H 2 O (l) + Zn +2 (aq)

27. Half Reactions We can use half-reactions to show the change Zn (s)  Zn +2 (aq) + 2 e - 2 e - + 2 H 3 O + (aq)  H 2 (g) + 2 H 2 O (l)

28. Half Reactions By adding the two half reactions together, we get the full reaction. – Of Note: The electrons are not just floating around aimlessly, they are handed off from one atom to the other. With half reactions you can tell which atom gained or lost electrons

29. Which of the below is Redox? HCl + H 2 O  H 3 O + + Cl - C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O

30. Balancing Redox Reactions Try to balance the below equation in acidic solution Cr 2 O 7 2- (aq) + Fe (s)  Cr +3 (aq) + Fe +3 (aq) Need to balance atoms and charge. You can add Water and H + to either side

31. Balancing Redox Reactions 1. Separate reaction into two half reactions. Cr 2 O 7 2- (aq)  Cr +3 (aq) Fe (s)  Fe +3 (aq)

32. Balancing Redox Reactions 2. Balance all atoms except for H and O Cr 2 O 7 2- (aq)  2 Cr +3 Fe (s)  Fe +3 (aq)

33. Balancing Redox Reactions 3. Balance O by adding H 2 O Cr 2 O 7 2- (aq)  2 Cr +3 (aq) + 7 H 2 O (l) Fe (s)  Fe +3 (aq)

34. Balancing Redox Reactions 4. Balance H by adding H + Cr 2 O 7 2- (aq) + 14 H + (aq)  2 Cr +3 (aq) + 7 H 2 O (l) Fe (s)  Fe +3 (aq)

35. Balancing Redox Reactions 5. Balance charge by adding electrons to the more positive side (the charge should then be the same on both sides of the equation) Cr 2 O 7 2- (aq) + 14 H + (aq)  2 Cr +3 (aq) + 7 H 2 O (l) – Reactants = +12Products = +6 Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 e -  2 Cr +3 (aq) + 7 H 2 O (l)

36. Balancing Redox Reactions Fe (s)  Fe +3 (aq) – Reactants = 0 Products = +3 Fe (s)  Fe +3 (aq) + 3 e -

37. Balancing Redox Reactions 6. Multiply half-reactions, if necessary, by integers to make number of electrons lost equal to number of electrons gained 2 Fe (s)  2 Fe +3 (aq) + 6 e -

38. Balancing Redox Reactions 7. Add half-reactions and cancel species where needed Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 e - + 2 Fe (s)  2 Cr +3 (aq) + 7 H 2 O (l) + 2 Fe +3 (aq) + 6 e - In this case, the only species which will cancel are the electrons. Other reactions may be different.