2. Oxidation Numbers Positive oxidation number Negative oxidation number - Loses partial or total control of electrons in a bond - Gains partial or total control of electrons in a bond Example Mg in the +2 state (Mg +2 ) Has lost partial or total control of 2 e- - Used to tell how many electrons an atom has lost or gained in a chemical reaction Redox - When one atom loses an electron another must receive it

3. Na  Na + + 1e - Cl + 1e -  Cl - 1) Oxidation- The lose of an electron by an atom - Causes an increase in oxidation number 2) Reduction- The gain of an electron by an atom - Causes a decrease in oxidation number If Fe loses three electrons, we write this as Fe  Fe +3 +3e - If the electrons appear on the products side, they are given off oxidation If Fe +3 gains three electrons, we write this as Fe +3 +3e -  Fe If the electrons appear on the reactants side, they are gained reduction

4. Lose Electrons Oxidation Gain Electrons Reduction LEOgoesGER LEO GER Indicate if the atoms shown have lost or gained electrons Then write down if the atom is oxidized or reduced A. H  H + + e - B. Cl 2 + 2e -  2 Cl - C. Fe +2 + 2e -  Fe D. Cl +5 + 6e -  Cl- E. S -2  S +4 + 6 e - Lost or gained e-Oxidized or reduced Lost Gained oxidized reduced Lost Gained oxidized reduced

5. Lightening Underwater pg 207

7. Oxidizing Agent- Causes another substance to be oxidized - Is reduced Ex. Sn +4 + 2 e-  Sn +2 Sn +4 is the oxidizing agent Reducing Agent- Causes another substance to be reduced - Is oxidized Ex. Na  Na +1 + 1 e- Na is reducing agent

8. Blue bottle demo Blue is oxidized = lost electrons Colorless is reduced = more electrons

9. Finding Oxidation States The most common oxidation numbers are in the upper right corner of your periodic table Pb +2 +4 Oxidation states Using your periodic table, list the oxidation states of the following Ca Na O N Examples +2 +1 -3, -2, -1 and a bunch more! -2

10. Rules for assigning oxidation numbers 1. Uncombined elements have an oxidation number of zero Cu 0 Mg 0 S0S0 2.) The oxidation state of an ion is the same as its charge Oxidation state of zero means the atom is not losing or gaining any electrons Cu + Mg +2 S -2 +1 oxid. state+2 oxid. state-2 oxid. state Cl 2 0

11. 3) Group 1 metals in a compound always have a +1 oxidation state 4) Group 2 metals in a compound always have a +2 oxidation state MgCl 2 LiBrNa 2 O +2+1 5) Hydrogen always has a +1 oxidation state in a compound Exception - When H is attached to a group 1 or 2 metal, it has a -1 oxidation state HCl H2OH2O +1 NaHCaH 2

12. 6) Oxygen in a compound always has a -2 oxid. state Exception - In a peroxide, oxygen becomes -1 Na 2 O 2 H2O2H2O2 H2OH2OHNO 3 -2 7) Halogens are usually -1 in compounds SO 4 -2 +6-2 -8+6 = -2 The sum of the one sulfur and the 4 oxygens should be -2 8) The sum of all oxidation numbers in a polyatomic ion must equal the charge of the ion

13. Examples - Assign the oxidation numbers to all elements in the following ions NO 3 - CrO 4 -2 +6 -6+5 -2 +6-8 = -1 = -2

14. Assigning Oxidation Numbers - All of the oxidation numbers of all atoms in a neutral compound must add up to zero For example, in H 2 O, H2OH2O +1-2 The two hydrogens add up to +2, +2 + -2 = 0 Since water is a neutral compound, all oxidation states must add up to zero Neutral compounds

15. Practice - Assign oxidation numbers to the following MgCl 2 LiOH Na 2 S H 2 SO 4 NaNO 3 +2 -1 +1 -2 +1+1 -2+1 +6 -2 +1 +5 -2 +2 -2+1 -2 +2-2+2-8+6 +1+5-6 0000 0 Cu(NO 3 ) 2 NaH NO 2 +2 +5 -2 +1 -1 +4 -2 -12+10+2 +1+4-4

17. Recognizing Redox Equations - Not all reactions are redox - Double replacement reactions are NOT - Single replacement reactions are - If a reaction is redox then the oxidation numbers of some of the elements must be different on either side of the equation Is the reaction redox? LiOH + HCl  H 2 O + LiCl 2H 2 + O 2  2 H 2 O Mg + CuSO 4  MgSO 4 + Cu

19. In the reaction Mg + Cl 2 --> MgCl 2 There are 2 steps occurring Mg loses 2 electrons Each chlorine gains an electron Mg Cl Electrons are moving. If e- could move through a wire, this would be an electric current

20. Electrochemistry Half reaction- Either the reduction or the oxidation portion of a redox reaction Al +3 + 3e- --> Al Ca --> Ca +2 + 2e- reduction oxidation The reduction ½ reaction shows an atom or ion gaining e- and the oxidation number decreasing The oxidation ½ reaction shows an atom or ion losing e- and the oxidation number increasing

21. Steps in Writing Half Reactions 2 Li + CaBr 2 --> 2 LiBr + Ca 1. Assign oxidation numbers +2 +1 0 0 2. For each atom that changes its state, write down the starting and ending oxidation states 3. Add electrons to balance the equation Ca +2 + 2e - Ca 0 Li 0 Li + + e - -Notice that each set of half reactions contains oxidation and reduction -For each half reaction, both sides have the same charge OxidationReduction Ca +2 --> CaLi --> Li +

22. Which reaction shows conservation of charge? Fe(s)  Fe 2+ (aq) + e- Fe + 2 e-  Fe 2+ Fe + 2 e-  Fe 3+ Fe(s)  Fe 2+ (aq) + 2 e-

23. Examples Write out the half reactions A. KBr + Na  K + NaBr B. Sr + MgO  Mg + SrO C. NiCl + CuO  NiO + CuCl K + +e- --> K Na --> Na + + e- Sr --> Sr +2 + 2e- Mg +2 + 2e- --> Mg Ni +1  Ni +2 + 1 e - Cu +2 + 1e -  Cu +1

24. Mg + CuSO 4  MgSO 4 + Cu Write the ½ reaction representing oxidation. Given the reaction Write the reduction ½ reaction for the following 2 Al + 3 Cu +2  2 Al 3+ + 3 Cu Mg  Mg +2 + 2 e - Cu +2 + 2e -  Cu

25. For the following Zn + Cr 3+  Zn 2+ + Cr a.) Write the ½ reaction for the reduction. b.) Write the ½ reaction for the oxidation c.) Which species loses electrons? d.) What happens to the number of protons in a Zn atom when it changes to Zn 2+ as the redox reaction occurs.

26. If one of the diatomic elements move to the zero oxidation state, we must write the reaction with two atoms Example We cannot write Cl -1 --> Cl + e - We must write 2 Cl - --> Cl 2 + 2e- Other examples Br 2 + 2 e- --> 2 Br - 2 H + + 2e - --> H 2

28. B. Electrochemical Cells electricity Electrochemical cell - Produces electricity from a divided redox reaction - Electrons are moved from one atom to another through a wire Voltaic Cell-(battery)-A spontaneous chemical reaction produces a flow of e- Anode- Is the metal in a voltaic cell that is oxidized( It will dissolve) Cathode- The metal in a voltaic cell where reduction will occur - Positive ions in solution will be reduced an collect on the cathode External conductor (Wires)-permit flow of e- Salt Bridge- U-tube containing electrolytic solution of + and - ions - Allows migration of ions - Keeps compartments neutral These two jars are called ½ cells

30. How to identify the anode and cathode and the direction of e- flow 1. Find out which metal is oxidized / reduced An OxAnode is Oxidized Table J: Best RA will lose e- so is the oxidizing ½ reaction Red CatReduction happens at the Cathode A reaction will be spontaneous if a pure metal is on the top, and the ion is below

32. Examples - Determine which atom/ion pairs will react spontaneously A. Cu +2 and Ni B. Mn +3 and Zn C. K and Ni +2 D. Co and Al +3 YES NO YES NO As long as the solid metal is on top, the redox will be spontaneous.

33. Magnesium and silver nitrate flash reaction

34. Thermite Reaction

35. 2. Write ½ reactions LEO- Loss of electrons is oxidation GER- Gain of electrons is reduction LEO An Ox GER RED CAT Mg 0  Mg +2 + 2e- Ni +2 + 2e  Ni 0

36. 3) Choose sign for anode / cathode - ( e- flow from – to + ) - (-) lost from anode  gained by cathode (+) - e- flow from Mg to Ni As the reaction continues Mg  Mg +2 + 2e - Ni +2 + 2e -  Ni Anode Cathode Will decrease in massWill increase in mass

37. Problem - Solutions must always be electrically neutral As soon as we add -, we must add + charges Zn Zn +2 Cu +2 Cu SO 4 -2 Each solution has a total charge of zero (neutral) If we want to add more Zn +2 to solution, we need to add some - ions If we want to remove some Cu +2 from solution, we need to add some + charges d. Salt BridgeTube containing a salt solution Na+ Cl- Porous plugs allow ions to move out of the tube

38. As Zn +2 goes into solution, 2 Cl - come out of the salt bridge As Cu +2 is removed from the solution, 2 Na + come out of the salt bridge Solutions are kept neutral Zn Zn +2 Cu +2 Cu Cu +2 Zn e- 2e- SO 4 -2 Na + Cl- Zn +2 Cl- Now the Zn +2 solution is neutral Na+ Now the Cu +2 solution is neutral

41. E. Electrolytic Cell Electrolytic cell An electric current is used to force a chemical reaction to occur Electrolysis A chemical reaction which uses electricity to break apart a compound – nonspontaneous 1. Description Only one cell is needed Cell contains a power source Battery pulls electrons off of one electrode And puts them on another electrode + e- + - -

42. The battery decides what is + and - electrodes If we add melted NaCl (MOLTEN NaCl) to the cell Na+ and Cl- will be free to move Na + move to the negative electrode and gains electrons This is reductionThe negative electrode is the cathode Cl- move to the positive electrode and loses electrons This is oxidationThe positive electrode is the anode + + - - Na + Cl- Na + + e- --> Na2 Cl- --> Cl 2 + 2e- reduction CATHODE oxidation ANODE

43. For electrolytic cells, the charges are opposite from galvanic cells The “red cat” and “an ox” statements still work 2. Electroplating - Electrolytic cell is used to produce pure metals (Na, Mg) ElectroplatingLayering a metal onto a surface using an electrolytic cell Cell looks the same (one cell with a power source + + - - Put the object to be plated on the – electrode ( cathode) Put the layering metal on the + electrode Ag This is supposed to be a copper ring!

44. + + - - Ag The battery will pull electrons off of the Ag, turning it into Ag + Ag --> Ag+ + e- The Ag + goes into solution and the e- go through the wire e- Ag + The Ag + now move over to gain its own e- on the ring The Ag + gets metallically bonded to the copper Ag+ + e- --> Ag and reduction occurs at the cathode, which is negative Again, oxidation occurs at the anode, which is positive Reduction Oxidation In electroplating, the same atom is oxidized and then reduced

45. Multicolored Electrolysis pg 240