1. 1 Oxidation- Reduction Chapter 16 Tro, 2 nd ed. 1.1

2. 2 Oxidation Number The oxidation number (oxidation state) of an atom represents the number of electrons lost, gained, or unequally shared by an atom. (IT IS NOT THE SAME AS CHARGE!!) Oxidation numbers can be zero, positive or negative. An oxidation number of zero means the atom has the same number of electrons assigned to it as there are in the free neutral atom, WHICH ARE ALWAYS ZERO. A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom. A negative oxidation number means the atom has more electrons assigned to it than in the neutral atom.

3. 3 Rules for Assigning Oxidation Numbers (similar to page 563) Apply rules from top down; first one you reach has priority. 1. All elements in their free state have an oxidation number of zero. (e.g., Na, Cu, H 2, O 2, Cl 2, N 2 ). 2. a. H is +1, except in metal hydrides, where it is -1 (e.g., NaH, CaH 2 ). b. F is always -1, except when it is F 2. 3. O is -2, except in peroxides, where it is -1, and in OF 2, where it is +2. 4. The metallic element in an ionic compound has a positive oxidation number. (Group IA is always +1 and Group IIA is always +2.) The nonmetallic element in an ionic compound has a negative oxidation number. (Halogens -1, Group VIA -2, Group VA -3.) 5. In covalent compounds the negative oxidation number is assigned to the most electronegative atom 6. The algebraic sum of the oxidation numbers of the elements in a compound is zero. 7. The algebraic sum of the oxidation numbers of the elements in a polyatomic ion is equal to the charge of the ion.

4. 4 Oxidation Number For the seven diatomic elements which exist as covalent molecules: each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity. H-H and Cl-Cl: both atoms in each molecule have an oxidation number of 0. Other elements: Cu, Ne, C, et., will all be assigned a 0.

5. 5 Oxidation Number The oxidation number of an atom that has gained or lost electrons to form a monatomic ion is the same as the positive or negative charge of the ion. NaCl Na + has an oxidation number of +1. Cl - has an oxidation number of -1.

6. 6 Oxidation Number Given a compound like HCl, use the rules to assign oxidation number. Rule 2 says H is +1 unless with a metal: therefore H is +1. Rule 6 says the sum of the oxidation numbers must be zero, therefore Cl will be assigned a -1. In binary compounds, halogens are often a -1, like their monatomic ions would be. However, if a halogen is part of a polyatomic ion, then its ON must be determined. HCl +1,-1

7. 7 N2N2 N2ON2ONON2O3N2O3 NO 2 N2O5N2O5 N oxidation number 0+1+2+3+4+5 Many elements have multiple oxidation numbers

8. 8 Notice that H and group I metals have O.N. of +1 Notice that group II metals have O.N. of +2 Halogens are typically assigned O.N. of -1 Oxygen is usually -2

9. 9 Step 1 Write the oxidation number of each known atom below the atom in the formula. Step 2 Multiply each oxidation number by the number of atoms of that element in the compound. Step 3 Write an expression indicating the sum of all the oxidation numbers in the compound. Remember: The sum of the oxidation numbers in a compound must equal zero. Rules for Determining the Oxidation Number of an Element Within a Compound

10. 10 Determine the oxidation number for sulfur in sulfuric acid. H 2 SO 4 Step 1 -2+1 Write the oxidation number of each known atom below the atom in the formula. Multiply each oxidation number by the number of atoms of that element in the compound. Step 24(-2) = -82(+1) = +2 Step 3+2 + S + (-8) = 0 Write an expression indicating the sum of all the oxidation numbers in the compound. Step 4S = +6 (oxidation number for sulfur)

11. 11 Determine the oxidation number for carbon in the oxalate ion. Step 1 -2 Write the oxidation number of each known atom below the atom in the formula. Multiply each oxidation number by the number of atoms of that element in the compound. Step 24(-2) = -8 Step 32C + (-8) = -2 (the charge on the ion) Write an expression indicating the sum of all the oxidation numbers in the compound.

12. 12 ASSIGNING OXIDATION NUMBERS PRACTICE: FeCl 2, MnO 2, F 2, Al, KI, IF 5, H 2 O, OF 2, SO 2, SO 4 2-, OH -, HCl, ClO -, ClO 4 - +2 and -1, +4 and -2, zero, zero, +1 and -1, +5 and -1, +1 and -2, +2 and -1, +4 and -2, +6 and -2, -2 and +1, +1 and -1, +1 and -2, +7 and -2

13. 13 Oxidation-Reduction Oxidation-reduction (redox) is a chemical process in which the oxidation number of an element is changed. Redox may involve the complete transfer of electrons to form ionic bonds or a partial transfer of electrons to form covalent bonds. Loss of electrons = oxidation Gain of electrons = reduction LEO says GER

14. 14 Oxidation-Reduction Oxidation occurs when the oxidation number of an element increases as a result of losing electrons. Reduction occurs when the oxidation number of an element decreases as a result of gaining electrons. In a redox reaction oxidation and reduction occur simultaneously, one cannot occur in the absence of the other.

15. 15 Oxidation-Reduction Oxidizing agent The substance that causes an increase in the oxidation state of another substance. The oxidizing agent is reduced in a redox reaction. Reducing agent The substance that causes a decrease in the oxidation state of another substance. The reducing agent is oxidized in a redox reaction.

16. 16 Oxidation-Reduction Electron-transfer reactions: Zn (s) + 2 HCl (aq)  ZnCl 2(aq) + H 2(g) Zn: from atom to ion; gives up e-s to H H: from ion to atom; accepts e-s from Zn by electron transfer! Zn atom becomes Zn 2+ ion and is oxidized. H + ion becomes H 2 atoms = reduced. Zn is the reducing agent and H + is the oxidizing agent. You must always have both oxidation and reduction in the reaction.

17. 17 17.1 SCALE OF OXIDATION NUMBERS AND MEANING OF OXIDATION AND REDUCTION

18. 18 Redox Reactions Practice: find oxid # for all elements in these reactions to see if electron-transfer occurs. 1. 2 Mg + O 2  2 MgO 2. 2 KClO 3  2 KCl + 3 O 2 3. Cu (s) + 2 AgNO 3(aq)  2 Ag (s) + Cu(NO 3 ) 2(aq) 4. HCl + NaOH  H 2 O + NaCl 5. CH 4 + O 2  CO 2 + 2 H 2 O 1. YES2. YES3. YES (Single Replacement reactions are ALWAYS redox.) 4. NO5. YES

19. 19 Balancing Oxidation Reduction Equations: The Half-Reaction Method (acidic solution) Step 1: Determine the oxidation number of every atom/ion in the reaction. Figure out which is being oxidized and which reduced. Connect them with arrows and determine the number of electrons lost or gained per atom. Then write the two half-reactions that contain the elements being oxidized and reduced. Step 2: Balance the elements other than hydrogen and oxygen. (This is the step that students forget!) Step 3: Balance oxygen by adding H 2 O, then balance H by adding H +. Step 4: Add electrons (e-) to each half-reaction to bring them into electrical balance. Step 5: Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number (using least common denominator) to make the number of electrons the same in each half- reaction. Step 6: Add the two half-reactions together, canceling electrons and any other identical substances (like H + and H 2 O) that appear on opposite sides of the equation. In ionic redox equations both the numbers of atoms and the electrical charge on both sides of the equation must be the same.

20. 20 Balance the equation Step 1: MnO 4 - + S 2-  Mn 2+ + S +7, -2 -2 +2 0 Mn reduced, gains 5 e-/Mn; S oxidized, loses 2 e-s/S Balance elements other than oxygen and hydrogen. (Step 2 is unnecessary, since these elements are already balanced). Step 2 Oxidation half- reaction Reduction half- reaction

21. 21 Step 3 The oxidation half-reaction requires neither O nor H, but the reduction equation needs 4 H 2 O on the right and 8 H + on the left. 4 water molecules are necessary to balance the 4 oxygens in MnO 4. 8H + balance the 8 hydrogens of 4 water molecules. MnO 4. Balance the equation

22. 22 Balance the equation Step 4 Balance each half-reaction electrically with electrons. You have already determined the number of electrons for each. net charge = +2 on each side net charge = -2 on each side balanced oxidation half-reaction balanced reduction half-reaction

23. 23 Balance the equation Step 5 Equalize loss and gain of electrons. In this case, multiply the oxidation equation by 5 and the reduction equation by 2.

24. 24 Balance the equation Step 6 Add the two half-reactions together, canceling the 10e - from each side, to obtain the balanced equation. The charge on both sides of the balanced equation is +4

25. 25 Balancing Redox Equations: The Half-Reaction Method (basic solution) For reactions in alkaline (basic) solutions, first balance as though the reactions were in an acid solution, using Steps 1-6 as before. Add OH - to each side equal to the number of H + on one side. Where both OH - and H + appear on same side of reaction, combine them to make H 2 O. Rewrite the equation, canceling equal numbers of water molecules that appear on opposite side of the equation.

26. 26 PRACTICE WITH BASIC SOLUTION ___Cl 2 + ___IO 3 -  ___Cl - + ___IO 4 - After steps 1-6: Cl 2 + IO 3 - + H 2 O  2 Cl - + IO 4 - + 2 H + Add 2 OH - to both sides: Cl 2 + IO 3 - + H 2 O + 2 OH -  2 Cl - + IO 4 - + 2 H + + 2 OH - Combine H + and OH - to make water: Cl 2 + IO 3 - + H 2 O + 2 OH -  2 Cl - + IO 4 - + 2 H 2 O Cancel one water on each side for finished equation: Cl 2 + IO 3 - + 2 OH -  2 Cl - + IO 4 - + H 2 O

27. 27 AND THAT’S ALL WE NEED FROM CHP 16! There is a lab on redox and practice handouts and homework! Do it!