1. Chapter 20 Oxidation – Reduction Reactions

2. What are they? A family of reactions that are concerned with the transfer of electrons between species Redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time

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4. 2Fe 2 O 3 + 3C  4Fe + 3CO 2 Oxidation refers to the complete or partial loss of electrons or the gain of oxygen Carbon is oxidized Reduction refers to the complete or partial gain of electrons or the loss of oxygen Fe 2 O 3 is reduced

5. Oxidation Numbers Def - A positive or negative number assigned to an atom to indicate its degree of oxidation or reduction General rule – a bonded atom ’ s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element Water – H 2 O (H = +1) (O = -2)

6. Some rules to follow Free elements (non combined) are assigned an oxidation state of zero (ie. Al, Na, Fe, H 2, O 2, N 2 ) The oxidation state for any simple one-atom ion is equal to its charge (ie. Na + = +1) Group 1A metals in compounds are always +1 (H, Li, Na, K) Group 2A metals in compounds are always +2 (Be, Mg, Ca)

7. Oxygen in compounds is assigned an oxidation state of -2, exception is peroxide where it is -1 (eg. H 2 O 2 ) Hydrogen is +1 unless it is in a metal hydride such as NaH; then it is -1 The sum of the oxidation states of all atoms in a species must be equal to the charge on the species eg. HClO 4 = 0  [+1(H)+7(Cl)-2*4=-8(O)]

8. Determine the oxidation number of chlorine in each of the following substances KClO 3, Cl 2, Ca(ClO 4 ) 2, Cl 2 O KClO 3 (Cl = +5) Cl 2 (Cl = 0) Ca(ClO 4 ) 2 (Cl =+7) Cl 2 O (Cl = +1)

9. Assign the following oxidation # ’ s S 2 O 3, Na 2 O 2, P 2 O 5, NO 3 - S = +3, O = -2 Na = +1, O = -1 P = +5, O = -2 N = +5, O = -2

10. 10 Examples - assigning oxidation numbers Assign oxidation states to all elements:

11. H 2 = 0 SO 3 = S +6, O -2 SO 4 -2 = S +6, O -2 K + = +1 NH 3 = N -3, H +1 MnO 4 - = Mn +7, O -2 Cr 2 O 7 -2 = Cr +6, O -2 CH 3 OH = C -2, O -2, H +1 PO 4 3- = P +5, O -2 ClO 3 - = Cl +5, O -2 HSO 3 - = H +1, S +4, O -2 Cu = 0

12. 12 an increase in oxidation number of an atom signifies oxidation a decrease in oxidation number of an atom signifies reduction +2 to +4 0 to -1 Changes in oxidation number

13. 13 SnCl 2 + PbCl 4 SnCl 4 + PbCl 2 CuS + H + + NO 3 - Cu +2 + S + NO + H 2 O +2+4 +4+2 +2 = Oxidation -2 = Reduction -2+2 +1-2+5+2 0 -2+2 -2+1 -3 = Reduction +2 = Oxidation

14. Oxidizing and Reducing Agents Now the confusing part… CuO + H 2  Cu + H 2 O Cu goes from +2 to 0 Cu is reduced, therefore the compound it comes from, CuO, is called an oxidizing agent because it causes some other substance to be oxidized H goes from 0 to +1 H is oxidized, therefore H 2 is called a reducing agent because it causes some other substance to be reduced.

15. Identifying Agents in an Equation CuO + H 2  Cu + H 2 O Reduction: CuO is the oxidizing agent Oxidation: H 2 is the reducing agent

16. 16 n In each reaction, look for changes in oxidation state. n If changes occur, identify the substance being reduced, and the substance being oxidized. n Identify the oxidizing agent and the reducing agent. Examples - labeling redox reactions H 2 + CuO  Cu + H 2 O 0 -2+2 0 -2+1 = +1 (H is oxidized) = -2 (Cu is reduced) (H 2 is the reducing agent) (oxidizing agent is CuO)

17. SnCl 2 + PbCl 4  SnCl 4 + PbCl 2 Sn +2 to +4 so oxidized so SnCl 2 is reducing agent Cl -1 to -1 no change Pb +4 to +2 so reduced so PbCl 4 is oxidizing agent 2Sb + 3 Cl 2  2SbCl 3 Sb 0 to +3 so oxidized = reducing agent Cl 0 to -1 so reduced so Cl 2 is oxidizing agent CuS + H + + NO 3 -  Cu 2+ + S + NO + H 2 O Cu +2 to +2 no change S -2 to 0 so oxidized so CuS is reducing agent H+ +1 to +1 no change N +5 to +2 so reduced so NO 3 - is oxidizing agent O -2 to -2 no change Assign OX#’s & determine what is oxidized/reduced/OA/RA

18. NaI + H 2 SO 4  H 2 S + I 2 + Na 2 SO 4 + H 2 O Na +1 to +1 no change I -1 to 0 oxidized so NaI is reducing agent H +1 to +1 no change S +6 to -2 so reduced so H 2 SO 4 is oxidizing agent O -2 to -2 no change SO 3 + H 2 O  H 2 SO 4 S +6 to +6 no change 0 -2 to -2 no change H +1 to +1 no change Not a redox reaction Assign OX#’s & determine what is oxidized/reduced/OA/RA

19. Redox rxn ’ s that form ions writing half reactions 2 Mg(s) + O 2 (g)  2 MgO(s) Oxidation: Mg  Mg 2+ + 2 é Mg is the reducing agent Reduction: O 2 + 2 é  O 2- O 2 is the oxidizing agent

20. Writing Half-Reactions Ca 0 + 2 H +1 Cl -1  Ca +2 Cl -1 2 + H 2 0 Oxidation: Ca 0  Ca +2 + 2e - Reduction: 2H +1 + 2e -  H 2 0 The two electrons lost by Ca 0 are gained by the two H +1 (each H +1 picks up an electron). PRACTICE SOME!

21. Practice Half-Reactions Don’t forget to determine the charge of each species first! Mg + S  MgS Oxidation Half-Reaction: Mg 0  Mg +2 + 2e - Reduction Half-Reaction: S 0 + 2e -  S -2 Al + P  AlP Oxidation Half-Reaction: Al 0  Al +3 + 3e - Reduction Half-Reaction: P 0 + 3e -  P -3

22. Ni +2 + Fe  Fe +2 + Ni Ni +2 + 2é  Ni 0 Fe 0  Fe +2 + 2é Ca + Se  CaSe Ca 0  Ca +2 + 2é Se 0 + 2é  Se -2 Be + Se  BeSe Be 0  Be 2+ + 2é Se 0 + 2é  Se -2 Write the half-reactions for the following equations

23. Identify these processes as either oxidation or reduction 2I -  I 2 + 2 é Oxidation – loss of electrons Iodine charge changes from - 1 to 0 Zn 2+ + 2 é  Zn Reduction – gain of electrons Zinc charge changes from + 2 to 0

24. Identify as either oxidation or reduction I 2 + 2é  2I - Reduction iodine goes from 0 to -1 Cu  Cu 2+ + 2é Oxidation copper goes from 0 to +2 O 2 + 4é  2O 2- Reduction oxygen goes from 0 to -2 Al  Al 3+ + 3é Oxidation aluminum goes from 0 to +3

25. Corrosion rusting – spontaneous oxidation Disintegration of a metal into its constituent atoms Water speeds up the process Most structural metals have reduction potentials that are less positive than O 2 so in the presence of O 2 they will rust

26. Water Rust Iron Dissolves- Fe  Fe +2 e-e- salt speeds up process by increasing conductivity making electron transfer easier O 2 + 2H 2 O +4e -  4OH - Fe 2+ + O 2 + 2H 2 O  Fe 2 O 3 + 8 H + Fe 2+

27. Preventing Corrosion Galvanizing – putting on a zinc coat Alloying – bonding with another metal that forms an oxide coat Coating – covering a metal with oil, paint or plastic to prevent corrosion Electroplating – putting a thin metal coat over another metal to protect it

28. Not all metals corrode, some form a protective coating to prevent it (Al, Cr) Other metals (gold & platinum) resist losing their electrons; therefore, don ’ t corrode