1. Chapter 17 Oxidation and Reduction Redox Oxidation Numbers Balancing by Redox Quiz Corrections Electrolysis Lab A-Electrolysis Electrolysis Lab Results Concept Check Assignment Quiz-Balancing/Electrolysis Lab C Faraday’s Lab Electrolysis Quiz Correction Lab B Voltaic Cell Pages 344-345

2. Notes One Unit Twelve Redox Oxidation Numbers Balancing by Redox

3. Redox Burning: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O+ Heat Rusting Iron: 4Fe + 3O 2  2Fe 2 O 3 + Heat Oxidation - Loss of e -1. Na(s)  Na +1 +1e -1 Reduction - Gain of e -1. Cl 2 + 2e -1  2Cl -1 Number line (Oxidation…Left or right?)

4. Demonstration combustible blown through a burner.

5. Key Elements (99%) H +1 H -1 (99%)O -2 O -1 (Always) Li +1, Na +1, K +1, Rb +1, Cs +1, Fr +1 (Always) Be +2, Mg +2, Ca +2, Ba +2, Sr +2, Ra +2 (Always) Al +3 (with only a metal) F -1, Cl -1, Br -1, I -1 (NO 3 -1 ) ion is always +5 (SO 4 -2 ) ion is always +6

6. Finding Oxidation Numbers +1-2 H2OH2O 2(+1)+1(-2)=Zero The sum of the oxidation numbers must be equal to _____ for a compound. Find Ox#’s for H 2 O? zero 2(H)+1(O)=Zero +1 -2 H 3 PO 4 Find Ox#’s for H 3 PO 4 ? 3(H)+4(O)=Zero +5 1(P)+ 3(+1)+4(-2)=Zero1(+5)+

7. Finding Oxidation #’s for Compounds +1-2 +1+5-2 H 3 PO 4 H2OH2O HNO 3 +1+5-2 H 2 SO 4 +1-2 +6 Hg 2 SO 4 +6+1-2 Na 2 Cr 2 O 7 +1 +6 -2 H 2 CO 3 +1 -2 +4 (NH 4 ) 2 CO 3 -3+1+4-2 Ca 3 (AsO 4 ) 2 +2+5 -2 Fe 2 (SO 4 ) 3 +6 +3 -2 Ba(ClO 4 ) 2 +2 +7 -2 Al 2 (CO 3 ) 3 +3+4 -2

8. Balancing By Redox Example One H 2 O + P 4 + H 2 SO 4  H 3 PO 4 + H 2 S #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1-20+1-2+6+1 -2 +5+1-2 5e -1 lost 8e -1 Gained X8 X5 285512 Multiply by 1. 12H 2 O +2P 4 +5H 2 SO 4  8H 3 PO 4 +5H 2 S

9. Balancing By Redox Example Two K 3 PO 4 + Cl 2  P 4 + K 2 O+ KClO 2 #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1-2 0+1-2+5+1-2 +3 0 5e -1 Gained 3e -1 Lost X3 X5 5/223/453 Multiply by 4. 12K 3 PO 4 +10Cl 2  3P 4 +8K 2 O+20KClO 2

10. Notes Two Unit Twelve Quiz Corrections Electrolysis Lab A-Electrolysis

11. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Reducing Agent Stronger Reducing Agent Loses e - Oxidizing Agent Weaker Oxidizing Agent Gains e - StrongerWeaker

12. Electrolysis  anode Cathode  e -1  Electrolysis Cathode Anode Electron flow? Mass Gain=? Which is the… Cathode=? Anode=? Cathode(spoon) Mass Loss=? Copper -reduction -oxidation -electric current produced chemical reaction

13. Electrolysis Lab- Demo KI (aq) What is available to react? K +1 I -1 H2OH2O Anode Reaction lowest reaction on right. Cathode Reaction highest reaction on left.

14. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 K +1 I -1 H2OH2O K + +e -  K(s) -2.92 An: lowest on right. Cath: highest on left. 2I -  I 2 (s)+2e - We see brown:I 2 (s) We see pink. We saw bubbles KI(aq)  

15. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Na + +e -  Na(s) -2.71 An: lowest(right) Cath: highest(left) H 2 O  1/2O 2 (g)+2H + (pH=7)+2e - We saw bubbles! We saw copper on the pencil tip! Cu +2 SO 4 -2 H2OH2O CuSO 4 (aq)  

16. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Na + +e -  Na(s) -2.71 An: lowest(right) Cath: highest(left) H 2 O  1/2O 2 (g)+2H + (pH=7)+2e - We saw bubbles. We saw pink. We saw bubbles Na +1 SO 4 -2 H2OH2O Na 2 SO 4 (aq)  

17. Electrolysis lab A

18. Notes Three Unit Twelve Electrolysis Lab Results Concept Check Assignment Quiz-Balancing/Electrolysis

19. Notes Four Unit Twelve Faraday’s Law Lab B This as a chemical process that uses electricity to produce industrial quantities of specific chemicals.

20. Application of Faraday’s law F = 96500 C/mol e - ) A x s = C

21. Faraday’s Law: Lab B Fe(s) NO 3 -1 e -1 Anode? Cathode? Lose Mass? Gain Mass? Fe +3 3e -1 Fe +3 NO 3 -1 F = 96500 C/mole - Amp x second = C

22. Faraday’s Law Calculation One 3.0 amp x Au +3 +3e -1  Au(s) 60 min 1 hour 1.5 Hour x 60 Sec 1 min =16000C 16000C x 1mole e -1 96500C = 0.17 mol e -1 0.17 mol e -1 x 1 m Au(s) 3mol e -1 =0.056mol Au(s) 0.056mol Au(s)x 197.0gAu 1mol Au = 11g Au 1. Balanced Equation 2. Calculate Coulombs. 3. Calculate moles e -1. 4. Calculate moles of substance. 5. Calculate grams. How many grams of Gold will be plated, using a current of 3.0 amps for 1.5 hours? x

23. Faraday’s Law Calculation Two 2.0 amp x Ag +1 +1e -1  Ag(s) 45 Hour x 60 Sec 1 min =5400C 5400C x 1mole e -1 96500C = 0.056 mol e -1 0.056 mol e -1 x 1 m Ag(s) 1mol e -1 =0.056mol Au(s) 0.05596mol Au(s)x 107.9gAg 1mol Ag = 6.0 g Ag 1. Balanced Equation 2. Calculate Coulombs. 3. Calculate moles e -1. 4. Calculate moles of substance. 5. Calculate grams. How many grams of Silver will be plated, using a current of 2.0 amps for 45 minutes?

24. Salt Bridge Cu +2 SO 4 -2 Cr +3 SO 4 -2 Cr +3 SO 4 -2 Cathode Anode Cu +2 + 2e -1  Cu Cr  Cr +3 +3e -1 Overall rxn: 2Cr+ x3 x2 3 6 3 2 6 2 3Cu +2  2Cr +3 + 3Cu +0.34 +0.74 emf=1.08volts 1.08 Cr +3 3e -1 SO 4 -2 Na +1 Cu(s)Cr(s) Salt Bridge e -1 ? rxn:

25. Cathode Verses Anode

26. Formation of Rust Fe(OH) 2 (s) Fe +2 H2OH2O e -1 Fe +2 e -1 H2OH2O O2O2 OH -1 Fe(OH) 2 (s) Fe +2 H2OH2O e -1 Fe +2 e -1 H2OH2O O2O2 OH -1 Fe(OH) 2 (s) O2O2 H2OH2O H2OH2O Fe(OH) 3 (s)

27. Ion-Electron Method for Balancing

28. UO 2 +2 + I 2  U +4 + IO 3 -1 (Acid) UO 2 +2 I 2  U +4 + H 2 O #2. Bal Non-O Elem. #3. + H 2 O. #4. + H +1. #5. + e -1 to bal +/-. 2 + H +1 4+ e -1 2 #1. Separate Half-rxn.  IO 3 -1 2 + H 2 O 6 + H +1 12 11 1 + e -1 10 5X 1X UO 2 +2 I 2  U +4 + H 2 O10 + H +1 20+ e -1 10  IO 3 -1 2 + H 2 O 6 + H +1 12 5 5 1 + e -1 10 84 UO 2 +2 5 + U +4 5 + H 2 O4+ H +1 8 + I 2 1  IO 3 -1 2

29. Ion-Electron Method for Balancing IO 3 -1 + Ti +3  I 2 + TiO 2 +1 (Acid) IO 3 -1 Ti +3  I 2 + H 2 O #2. Bal Non-O Elem. #3. + H 2 O. #4. + H +1. #5. + e -1 to bal +/-. 6 + H +1 12+ e -1 10 #1. Separate Half-rxn.  TiO 2 +1 1 + H 2 O 2 + H +1 4 1 2 1 + e -1 2 1X 5X IO 3 -1 Ti +3  I 2 + H 2 O6 + H +1 12+ e -1 10  TiO 2 +1 5 + H 2 O 10 + H +1 20 1 2 5 + e -1 10 8 4 IO 3 -1 2 + I 2 1 + H +1 8+ H 2 O4 + Ti +3 5  TiO 2 +1 5

30. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Cl 2 (g)+2e -  2Cl - +1.36 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 K + +e -  K(s) -2.92 An: lowest(right) Cath: highest(left) H 2 O  1/2O 2 (g)+2H + (pH=7)+2e - We would see bubbles and no pink. We would see solid metal Ni +2 Cl -1 H2OH2O NiCl 2 (aq)  

31. Balancing By Redox Example Four Cs 3 AsO 4 + Cl 2  As+ Cs 2 O+ CsClO 2 #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1-2 0+1-2+5+1-2 +3 0 5e -1 Gained 3e -1 Lost X3 X5 5/2 2 353 Multiply by 2. 6Cs 3 AsO 4 + 5Cl 2  6As+4Cs 2 O+10CsClO 2