2. Redox comes from the two words oxidation and reduction. Oxidation: Loss of electrons. When things lose electrons, they become positive. We learned that metals easily give up their electrons. This means metals are easily oxidized. Reduction: Gain of electrons. When things gain electrons, they become negative. We learned the non-metals easily gain electrons. This means non-metals are easily reduced.

3. When an atom is oxidized, it gives up its electrons and causes another atom to gain electrons. Therefore, anything that is oxidized can also be called a “reducing agent” When an atom is reduced, it gains electrons and forces another atom to give up electrons. Therefore, anything that is reduced can also be called an “oxidizing agent”

4. How can we tell if something is oxidized or reduced? We use oxidation numbers When something is oxidized, it loses electrons so the oxidation number increases…(gets more positive). When something is reduced, it gains electrons so the oxidation number decreases… (gets more negative).

5. 1.Uncombined elements have a charge of ZERO. Even the BrINClHOF’s. 2.If an element has only 1 charge on PT, that is the oxidation #. 3.If a non-metal atom is the negative ion in a binary ionic compound then the top charge is the oxidation #. (Add: Hydrogen can be a -1 if with a Group 1 metal; Oxygen can be -1 or +2 in certain situations) 4.If an element has more than 1 charge, use other charges to figure it out. 1.The sum of all charges in a compound = ZERO 2.The sum of all charges in a PAI = charge on PAI.

6. Example: Determine the charge on “S” in ZnSO 4 Use the formula and your ref. tables to figure out what we know: ZnOS

7. Example: Determine the charge on “N” in Fe(NO 3 ) 2 Use the formula and your ref. tables to figure out what we know: Before we do anything, we have to use the nitrate ion to determine the charge on Fe. Remember, Fe has multiple oxidation numbers on the PT. Think of it as if you are trying to name the compound using roman numerals. Fe(NO 3 ) 2 Charge listed on table E. +2

8. Example: Determine the charge on “N” in Fe(NO 3 ) 2 Use the formula and your ref. tables to figure out what we know: FeON 1, +2 charge 6, -2 charge 2 +2-12+10 +5 The +10 charge has to be split between 2 Nitrogen atoms.

9. Reminder: When determining the oxidation number… We are only concerned with charge assigned to the individual atom.

10. Determine the charge on S in H 2 SO 3 Fe in FeCl 3 O in OF 2 Cr in Na 2 Cr 2 O 7 P in (PO 4 ) -3 + 4 + 3 +2 +6 +5 Tricky, Tricky

11. Once we are able to assign oxidation numbers to element… we can look at the oxidation numbers before and after reaction. If oxidation number increases If oxidation number decreases OXIDATION: LOSS OF ELECTRONS REDUCTION: GAIN OF ELECTRONS From left to right:

12. For the reaction Cu + AgNO 3  Cu(NO 3 ) 2 + Ag Step 1: Assign all atoms oxidation numbers. *HINT*: If a polyatomic ion remains the same on both sides of the equation, you do not have to assign charges to each element. Just assign the PAI a charge as a whole.

13. Step 2: Compare oxidation numbers from the left to the right side of the equation. INCREASE: Oxidized DECREASE: Reduced Increase in oxidation number: Cu 0 was oxidized Decrease in oxidation number: Ag +1 was reduced

14. Agents: The species that was oxidized, is known as the reducing agent. The species that was reduced, is known as the oxidizing agent. Cu 0 was oxidized, known as reducing agent. Ag +1 was reduced, known as oxidizing agent.

16. Identify the species that has been oxidized and reduced. F 2 + CaCl 2  CaF 2 + Cl 2 Reaction after assigning oxidation numbers: Reduced Oxidized

17. 2 Mg + O 2  2 MgO 2 Mg 0 + O 2 0  2 Mg +2 O -2 Mg goes from 0  +2… lost 2 electrons = oxidized (reducing agent) O 2 goes from 0  -2…. Gained 2 electrons = reduced (oxidizing agent)

18. 2 Na + 2 H 2 O  2 Na + + 2OH - + H 2 2Na 0  2Na + : Lost 1 electron: oxidized 2 H 2 +1  H 2 0 : Gained 1 electron: reduced

19. To identify a redox rxn: Look for reactions that have a free, uncombined element. This is usually a redox reaction. Or Go through and assign oxidation numbers to every single element in every single equation. If the oxidation number of an element changes = redox reaction.

20. Ex. Which is a redox reaction? A. H + + Cl -1  HCl B. NaOH + HCl  NaCl + H 2 O C. Fe + 2 HCl  FeCl 2 + H 2 D. MgO + H 2 SO 4  MgSO 4 + H 2 O Answer: C – Iron loses 2 e- and hydrogen gains 2 e-

21. Ex. Which equation represents a redox reaction: A. 2Na+ + S -2  2Na 2 S B. H + + C 2 H 3 O 2 -1  HC 2 H 3 O 2 C. NH 3 + H +1 + Cl -1  NH 4 + + Cl - D. Cu + 2 Ag +1 + 2NO 3 -  2 Ag + Cu +2 + 2NO 3 - Answer: D

24. We balance redox reactions to achieve a conservation of both mass and charge. Conservation of charge means that the number of electrons lost must be equal to the number of electrons gained. Conservation of mass means there are equal number of each element on each side.

25. Balancing ½ rxns… Balancing redox reactions can be very difficult. Instead, we split them into their respective ½ rxns and balance them separately. We write one half reaction to show oxidation, and a separate half reaction to show reduction.

26. The oxidation ½ rxn: The reduction ½ rxn: We put the electrons on the right hand side to show the Cu0 becomes oxidized by giving up electrons. It is specifically 2 electrons to balance the charge. We put the electrons on the left hand side to show the two Ag +1 become reduced by gaining electrons. It is specifically 1 electron per Ag +1 to balance the charge. 2 2 

27. Zn + HCl  ZnCl 2 + H 2 00+2+1 Oxidation Half Reaction: Reduction Half Reaction: Zn  Zn 0 +2 + 2 electrons H +1  H 2 0 2 electrons +

28. Example: Which of the following ½ rxns represents reduction? A. O 2 + 4e-  2O 2- B. 2 O -2 + 4e  O 2 C. 2 O -2  O 2 + 4e- D. O 2  2O -2 + 4e- Answer: a

29. When writing half reactions, we add electrons to one side or the other, to maintain charge balance. Oxidation Rxns: electrons are on the product side Reduction Rxns: electrons are on the reactant side *ADD ENOUGH ELECTRONS so that the total charges on both side are equal.

30. Example… Balance the following rxn: ___ Mg + ___ Cr 3+  ___ Mg 2+ + __ Cr At first glance, it appears balanced. However, we have to remember that the total charges on both sides must be equal. (Electrons Gained = Electrons Lost) Break up into ½ rxns: Oxidation: ___ Mg  ___ Mg 2+ + Reduction: ___ Cr 3+ +  ___ Cr Now, balance the charges so that the electrons lost = electrons gained. Make charges equal! 2e- 3e- 3 2 6 e-

31. To complete… Add the two half reactions together, to get one complete reaction: 3 Mg + 2Cr 3+ +  3 Mg 2+ + 2 Cr + We get rid of the electrons because they cancel each other out. We are left with: 3 Mg + 2 Cr 3+  3 Mg 2+ + 2 Cr 6e-

32. Step 1: Write ½ rxns and balance for mass Step 2: Put coefficients in front of oxidation and reduction equations so that the loss of of electrons = gain of electrons. Step 3: Add oxidation and reduction rxns together to get one net equation. Step 4: Add in any other species in the original equation and balance them the way we would ordinarily.

33. Balance: __Na 0 + ___Fe +3  ___ Na +1 +___ Fe 0 Oxidation: Na 0  Na +1 + 1 e- Reduction: Fe +3  Fe 0 3 e- + Electrons gained must equal electrons lost. Multiply equation(s) by an integer! ()3 Final:

34. Balance: __Au 0 + ___H +1  ___ Au +3 +___ H 2 0 Oxidation: Au 0  Au +3 + 3e- Reduction: H +1  H 2 0 2 Balance for mass before you ever balance for charge! Final: 2e- + x 2 x 3

35. Balance the following…. 0 0 +3 -2 Al + O 2  Al 2 O 3 Write the half reactions. You have to balance the atoms before you can balance the electrons! Al 0  Al 2 +3 O 2 0  O 3 -2 2 32 + 6e- + 12e- Now, balance the electrons! ( ) 2 ( )1

36. Mg + AlBr 3  Al + MgBr 2 ID oxidation and reduction rxns: Mg  Mg +2 + 2e Al +3 + 3e-  Al Multiply both equations by integer to get charges equal! 0 +3 -1 0 +2 -1 Oxidation: Oxidation Number Increases Reduction: Oxidation number decreases 3 (Mg  Mg+2 + 2e-) 2 (Al+3 + 3e-  Al)

37. Now insert the coefficients into the main equation… 3 Mg + 2 AlBr 3  2Al + 3 MgBr 2 Which species was oxidized? Which species was reduced? Which was the reducing agent? Which was the oxidizing agent? Mg Al +3 Mg

39. Two types of Electrochemical cells: 1.Voltaic cells: Commonly known as batteries. A spontaneous redox reaction that produces an electric current. 2.Electrolytic cells: An electric current is used to force a reaction to occur that would not happen spontaneously. 1.This reaction is called electrolysis.

40. also known as

41. Determining Spontaneity We use Table J to determine what elements are the most reactive. The higher up on Table J, the more reactive the metal is. In electrochemistry, the more reactive is the more likely to be OXIDIZED. For non-metals, the higher up on Table J, the more reactive is more likely to be REDUCED.

42. Metals: higher on table J, more likely to be oxidized. Non metals: higher on table J, more like to be reduced. Elements HIGHER on Table J will spontaneously replace elements lower on Table J.

43. Which reaction is more likely to take place? 1) Cu +2 + Al 0  Cu 0 + Al +3 2) Cu 0 + Al +3  Cu +2 + Al 0 RED: Cu +2 OX: Al 0 OX: Cu 0 RED: Al +3 If this reaction is spontaneous: then Al has to be higher up than Cu on Table J. Does this agree with Table J? YES If this reaction is spontaneous: then Cu has to be higher up than Al on Table J. Does this agree with Table J? NO Reaction 1 will take place

44. Which rxn is more likely to occur? For non- metals, we need to identify which is more likely to be reduced… Cl 2 0 + 2 F -1  2Cl -1 + F 2 0 OX: F -1 RED: Cl 2 F 2 0 + 2 Cl -1  2F -1 + Cl 2 0 OX: Cl -1 RED: F 2 Does this agree with Table J. Is Cl 2 higher than F? NO! Cl is less active. Does this agree with Table J. YES! F is more active than Cl RXN 2 will take place

45. Voltaic Cell: Uses redox reactions that are spontaneous in nature to produce electricity. Each half reaction is carried out in half- cells. The electrons given off by oxidation rxn travel through a wire and given to the reduction rxn. REMEMBER: LEO goes GER

46. PARTS OF A VOLTAIC CELL: 1. Two half-cells: In each half cell, there is metal strip called an electrode. The electrode is submerged in a solution of the ion of the element. 2. The electrodes are connected to each other with a wire. This wire allows electrons to travel from oxidation rxn  reduction rxn. 3. A salt bridge also connects the two half cells. This allows positive and negative ions to travel back and forth between cells to keep everything neutral.

48. Labeling an electrochemical cell: The Oxidation Half-Cell is called the anode (neg) The Reduction Half-Cell is called the cathode (pos) Electrons always travel from anode  cathode Anions from salt bride travel from cathode  anode

49. Zn because it is the more active metal undergoes oxidation. Zn 0  Zn +2 + 2e- Anions from salt bridge flow to make up for loss of negative charges. Electrons given up by Zn, travel through the wire to the Cu half cell. Cu +2 ions undergo reduction. Cu +2 + 2e-  Cu Cations from salt bridge flow to make up for lost positive charges. Cathode gets bigger Anode gets smaller

52. Non-spontaneous redox reactions require an electricity source in order to force electrons to flow from anode to cathode. Using electricity to force these reactions to occur are called electrolysis reactions. Electrolysis is typically used to decompose compounds or to electroplate items.

53. Remember way back when… When we discussed Groups 1,2, and 17 elements, we said they are never found alone in nature. They are often tied up in compounds called fused salts. When we want to get one of those elements, by itself from a fused salt, we used electrolysis.

54. 1. Electrons flow out of DC power supply and down the wire. This gives the wire connected to the negative end of the battery a negative charge. 2. The Na +1 ion from the solution, migrates to the negative electrode. The following reaction occurs: Na +1 + 1e-  Na 0 3. Because this is a reduction reaction, this electrode is designated as the cathode. However note, the cathode is negatively charged in electrolytic cells. Loaded with electrons = negative charge Na+ Cl- Free, elemental sodium begins to build up on the electrode. The cathode gets bigger in size.

55. 1. The other anode assumes a positive charge. 2. The Cl -1 ion from the solution, migrates to the positive electrode. The electrons are being yanked away. The following reaction occurs: 2 Cl -1  Cl 2 0 + 2e- 3. Because this is a oxidation reaction, this electrode is designated as the anode. However note, the anode is positively charged in electrolytic cells. Positive charge Na+ Cl- Free, elemental chlorine begins to bubble on the electrode. The cathode usually gets smaller in size.

56. Electrolysis of a Salt… The positive metal ion is always reduced at the cathode, which is negative. –Ex. K +1 + 1e-  K 0 (The K will build onto the electrode) The negative non-metal ions is always oxidized at the anode, which is positive. –Ex. 2F -1  F 2 0 + 2e- (The fluorine gas will bubble at the electrode)

57. Electrolysis of Water…

58. You should know how to set up an electrolytic cell in order to electroplate something. This is very hard to understand and make sense of. Study this. It is always on the regents.

59. Electroplating… The silver anode is hooked up to the positive end of the battery. Silver is being stripped of its electrons because they want to travel to the pos. end of the battery. Silver is being oxidized… Ag 0  Ag +1 + 1e-

60. On the other side of the cell, the power supply is giving the ring a negative charge. The Ag+1 ions are attracted to the negative charge. Once they migrate there… Ag +1 + 1e-  Ag 0 The ring gets a silver coating… hence silver-plated jewlery.

61. Setting up Electroplating To set up a electrolytic cell to silver plate a spoon: Hook up the spoon to the negative electrode, the cathode. The bar of silver needs to be hooked up to the positive electrode, the anode. The silver ions that are formed by oxidation, will migrate to the negative electrode and coat the spoon. Put the spoon in a solution of AgNO 3, to provide more Ag + ions.

62. Check for understanding: Application Questions When KF is decomposed, which element will form at the anode? Decomposed = electrolytic cell! The anode is always the site of oxidation. In electrolytic cells, the anode is positive charged. Therefore the F -1 ions will migrate to the anode and become oxidized to form F 2.

63. Check for understanding: Application Questions When KF is decomposed, which element will form at the cathode? Decomposed = electrolytic cell! The cathode is always the site of reduction. In electrolytic cells, the cathode is negatively charged. Therefore the K +1 will migrate to the cathode and be reduced to form K 0.

64. Application Questions: If you want to chrome plate an aluminum tailpipe, what gets hooked up to what electrode of the DC power supply? You want to hook up the tailpipe to the negative end. Why? You need the positive chromium ions to migrate over to the tailpipe. They have just lost their electrons at the anode, and will migrate through the solution to coat the tailpipe. Hence, you want your object to be negatively charged.

65. Electrolytic vs. Voltaic VOLTAIC Produce electricity Spontaneous Reaction Cathode is positive Anode is negative ELECTROLYTIC Require electricity Non-spontaneous rxn Cathode is negative Anode is positive