Chapter 20: Electrochemistry

1. Chapter 20: Electrochemistry

2. 20.1Oxidation Reactions • Oxidation • A. Increase oxidation number • B. Loses electrons • C. Is a reducing agent (a reductant) • II. Reduction • A. Decrease oxidation number • B. Gains electrons • C. Is an oxidizing agent (an oxidant)

3. Sample 20.1 Cd + NiO2 + 2 H2O  Cd(OH)2 + Ni(OH)2 Oxidation Reduction Oxidation Agent Reduction Agent Gains Electrons Loses Electrons Spectator Ions NiO2 Cd Ni(2) Cd(2)

4. III. Oxidation Number Exceptions: H2O2 NH3 Hg22+ (Mercury I)

5. 20.2Balancing Red-ox Reactions • I. ½ Reaction • Shows oxidation or reduction, NOT BOTH! • II. Balancing ½ reactions • A. Acid… use H+, H2O, e- • B. Base… use OH-, H2O, e-

6. Extremely painful to type example 20.2 MnO4- + C2O4-2 Mn2+ + CO2 Ox: (C2O4-2  2CO2 + 2e-)5 Red: (5e- + 8H+ + MnO4-1  Mn+2 + 4H2O)2 16H+ + 2MnO4- + 5C2O42- 10CO2 + 2Mn2+ + 8H2O 1.Assign Oxidation # 2. What’s oxidation, reduction 3. Balance ½ Reaction 4. Cancel electrons 5. Add

7. ACID • Add… • H+ • H2O • and e- • BASIC • Add… • OH- • H2O • and e-

8. Ex: Acid: Fe2+Fe3+ SO42-SO32- NO2-NO3- SO2SO42- Cl2ClO4- O2H2O Base: Cr2O72-Cr3+ Bi(OH)3Bi MnO4-MnO2

9. Sample 20.3: WHY AGAIN????? CN- + MnO4- CNO- + MnO2 Ox: 3(2OH- + CN-  CNO- + H2O +2e-) Red:2(3e- +2H2O+ MnO4  MnO2 +4OH-) H2O + 2MnO4- + 3CN-  3CNO-+ 2MnO2+ 2OH-

10. 20.3 Cel EMF • Potential Difference– Difference in energy per elect charge; measured in VOLTS! • 1 VOLT= Energy=J = Joule • Charge C Coulomb • II. EMF: Electromotive Force • Ecell = Cell Potential • = Cell Voltage • Change in E = Positive for spontaneous • ∆E = +

11. III. Standard EMF= Standard cell potential = Ecell 1M + 25oC + 1atm Table 20.1 page 1117 IV. Standard Reduction Potentials Eocell = Eo(cathode) – Eo(anode) Eo = Eo(red) – Eo(ox) V. Oxidizing and Reducing Agents Figure 20.14 F2– Strongest Oxidizing Agent Li- Strongest Reducing Agent

12. p. 857

13. p. 857 p. 1117 in back of book.

14. p. 861

15. Ox: Zn  Zn2+ + 2e- -0.76V • Red: Cu2+ + 2e-  Cu 0.34V • Overall Zn + Cu2+  Zn2+ + Cu • 1.10V

16. Galvanic cell or Voltaic cell (p.852)

17. Auto-Car • Anode-oxidation loses e- • Cathode-reduction gains e-

18. Count Alessandro Volta Luigi Galvani

19. Faraday’s Law Cu  Cu+2 + 2e- = 2 Faradays 1 Faraday= number of electrons needed to balance half of a reaction.

20. What is a Faraday? F • a.) number of electrons needed to balance half of a reaction. • b.) = 96500 C • c.) the charge on a mole of e- • d.) that day in September we get off school. • Perhaps that joke was a little too hippie-dippy for this crowd.

21. The charge on an e- is -1.60217657 x 10-19 C. • This number times a mole of e- = … • 6.02214129 x 1023 e- / mol • 96485 C = 96500 C • Faraday is also credited to have coined the term electrolyte.

22. Faraday’s Law 1.) What mass of copper will be deposited by a current of 7.89 Amps flowing for 1200. seconds?

23. Faraday’s Law C= Amp x seconds 7.89 A x 1200 s = 9468 C 9468 C x 1 mol e-x1 mol Cux63.5 g Cu= 96500 C 2 moles e- 1 mol Cu 3.11 g F

24. Faraday’s Law 2.) 14 H+ + 6 Fe+2 + Cr2O7-2 6 Fe+3 + 2 Cr+3 + 21 H2O

25. Faraday’s Law 1.64g Cr3+ is plated out in 3.556 hours. How many Amps are needed?

26. Faraday’s Law 1.64 g Cr+3 x 1 mol Cr+3 x 3 F x 96500 C = 52.0 g Cr+3 1 mol 1 F 9130 C 3.556 hr x 3600 s = 12802 s 1 hr Amps=C=9130 C= 0.713 A s 12802 s

27. Cell potential and Gibb’s Free NRG • ∆Go = - nFEo • J = -(mol)(C/mol)(V) • Volt = J/C • n = moles of e- transferred.

28. Nernst Equation

29. Where n is the number of electrons transferred in the RED-OX equation. • Q like K = Products/Reactants

31. Electrolysis of water. • anode+ - cathode ∆G always positive

32. reaction • Reduction at cathode: 2 H+(aq) + 2e− → H2(g) • doubled 4 H+(aq) + 4e− → 2H2(g) • Anode (oxidation): • 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− • 2 H2O(l) → O2(g) + 2H2(g)

33. Electrolysis of molten sodium chloride (Down's cell.) • Anode (oxidation): 2 Cl– → Cl2(g) + 2 e– • Cathode (reduction): 2 Na+(l) + 2 e– → 2 Na(l) • Overall reaction: 2Na+ + 2Cl–(l) → 2Na(l) + Cl2(g)

34. The emf for this process is approximately −4 V indicating a (very) non-spontaneous process. In order for this reaction to occur the power supply should provide at least a potential of 4 V. However, larger voltages must be used for this reaction to occur at a high rate.

35. Electrolysis of a solution of sodium chloride • 1.) Cathode: Na+(aq) + e– → Na(s) E°red = –2.71 V • 2.) Anode: 2Cl–(aq) → Cl2(g) + 2 e–E°red = +1.36 V • 3.) Cathode:2H2O(l) + 2 e– → H2(g) + 2OH–(aq)     E°red = –0.83 V • 4.) Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e–E°red = +1.23 V

36. The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.

37. Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process. • When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.

38. Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage).[27]

39. Finally, reaction 3 is favorable because it describes the proliferation of OH- ions thus letting a probable reduction of H+ ions less favorable an option. • The overall reaction for the process according to the analysis would be the following:[27]

40. Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e– • Cathode (reduction):2H2O(l) + 2 e– → H2(g) + 2 OH–(aq) • Overall reaction: • 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq) • OR • 2NaCl + 2 H2O → H2(g) + Cl2(g) + 2NaOH

41. As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.

42. Disproportionation of Cl • Bleach equation • NaOH + Cl2 NaClO + H2O + NaCl

43. Electrolytic cells / voltaic cells

44. Exp 44

45. Exp 44 • It's more pinkish in person. It looks exactly like the lab though. If they go past the end point, it looks more like apple cider vinegar.  • Descriptive chemistry win.  • Jessy  • “When you want to succeed as bad as you want to breathe, then you will be successful.” • –Eric Thomas 

46. F