1. Electrochemistry Lesson 5 Predicting Spontaneous Reactions Balancing Half Reactions

2. Predicting Spontaneous Reactions Using the Standard Reduction Chart Does Au 3+ react with Cl - ? If it does write the spontaneous reaction.

3. The top reaction is written forward- reduction The bottom reaction is written in reverse- oxidation (Au 3+ + 3e -  Au (s) ) (2Cl -  Cl 2(g) +2e - ) 2 3 2Au 3+ + 6Cl -  3Cl 2(g) + 2Au (s)

4. Does Ag + react with Br - ?

5. nonspontaneous

6. Some ions are on both sides of the table and are oxidizing or reducing agents. For example: Fe  Fe 2+  Fe 3+ Sn  Sn 2+  Sn 4+ Cu  Cu +  Cu 2+

7. Does Sn 2+ react with Cr?

8. spontaneous Check both (Sn 2+ + 2e -  Sn (s) ) (Cr (s)  Cr 3+ + 3e - ) 3 2 2Cr (s) + 3Sn 2+  2Cr 3+ + 3Sn (s)

9. Can you keep HCl in a Cu container? Explain!

10. nonspontaneous

11. Can you keep HCl in a Cu container? Explain! Yes, nonspontaneous. H + is a weaker oxidizing agent than Cu 2+ and Cu +.

12. Can you keep HCl in a Zn container? Explain!

13. Spontaneous- no H + is a stronger oxidizing agent than Zn 2+

14. Write the spontaneous reaction. 2H + + 2e -  H 2(g) Zn (s)  Zn 2+ + 2e - 2H + + Zn (s)  H 2(g) +Zn 2+

15. Can you keep HNO 3 in a Au container? Explain!

17. Yes, nonspontaneous. HNO 3 is a weaker oxidizing agent than Au 3+.

18. Can you keep HNO 3 in a Cu container? Explain!

19. Take the strongest ox agent- higher Take the strongest red agent- lower

20. Can you keep HNO 3 in a Cu container? No, spontaneous. HNO 3 is a stronger oxidizing agent than Cu 2+. Write the spontaneous reaction. NO 3 - + 4H + + 3e -  NO (g) + 2H 2 Ox2 Cu (s)  Cu 2+ + 2e - x3 2NO 3 - + 8H + + 3Cu (s)  2NO (g) + 4H 2 O + 3Cu 2+

21. nonspontaneous- both on same side Does Au 3+ react with Fe 3+ ?

22. spontaneous- downhill Does Fe 3+ react with Pb?

23. nonspontaneous- uphill Does Cl 2 react with F - ?

24. spontaneousdownhill Does S 2 O 8 2- react with Pb?

25. nonspontaneous- uphill Does Cu 2+ react with Br - ?

26. spontaneous- downhill Does Br 2 react with Sn 2+ ?

27. nonspontaneous- uphill Does Pb 2+ react with Fe 2+ ?

28. spontaneous- downhill Can you keep HCl in a Fe or a Ag container? HCl → H + + Cl - 2H + + 2e - → H 2(g) Fe → Fe 2+ + 2e - 2H + + Fe → H 2(g) + Fe 2+ Yes No

29. Can you keep HNO 3 in a Au or a Ag container? NO 3 - + 4H + + 3e - → NO (g) + 2H 2 O Ag → Ag + + e - NO 3 - + 4H + + 3Ag → NO (g) + 2H 2 O + 3Ag + Yes No x 3

30. Core Concepts What is reduced and oxidized? What gains or loses electrons? What are the oxidizing and reducing agents? +3 +5 +5 +3 2H + + As 2 O 3 + 2NO 3 - + 2H 2 O  2H 3 AsO 4 + N 2 O 3 oxidizedreduced loses e'sgains e's red agentox agent

31. Core Concepts Is the following a redox reaction? 1 6 -2 1 -2 1 1 6 -2 1 -2 H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O Not Redox- no change in Oxidation number

32. PdCl 2 reacts with Mn but not Br -. Pd 2+ Mn Br - Br 2 Mn 2+ Pd Br 2 Pd 2+ Mn 2+ Rank the oxidizing agents in decreasing strength. Rank the reducing agents in decreasing strength. MnPdBr -

33. Balancing Redox Half Reactions in Acid Solution Cr 2 O 7 2- →Cr 3+ 1. Balance the central atom 2.Balance O by adding H 2 O 3.Balance H by adding H + 4.Balance the charge by adding electrons 5.Check charge 2+ 7H 2 O+ 14H + +12 +6 6e - + most positive side

34. Balancing Redox Half Reactions in Acid Solution HV 2 O 5 3- →H 2 VO 3 + 1. Balance the central atom 2.Balance O by adding H 2 O 3.Balance H by adding H + 4.Balance the charge by adding electrons 5.Check charge 2H 2 O ++ H + -2+2 + 4e - -2

35. Balancing Redox Half Reactions in Acid Solution HBr 2 O 7 2- → H 2 BrO 2 - 2+ 3H 2 O+ 9H + 9e - + +7-2

36. Balancing Redox Half Reactions in Acid Solution 9e - + HBr 2 O 7 2- + 9H + → 2H 2 BrO 2 - + 3H 2 O To balance in base solution………. 1.Balance in acid 2.Add OH - to each side to neutralize H + 3.Simplify 9H2O9H2O 9OH - 6

37. Balancing Redox Half Reactions in Acid Solution 9e - + HBr 2 O 7 2- + → 2H 2 BrO 2 - +-11 6H2O6H2O9OH -

38. Balance in alkaline solution. Au+ ClO - →HAuCl 4 2- 4+ 4H 2 O9H + + +5-2 7e - + -2

39. Balance in alkaline solution. + 7e - + Au+ 4 ClO - →HAuCl 4 2- + 4H 2 O 9H 2 O 9OH - 9H +

40. Balance in alkaline solution. 5H 2 O + 7e - + Au + 4ClO - →HAuCl 4 2- + 9OH - -11