1. IIIIIIIVV Ch. 15 – Oxidation Reduction Equations I. Oxidation and Reduction Defined

2. Redox Reactions n Historically meant reacting with oxygen n Oxidation : the loss of electrons:  Ca  Ca 2+ + 2 e – n The opposite reaction Reduction: the gain of electrons  Br 2 + 2 e –  2 Br 1- n Oxidation & Reduction are 2 opposite processes which usually occur at the same time.

3. Redox Reactions n When calcium and bromine get together redox ocurs: Ca + Br 2  Ca 2+ + 2 Br 1– n One atom is oxidized the other is reduced n Easy to remember which reaction gains and loses electrons my remembering: n O I L R I G n X S O E S A n I S D I n D S U N n A C n T T n I I n O O n N N

4. Oxidation State n One can determine how much oxidized by memorizing a simple set of rules to determine the ‘oxidation state’. Rules for Determining Oxidation State 1. Any neutral element is zero 2. Any alkali metal is + 1 3. Any alkaline earth in a compound is + 2 4. Oxidation state of oxygen in a compound is always -2, except in peroxides where it is -1 5. Hydrogen in a compound is +1, except in hydrides where it is -1.

5. Rules for Determining Oxidation State n 6. Variable Oxidation of atoms  Sum of oxidation states must equal zero for neutral molecules OR must equal the value of the charge for ions  Examples:  Sodium phosphate: Na 2 PO 4  Oxidation state: +1 ? 2-  Totals: +2? 8- =0  Charge on Phosphorous must be: +6  Sulfite: SO 3 2-  Oxidation state: ? 2-  Totals: ? 6- = 2-  Charge on Sulfur must be: +4

6. Rules for Determining Oxidation State n 7. More than one unknown oxdation state  Determine which unknown atom is closest to the upper right hand corner of the periodic table.  Assign that atom a negative oxidation state equal to the number of electrons it would need to have the electron configuration of a noble gas.  Example:  NaSClO 2  Oxidation state: +1 ? ? 2-  Most upper right: +1 ? 1- 2-  Totals: +1 ? 1- 4- = 0  Charge on Sulfur must be: +4

7. Examples: 1) sodium nitrate NaNO 3 2) ammonia NH 3 3) zinc oxide ZnO 4) chromate ion CrO 4 2- 5) calcium hydride CaH 2 6) carbon dioxide CO 2 7) nitrogen N 2 8) sodium sulfate Na 2 SO 4 9) aluminum hydroxide Al(OH) 3 10) nitric acid HNO 3 1) +1, +5, 2- 2) 3-, +1 3) 2+, 2- 4) +6, 2- 5) +2, 1- 6) 4+, 2- 7) 0 8) +1, +6, 2- 9) +3, -2, +1 10) +1, +5, 2-

8. Balancing Half Reactions n Half reactions are equations that represent either a loss or a gain of electrons from one compound. n Steps: 1) Balance atoms other than hydrogen or oxygen. 2) Balance the oxygen by adding water molecules to one side. 3) Balance the hydrogen by adding hydrogen ions to one side. 4) Balance the charge by adding electrons to one side.

9. Example 1 Balance the half reaction: n N 2 O 2  NH 4 + n 1 st atoms other than O & H: N 2 O 2  2 NH 4 + n 2 nd balance oxygen: N 2 O 2  2 NH 4 + + 2 H 2 O n 3 rd balance hydrogen: 12H + + N 2 O 2  2 NH 4 + + 2 H 2 O n 4 th balance the charge (1) add up the charge on both sides (2) add enough electrons to the most positive side to make the charge on that side equal the charge on the other side n 10 e - + 12H + + N 2 O 2  2 NH 4 + + 2 H 2 O +12 +2

10. Balance the half reactions ! n 1. SO 4 2-  S 2 O 3 2-  8 e - + 10H + + 2 SO 4 2-  S 2 O 3 2- + 5 H 2 O n 2. NH 4 +1  NO 2 1-  2 H 2 O + NH 4 +1  NO 2 1- + 8 H + + 6 e - n 3. HSO 3 1-  S 8  32 e - + 40H + + 8 HSO 3 1-  S 8 + 24 H 2 O n 4. Cl 2  ClO 3 1-  6 H 2 O + Cl 2  2 ClO 3 1- + 12 H + + 10 e - n 5. H 3 PO 4  PH 3  8 e - + 8H + + H 3 PO 4  PH 3 + 4 H 2 O

11. Balancing Redox Equations by Half Reactions n An oxidation-reduction equation may be balanced by breaking it up into two half reactions, balancing each half reaction, and putting the two half reactions back together again. n Sample Problem:  SO 2 + MnO 4 1- + H 2 O  SO 4 2- + MnO 2 + H + n Solution:  First pick the two half reactions. Each half reaction will generally contain an element other than hydrogen and oxygen. »SO 2  SO 4 2- »MnO 4  MnO 2

12. Balancing Redox Equations n Second, balance each half reaction as we did in the last section:  2 H 2 O + SO 2  SO 4 2- + 4 H + + 2 e -  3 e - + 4H + + MnO 4 1-  MnO 2 + 2 H 2 O n Next, multiply each half reaction by the lowest numbers that will make the electrons the same & add the 2 half reactions together.  ( 2 H 2 O + SO 2  SO 4 2- + 4 H + + 2 e - ) x 3  (3 e - + 4H + + MnO 4 1-  MnO 2 + 2 H 2 O) x 2 n 6 H 2 O + 3SO 2 + 6 e - + 8H + + 2MnO 4 1-  n 3SO 4 2- + 12 H + + 6e - + 2 MnO 2 + 4 H 2 O +

13. Balancing Redox Equations n Finally simplify by canceling like terms (notice H 2 O, e-, & H + on both sides) n 6 H 2 O + 3SO 2 + 6e - + 8H + + 2MnO 4 1-  3SO 4 2- + 12 H + + 6e - + 2 MnO 2 + 4 H 2 O Solution: n 2 H 2 O + 3SO 2 + 2MnO 4 1-  »3SO 4 2- + 4 H + + 2 MnO 2

14. Redox Problems n Balance each of these redox reactions by the half reaction method demonstrated in the previous sample n 1) Cu + H + + NO 3 -  Cu 2+ + NO 2 + H 2 O  ( Cu  Cu 2+ + 2 e - ) x 1  (1 e - + 2H + + NO 3 1-  NO 2 + H 2 O) x 2 n Cu +4H + +2e - + 2NO 3 -  Cu 2+ + 2e-+ 2NO 2 + 2H 2 O n Solution: n1n1 n Cu +4H + + 2NO 3 -  Cu 2+ + 2NO 2 + 2H 2 O oxidized reduced

15. oxidized Redox Problems n 2) Mn 2+ + H + + PbO 2  MnO 4 1- + Pb 2+ + H 2 O  ( 4H 2 O + Mn 2+  MnO 4 1- + 8H + + 5 e - ) x 2  (2e - + 4H + + PbO 2  Pb 2+ + 2H 2 O) x 5 n 8H 2 O + 2Mn 2+ + 10e - + 20H + + 5PbO 2  2MnO 4 1- + 16H + + 10e - + 5Pb 2+ + 10H 2 O n Solution: n 2Mn 2+ + 4H + + 5PbO 2  2MnO 4 1- + 5Pb 2+ + 2H 2 O

16. Redox Problems n 3) I 2 + HClO + H 2 O  IO 3 1- + Cl 1- + H +  ( 6H 2 O + I 2  2 IO 3 1- + 12H + + 10 e - ) x 1  (2e - + H + + HClO  Cl 1- + H 2 O) x 5 n 6H 2 O + I 2 + 10e - + 5H + + 5HClO  2IO 3 1- + 12H + + 10e - + 5Cl 1- + 5H 2 O n Solution: n H 2 O + I 2 + 5HClO  2IO 3 1- + 7H + + 5Cl 1- oxidized reduced