Redox Class #1: oxidation + reduction reactions made fun 1.2.3: Redox explains the chemistry behind batteries and electroplating precious metals onto strong base metals. It also explains how to break apart water into hydrogen and oxygen, and lots more. 4: These 2 reactions are always paired and balanced.

5. In the old days, Oxidation meant____________________________________ Reduction meant ___________________________________ Now, the real definitions are Oxidation is: Reduction is: 7. To remember this we’ll say LEO the lion goes GER!

Example 8 Magnesium and sulfur form magnesium sulfide (synthesis) Mg + S MgS Magnesium atom + sulfur atom make ionic magnesium sulfide All atoms are neutral so we can ADD to our drawing like this: Mg° + S° MgS 9. Those little circles means neutral

Mg° + S° MgS The magnesium and sulfur atoms are of course neutral, they have equal numbers of protons and electrons (all atoms do). The magnesium sulfide is neutral too, but it’s formed by the combination of Mg +2 and S -2 ions. It’s net neutral, but each ion has an individual charge. Even though we wouldn’t write out the charges, what really happens is this: 10. Mg° + S° [ Mg +2 S -2 ] net neutral

Mg° + S° Mg +2 S In this reaction 2 different things happened... The magnesium lost 2 electrons, the magnesium was oxidized. The sulfur gained 2 electrons, the sulfur was reduced. 12. Redox is the pair of reactions that allows for the balanced transfer of electrons. Redox in this case is synthesis. It can be single replacement, it can be decomp, it can be combustion, etc. Just watch those pesky electrons.

14. The magnesium is oxidized by the sulfur, the sulfur is reduced by the magnesium 15. Neutral atoms form a neutral ionic compound, but the OXIDATION NUMBERS change, when ions form. If the oxidation numbers change, it ’s redox. 16. Word EQ: silver nitrate solution + copper yields… 17. Balanced chemical equation for that is…

17. AgNO 3(AQ) + Cu (S) CuNO 3(AQ) + Ag (S) Let us re-write this with the oxidation numbers that are there only in our minds so we can see the redox reactions. 18. Ag NO 3 (AQ) + Cu (S) Cu NO 3 (AQ) + Ag (S)

Ag +1 NO 3 -1 (AQ) + Cu° (S) Cu +1 NO 3 -1 (AQ) + Ag (S) ° The ionic silver nitrate combines to the neutral atoms of copper. Copper changes into a +1 cation, jumping into solution to combine with the nitrate ions. The silver ions in solution get bumped out as a solid, and these new silver atoms are of course neutral The Silver Ions ___________ electrons, the Ag +1 are ________________ 20. The Copper atoms ___________ electrons, the Cu° are ____________

Ag +1 NO 3 -1 (AQ) + Cu° (S) Cu +1 NO 3 -1 (AQ) + Ag (S) ° The copper atoms lose electrons (LEO) so you can say, the copper atoms are oxidized into copper +1 cations. The silver +1 cations gain electrons to form into silver atoms, so you can say that the silver +1 cations are reduced into silver atoms. 21. The nitrates are just swimming around, available to bond to any cations, they do not participate in the reaction, they just are there. We call these kinds of non-participatory ions the Spectator ions. (They don’t change)

Redox in proper form… 22. Li + NaCl LiCl + Na ½ox: ½red: 23. Lithium is oxidized into the lithium cation The sodium cation is reduced into the sodium atom.

Redox in proper form… 22. Li + NaCl LiCl + Na ½ox: Li° Li e -1 ½red: Na e -1 Na° 23. Lithium is oxidized into the lithium cation The sodium cation is reduced into the sodium atom. 24. The spectator ion is the CHLORIDE Cl -1

24. show the half reactions for this single replacement/redox reaction. Mg + 2HCl MgCl 2 + H 2 ½ox: ½red: 25. Here, the Mg atoms are oxidized into _____________. The H +1 cations are reduced to ________________ 26. In this case, ____ electrons are oxidized, so ____ electrons must be reduced. The electron transfer MUST be in balance!

24. show the half reactions for this single replacement/redox reaction. Mg + 2HCl MgCl 2 + H 2 ½ox: Mg° Mg e -1 ½red: 2H e -1 H 2 ° 25. Here, the Mg atoms are oxidized into Mg+2 cations The H +1 cations are reduced to H 2 ° neutral atoms 26. In this case, 2 electrons are oxidized, so 2 electrons must be reduced. The electron transfer MUST be in balance!

Write the balanced chemical reaction and the half reactions for this last redox. 27. Sodium atoms + chlorine molecules synthesize into table salt Balanced reaction: ½ox: ½red: Tonight: Read the BASICS (no kidding). Go buy a review book too!

Redox Class #2 OB: assigning oxidation numbers plus how batteries (voltaic cells) work. If you read the BASICS this should be easy, if you didn’t read the BASICS yet, ask yourself why? Take better care of yourself, you’re worth it.

28. Ions have easy oxidation numbers, it’s just the charge of the ion. For examples: 29. The oxidation number for the sodium cation is _____ For the chloride anion it is ____ For the sulfate anion (table E), it’s ______ For magnesium cation it is _____ For all atoms (including the HONClBrIF twins) it is ____, because they have no charge. Inside molecules, like carbon dioxide (no ions) there are still oxidation numbers. Take your reference tables out now, periodic table, please.

31. What are the individual oxidation numbers for all of these species? (they better sum to zero!) CO 2 CO CaCl 2 NO 2 PCl 3 PCl 5 H 2 SO 4 Cr 2 O 7 -2 NbBr 5

31. What are the individual oxidation numbers for all of these species? (they better sum to zero!) CO 2 C +4 O -2 CO C +2 O -2 CaCl 2 Ca +2 Cl -1 Cl -1 NO 2 N +4 O -2 O -2 PCl 3 P +3 Cl -1 Cl -1 Cl -1 PCl 5 P +5 Cl -1 Cl -1 Cl -1 Cl -1 Cl -1 H 2 SO 4 H +1 H +1 S +6 and 4O -2 Cr 2 O Cr +6 and 7O -2 NbBr 5 Nb +5 and 5Br -1

single replacement (redox too) reaction Copper (II) sulfate solution plus lithium forms Lithium sulfate solution and copper 32. Write a balanced chemical equation now. 33. What species is oxidized? __________________ 34. What species is reduced? __________________ 35. Name the spectator ion_______________

36. Write the half reactions and the net ionic equation now (WHAT???) ½ ox: ______________________ ½ red: ______________________ NET: _______________________ CuSO 4(AQ) + 2Li (S) → Li 2 SO 4(AQ) + Cu (S)

single replacement (redox too) reaction Copper (II) sulfate solution plus lithium forms Lithium sulfate solution and copper Write a balanced chemical equation now. What species is oxidized? Li° What species is reduced? Cu +2 name the spectator ion SO 4 -2 Write the half reactions and the net ionic equation now. ½ ox: 2Li° 2Li e - ½ red: Cu e - Cu ° NET: 2Li° + Cu +2 2Li +1 + Cu ° CuSO 4(AQ) + 2Li (S) Li 2 SO 4(AQ) + Cu (S)

37. A battery is a voltaic cell. 38. Batteries spontaneously produces electricity. Electricity is the flow of electrons, which can do work – like light up your bulb, or start your car, or what ever. Batteries are cute, and neatly packaged, but to learn how they work we sort of have to take them apart to see the details.

bulb We’re going to label this now. It shows two beakers of solution, with a piece of metal in each one, connected by a wire. There’s also a glass tube with a solution in it connecting the 2 solutions, with cotton balls clogging the ends. At top is a bulb, which would light up if electricity goes through the wire. Go AHEAD

bulb

ZnSO 4(AQ) CuBr2 (AQ) Zn Cu Salt bridge KCl (AQ) 39. Label the PARTS, Take out table J now…

bulb ZnSO 4(AQ) CuBr 2(AQ) Zn Cu Salt bridge KCl (AQ) 40. First, decide which metal is going to oxidize and which will be forced to reduce, then we will label this diagram completely. 41. Next slide, write the half reactions on the diagram in place.

bulb ZnSO 4(AQ) CuBr 2(AQ) Zn Cu Salt bridge KCl (AQ) 42. Write the half reactions below with the net ionic equation as well. ½ ox: __________________________ ½ red: ______________________________ NET ionic equation: ______________________________________

42 ½ ox: Zn° Zn e - ½ red: Cu e - Cu° NET ionic equation: Zn° + Cu +2 Zn +2 + Cu°

bulb ZnSO 4(AQ) CuBr 2(AQ) Zn Cu Salt bridge KCl (AQ) OxidationReduction Zn +2 Cu + 2 Electron flow The solutions become charged immediately, stopping the flow of electrons. What will we do?

bulb ZnSO 4(AQ) CuBr 2(AQ) Zn Cu Salt bridge KCl (AQ) OxidationReduction Zn +2 Cu + 2 Electron flow + - Cl -1 K The ions of the salt bridge balance out the electrical potential that builds up, letting the electricity flow and flow!

bulb PbCl 2(AQ) Mg(NO 3 ) 2(AQ) Pb Mg Salt bridge NaCl (AQ) 46. Completely label this, OX side, RED side, anode, cathode, direction of electrons, directions of salt ions, and half reactions. 47. Name the 3 reasons that this battery will die.

bulb PbCl 2(AQ) Mg(NO 3 ) 2(AQ) Pb Mg Salt bridge NaCl (AQ)

The 3 Reasons that ALL Batteries Die? A. B. C.

The 3 Reasons that ALL Batteries Die? A. Run out of ANODE metal no more oxidation, no more electricity produced B. Run out of CATHODE SIDE CATIONS no more reduction, so no more oxidation either. C. Run out of salt ions solutions get charged, electrons cease their flow.

OB: redox class #3 Lots more practice with voltaic cells 48. Label this voltaic cell diagram completely. 49. Write both half reactions, then the net ionic equation. ½Oxidation: ____________________________________________ ½Reduction: ________________________________________________ Net Ionic Equation: __________________________________________ 50. State the 3 specific reasons that THIS voltaic cell will die. 51. Name the solutions and the salt solution properly

bulb RbF (AQ) CsCl (AQ) Rb Cs NaI (AQ) ½ ox: _________________________________ ½ red: __________________________________ NET IONIC EQ: ________________________________________ Also name the two solutions and the salt properly.

bulb RbF (AQ) CsCl (AQ) Rb Cs NaI (AQ) ½ ox: Rb° Rb e -1 ½ red: Cs e -1 Cs° NET IONIC EQ: Rb° + Cs +1 Rb +1 + Cs° Rubidium fluoride, cesium chloride solutions, with sodium iodide salt solution. This cell will die when it runs out of Rubidium anode, Cesium cathode side cations, or runs out of sodium iodide salt solution. RED Rb +1 OX Cs +1 anode cathode I -1 Na +1 e -1

bulb Ba(NO 3 ) 2(AQ) Sr(C 2 H 3 O 2 ) 2(AQ) Ba Sr (NH 4 ) 2 CO 3(AQ) 52. ½ ox: _________________________________ ½ red: __________________________________ NET IONIC EQ: ________________________________________ Also, name the two solutions and the salt properly. 53. Why will this cell die???

bulb Ba(NO 3 ) 2(AQ) Sr(C 2 H 3 O 2 ) 2(AQ) Ba Sr (NH 4 ) 2 CO 3(AQ) ½ ox: Ba° Ba e -1 ½ red: Sr e -1 Sr° NET IONIC EQ: Ba° + Sr +2 Ba +2 + Sr° Barium nitrate and strontium hydroxide solutions, ammonium carbonate salt solution. anode cathode bulb CO 3 -2 NH 4 +1 e -1 Ba +2 Sr +2 OXRED

bulb AgHCO 3(AQ) MgCrO 4(AQ) Ag Mg Li 3 (PO 4 ) (AQ) 54. ½ ox: _________________________________ ½ red: __________________________________ NET IONIC EQ: ________________________________________ Also name the two solutions, and the salt, properly. 55. Death of this cell????

bulb AgHCO 3(AQ) MgCrO 4(AQ) Ag Mg Li 3 (PO 4 ) (AQ) ½ ox: Mg ° Mg e -1 ½ red: 2Ag e -1 2Ag° NET IONIC EQ: Mg° + 2Ag +1 Mg Ag° Silver hydrogen carbonate and magnesium chromate solutions, with lithium phosphate salt solution REDOX Mg +2 Ag +1 anode cathode e -1 Li +1 PO 4 -3

OB: Redox Class #4 Voltaic vs. Electrolytic cells Voltaic cells has chemistry spontaneously produce electricity. The other kind of electrochemical cell, the electrolytic cell, uses electricity to make redox chemistry happen.

56. A voltaic cell will _______________________________ create ___________________________________ from ____________________. 57. The batteries always run until they stop. The three reasons every voltaic cell will die are: A. B. C. 58. The “opposite” of a voltaic cell, when electricity is used to force a non- spontaneous redox reaction is called an _______________________ cell. 59. There are 2 kinds of electrochemical cells, the _____________________ and the _____________________________ cells.

When you set up an electrochemical cell, called a voltaic cell, or battery, the chemistry happens and the electricity is produced until the cell dies out. It just happens spontaneously because it can happen. If you want a “reverse” redox to occur, one that should not happen spontaneously, you can force it to happen if you push it with electricity. For example: If we put a hunk of copper metal into a beaker containing silver nitrate solution, 60. Write the balanced spontaneous chemical equation. AgNO 3(AQ) copper Cu (S) + AgNO 3(AQ) CuNO 3(AQ) + Ag (S)

61 (read all of this)… The spontaneous reaction is that the copper ionizes into solution, replacing the silver ions. The silver ions pick up the electrons, forming into silver atoms, precipitating into solids. The copper is “higher on table J” than the silver is. But what if you wanted to put silver plating onto copper? What if you wanted to FORCE a non-spontaneous reaction to occur? Could you? With some smarts, and some equipment, and some electricity, you can force a redox reaction that would not occur spontaneously. This is called forming an ELECTROLTIC CELL, or electroplating. In this case we’d be silver plating the copper with silver. 62. By using electricity to PUSH a nonspontaneous redox reaction, we’ve created an _______________________ cell

AgNO 3(AQ) Ag (S) battery Let’s try to label the flow of electricity (electrons) through this whole system. 64. Battery, flow of electricity, anode, cathode, state clearly what metal is going to electroplate onto the other. 65. Why would this ever stop? Cu (S)

AgNO 3(AQ) Ag (S) battery Cu (S) The electricity pushes electrons onto the copper ring. The silver cations pick up these electrons, forming more silver on the copper. The silver bar oxidizes into more silver cations for the solution. The electrons from this oxidation replace the electrons from the battery, completing the redox in the battery. It only stops when the battery dies. Ag +1 e -1

AgNO 3(AQ) Ag (S) battery Cu (S) Reduction happens ON THE CATHODE METAL, so the ring is the cathode. CATHODE Oxidation happens ON the silver bar. Ag +1 e -1 ANODE

AgNO 3(AQ) Ag (S) battery Cu (S) CATHODE Ag +1 e -1 ANODE Let’s show the half reactions, and WEIRD Net Ionic Equation now…

AgNO 3(AQ) Ag (S) battery Cu (S) CATHODE Ag +1 e -1 ANODE ½OX: Ag ⁰ → Ag e - ½RED: Ag e - → Ag ⁰ NET: Ag ⁰ + Ag +1 → Ag ⁰ + Ag +1

69. In the next slide we will attempt to plate gold onto an aluminum spoon. Try drawing it from scratch in the blank space. Aluminum is more reactive (higher on table J) than gold, so Al should go into solution, gold cations should become gold atoms and precipitate. To avoid this, we’ll connect an outside electrical source to make the redox we want to happen, not the spontaneous redox that would otherwise occur. Then, we’ll work out the half reactions. FINALLY: which is the anode and cathode?? Is it the spoon, or the gold bar? Hint, Leo is STILL a red cat.

Au (S) battery AuCl (AQ) Aluminum spoon 69 Let’s try to label this up, electrons from the battery, to the spoon, what happens in the solution, what happens on the gold bar, etc. e -1

Au (S) battery AuCl (AQ) Aluminum spoon Electrons from the battery land on the spoon, which causes the reduction of the Au +1 cations, plating the spoon gold. The gold bar oxidizes gold atoms into cations, and the free electrons from this oxidation replace the electrons from the battery. ½ ox: Au° (S) Au +1 (AQ) + 1e -1 ½ red: Au +1 (AQ) + 1e -1 Au° (S) NET: Au°+ Au +1 Au° + Au Au +1

Au (S) battery AuCl (AQ) Au +1 Electrons are “lost” at the anode, so the gold bar is the anode. RED CAT… Reduction happens at the cathode, so the spoon is the cathode. ANODE

Redox Class #5 The electrolysis of water is redox, and showing this using a Hofmann Apparatus 73. Define electrolysis:

Fill with water here Add a bit of H 2 SO 4(AQ) to allow water to conduct electricity. 2 tubes filled with water, each with a spigot on top. 2 platinum electrodes, do not corrode, conduct electricity very well. Attach to power supply here. 74. The electrodes must be connected to a Direct Current power supply (like a strong battery). The electricity forces a non spontaneous redox reaction.

H 2 O (L) decomposes into H 2(G) + O 2(G) 75.The decomposition of water: 2H 2 O (L) 2H 2(G) + O 2(G) Rewrite this with oxidation numbers showing

H 2 O (L) decomposes into H 2(G) + O 2(G) 2H 2 O (L) 2H 2(G) + O 2(G) Rewrite this with oxidation numbers showing 2H 2 O 2H 2 ° + O 2 ° +1 -2

2H 2 O (L) 2H 2(G) + O 2(G) 2H 2 O 2H 2 ° + O 2 ° Write out the half reactions for oxidation and reduction now. ½ Oxidation: ________________________________________________________ ½ Reduction: ________________________________________________________

Electricity supply -+ Electrons go in here, which changes the 4H +1 into 2H 2 ° This is gaining electrons, or reduction. Electrons out here, where the 2O -2 become O 2 ° This is losing electrons, or oxidation. 2H 2 O 2H 2 + O ° °

The decomposition of water is called electrolysis (electricity breaking apart) To figure out how to do this without polluting the environment would put pure oxygen into our air, and allow you to bottle the hydrogen up to burn (synthesize) with oxygen in a car, and make the car go. The waste product from this car would be water. How cool is that?

77. Name the type of electrochemical cell that can spontaneously produce electricity from a chemical reaction. 78. Name the type of cell where electricity forces a redox reaction that would not be spontaneous. 79. What always happens at the anode? 80. What always happens at the cathode? 81. Does it matter what electrochemical cell you have to determine what’s happening at the anode + cathode?