Electrochemistry

1. Electrochemistry Chapter 19

2. 산화환원 반응과 전기화학 • 산화환원 반응과 산화수 • 갈바니 전지 • 표준환원전위 • 산화환원 반응의 자발성. 전위차와 자유에너지 • 전위차와 농도, Nernst 방정식 • 전지 • 부식 • 전기분해 • 전기야금

3. 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- • 전기화학 반응 • oxidation-reduction reactions in which: • 자유에너지의 감소 => electrical energy : 갈바니 전지 • electrical energy => 자유에너지의 증가 : 전해 전지 0 0 2+ 2- Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-)

4. 산화수(Oxidation number) The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. In H2O2 and O22- it is –1.

5. O = -2 H = +1 Oxidation numbers of all the atoms in HCO3- ? 3x(-2) + 1 + ? = -1 C = +4 • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3-

6. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ 산화-환원 반응식의 균형 맞추기 산성 용액에서 Fe2+가 Cr2O72-에 의해 Fe3+로 산화되는 반응 • 균형되지 않은 반응식을 이온형으로 쓴다. • 반응식을 두 개의 반쪽반응으로 나눈다. Oxidation: Reduction: • 각 반쪽반응에 대하여 O 와 H 를 제한 나머지의 계수를 맞춘다.

7. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O 산화-환원 반응식의 균형 맞추기 • 산성 용액의 반응에서, O 원자의 균형을 맞추기 위해 H2O을 첨가하고, H 원자의 균형을 맞추기 위해 H+ 를 첨가한다. • 전자를 첨가하여 각 반쪽 반응의 전하 균형을 맞춘다. • 필요하면 각각의 반쪽반응에 적절한 계수를 곱하여 두 반쪽 반응에서의 전자의 수를 같도록 한다.

8. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 산화-환원 반응식의 균형 맞추기 • 두 반쪽반응을 더하여 최종적으로 균형을 맞춘다. 양변의 전자수가 상쇄되어야 한다. Oxidation: Reduction: • 균형을 맞춘 반응식의 각 변의 원자와 전자의 수가 같은지 확인한다. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 • 염기성 용액에 대하여는, 최종 식에 나오는 H+ 마다 OH-를 양변에 모두 추가한다..

9. 산화-환원 반응식의 균형 맞추기-예제 구리를 질산에 넣으면 산화질소(NO)가발생한다. Cu(s) + HNO3(aq)  Cu2+(aq) + NO(g) • 반응식을 두 개의 반쪽반응으로 나눈다. 산화: Cu Cu2+ 환원: NO3– NO • 각 반쪽반응에 대하여 O 와 H 를 제한 나머지의 계수를 맞춘다. • 산성 용액의 반응에서, O 원자의 균형을 맞추기 위해 H2O을 첨가하고, H 원자의 균형을 맞추기 위해 H+ 를 첨가한다. 환원: 4H+ + NO3– NO + 2H2O • 전자를 첨가하여 각 반쪽 반응의 전하 균형을 맞춘다. 산화: Cu Cu2+ + 2e– 환원: 4H+ + NO3– + 3e– NO + 2H2O

10. 산화-환원 반응식의 균형 맞추기-예제 • 필요하면 각각의 반쪽반응에 적절한 계수를 곱하여 두 반쪽 반응에서의 전자의 수를 같도록 한다. 산화: 3Cu 3Cu2+ + 6e– 환원: 8H+ + 2NO3– + 6e– 2NO + 4H2O • 두 반쪽반응을 더하여 최종적으로 균형을 맞춘다. 양변의 전자수가 상쇄되어야 한다. 3Cu + 8H+ + 2NO3– 3Cu2+ + 2NO + 4H2O or 3Cu + 6H+ + 2HNO3 3Cu2+ + 2NO + 4H2O or 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

11. 산화-환원 반응식의 균형 맞추기-예제 1) H2O2 + Fe2+ → Fe3+ + H2O (acidic) 2) Cu + HNO3 → Cu2+ + NO + H2O (acidic) 3) CN- + MnO4-→ CNO- + MnO2 (basic, 힌트 C의 산화수는 양쪽에서 모두 +2) 4) Br2 → BrO3- + Br- (basic,힌트 Br2이 산화와 환원이 함께 일어난다) 5) Cr(OH)3(s) + ClO3−(aq) → CrO42−(aq) + Cl−(aq) (basic)

12. 갈바니 전지(Galvanic Cells) - + anode oxidation cathode reduction spontaneous redox reaction

13. Zn (s) + Cu2+(aq) Cu (s) + Zn2+(aq) 갈바니 전지(Galvanic Cells) • The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential Cell Diagram [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode 산화 환원

14. 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) Zn2+ (1 M) + 2e- Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) 표준전극전위(Standard Electrode Potentials) + - Anode: 산화극 Cathode: 환원극 Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Cathode (reduction):

15. 2e- + 2H+ (1 M) H2 (1 atm) 표준전극전위(Standard Electrode Potentials) Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction E0= 0 V Standard hydrogen electrode (SHE)

16. E0 = 0.76 V E0 = Ecathode - Eanode E0 = EH /H- EZn /Zn cell cell cell Standard emf (E0 ) cell 0 0 0 0 2+ + 2 0.76 V = 0 - EZn /Zn 0 2+ EZn /Zn = -0.76 V 0 2+ Zn2+ (1 M) + 2e- ZnE0 = -0.76 V 표준전극전위(Standard Electrode Potentials) Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

17. 0 0 0 Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 0 2+ 0 ECu /Cu = 0.34 V 2+ E0 = 0.34 V E0 = Ecathode - Eanode cell cell 0 0 H2 (1 atm) 2H+ (1 M) + 2e- 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M) 표준전극전위(Standard Electrode Potentials) 표준환원전극전위 Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): Cathode (reduction):

18. Standard Reduction Potentials at 25oC* Easier to oxidize. Stronger reducer Easier to reduce. Stronger oxidizer

19. E0 = Ecathode - Eanode cell 0 0 갈바니 전지에 대한 전위 계산 1. 2. 열역학 방법,using DG = -nFE

20. Cd2+(aq) + 2e- Cd (s)E0 = -0.40 V Cr3+(aq) + 3e- Cr (s)E0 = -0.74 V x 2 Cr (s) Cr3+ (1 M) + 3e- E0 = -0.40 – (-0.74) E0 = Ecathode - Eanode E0 = 0.34 V x 3 cell cell cell 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) 0 0 2e- + Cd2+ (1 M) Cd (s) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd is the stronger oxidizer Cd will oxidize Cr Anode (oxidation): Cathode (reduction):

21. 2Cr (s) + 3Cd2+ (1 M) → 3Cd (s) + 2Cr3+ (1 M) anotherapproach DG10 = -nF Eh0 = -6F(-0.40) V 3Cd2+(aq) + 6e- → 3Cd (s) DG20 = -nF Eh0 = -6F(-0.74) V 2Cr3+(aq) + 6e- →2Cr (s) DG0 = DG10 - DG20 = -6F(-0.40+0.74) V =-6F E0 cell E0 = 0.34 V cell

22. Cu2+(aq) + 2e- Cu (s)E0 = 0.340 V Cu+(aq) + e- Cu (s)E0 = 0.522 V Cu (s) Cu+ + e- E0 = Ecathode - Eanode E0 = 0.340– (0.522) = -0.184 V ? cell hcell Cu2+ + e- Cu+ 0 0 2e- + Cu2+ Cu (s) Cu2+(aq) + e- → Cu+(aq) E0 = ? Anode (oxidation): Cathode (reduction):

23. Cu2+(aq) + 2e- Cu (s) Cu+(aq) + e- Cu (s) Cu2+(aq) + e- → Cu+(aq) Eh0 = ? DG10 = -nF Eh0 = -2F(0.340) V DG20 = -nF Eh0 = -F(0.522) V DG0 = DG10 - DG20 = -F(2(0.340)-0.522) V =-F Eh0 Eh0 = 0.158 V 0.522 V 0.158 V Cu2+ → Cu+(aq) → Cu 0.340 V

24. Fe2+(aq) + 2e- Fe (s) Fe3+(aq) + e- Fe2+(aq) Fe3+(aq) + 3e- → Fe (s) Eh0 = ? Eh0 = -0.440 V Eh0 = +0.771 V Eh0 = ? -0.440 V +0.771 V Fe3+ → Fe2+(aq) → Fe ?

25. DG0 = -nFEcell 0 0 0 0 0 = -nFEcell Ecell Ecell Ecell F = 96,485 J RT V • mol ln K nF (8.314 J/K•mol)(298 K) ln K = 0.0257 V 0.0592 V log K ln K n (96,485 J/V•mol) = n n = = 산화환원 반응의 열역학 DG = -nFEcell n = number of moles of electrons in reaction = 96,485C/mol DG0 = -RT ln K

26. 산화환원 반응의 열역학

27. 0 2+ 0 + 2Ag 2Ag+ + 2e- 0 Ecell n = 2 2e- + Fe2+ Fe 0 0 2+ + E0 = -0.44 – (0.80) EFe /Fe= -0.44 V EAg /Ag = 0.80 V E0 = EFe /Fe – EAg /Ag What is the equilibrium constant for the following reaction at 250C? Fe2+(aq) + 2Ag (s) Fe (s) + 2Ag+(aq) E0 = -1.24 V E0 cell x n = exp K = 0.0257 V 0.0257 V -1.24 V x2 0.0257 V ln K K = 1.23 x 10-42 exp n = Oxidation: Reduction:

28. DG0 = -nFE 0 RT nF E = E0 - ln Q 0 0 E = E = E E 0.0257 V 0.0592 V log Q ln Q n n - - 전지의 전위에 미치는 농도의 효과 DG = DG0 + RT ln Q DG = -nFE -nFE = -nFE0+ RT ln Q Nernst equation At 298

29. Measurement of pH: the glass electrode The emf of the cell made up of the glass electrode and the reference electrode is measured with a voltmeter that is calibrated in pH units. Very thin glass membrane that is permeable to H+ ions. A potential difference develops between the two sides of the membrane.

30. Glass pH Electrodes + - a sensing part of electrode, a bulb made from a specific glass sometimes the electrode contains a small amount of AgCl precipitate inside the glass electrode internal solution, usually 0.1MHCl for pH electrodes internal electrode, usually silver chloride electrode or calomel electrode body of electrode, made from non-conductive glass or plastics. reference electrode, usually the same type as 4 junction with studied solution, usually made from ceramics or capillary with asbestos or quartz fiber. Ag/AgCl reference electrode Glass pH electrode

31. Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+(aq) + Cd (s) Fe (s) + Cd2+(aq) Cd Cd2+ + 2e- n = 2 2e- + Fe2+ 2Fe 0 0 2+ 2+ E0 = -0.44 – (-0.40) E0 = EFe /Fe – ECd /Cd E = -0.04 V E0 = -0.04 V 0.010 E = 0.013 0.60 0 E = E 0.0257 V 0.0257 V ln Q ln 2 n E > 0 Spontaneous - - 0 EFe /Fe = -0.44 V ECd /Cd= -0.40 V 2+ 0 2+ Oxidation: Reduction:

32. Zn (s) Zn2+ (aq) + 2e- + 2NH4(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O (l) Zn (s) + 2NH4+ (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s) 전지 Dry cell Leclanché cell Anode: Cathode:

33. Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e- HgO (s) + H2O (l) + 2e- Hg (l) + 2OH-(aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) 전지 Mercury Battery Anode: Cathode:

34. Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4 PbO2(s) + 4H+(aq) + SO2-(aq) + 2e- PbSO4(s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4 전지 Lead storage battery Lead-acid battery Anode: Cathode:

35. 전지 Solid State Lithium Battery

36. 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e- O2(g) + 2H2O (l) + 4e- 4OH-(aq) 2H2 (g) + O2 (g) 2H2O (l) 연료전지(Fuel Cell ) converts chemical energy into electricity. In contrast to storage battery, fuel cell does not need to involve a reversible reaction since the reactant are supplied to the cell as needed from an external source. This technology has been used in the Gemini, Apollo and Space Shuttle program. Half reactions: E°Cell = 0.9 V Advantage: Clean, portable and product is water. Efficient (75%) contrast to 20-25% car, 35-40% from coal electrical plant Disadvantage: Needs continuous flow of reactant, Electrodes are short lived and expensive. Anode: Cathode:

37. Relative Energy Density of Some Common Secondary Cell Chemistries

38. Current Science: microbial fuel cell (MFC) technology From: Andrew Kato Marcus, arizonaState Univ.,

40. 부동화(Passivation) Al2O3 Al Fe2+ + 2e- Fe Eo = -0.44 V Sn2+ + 2e- Sn Eo = -0.14 V Zn2+ + 2e- Zn Eo = -0.76 V galvanized steel tin steel Zn Sn Fe Fe

41. 성수대교 붕괴 1994년 10월 21일 아침 7시 38분

42. 부식(Corrosion) Fe2O3·xH2O and (FeO(OH), Fe(OH)3). • Additional reactions involved: • Fe2+ + 2 H2O ⇌ Fe(OH)2 + 2 H+ • Fe3+ + 3 H2O ⇌ Fe(OH)3 + 3 H+ • Fe(OH)2⇌ FeO + H2O • Fe(OH)3⇌ FeO(OH) + H2O • 2 FeO(OH) ⇌ Fe2O3 + H2 • O2 + 4 e- + 2 H2O → 4 OH- E° = 1.23 V • Because it forms OH-, it is strongly affected by acid. • Fe → Fe2+ + 2 e− E° = 0.44 V • Fe2+ → Fe3+ + e− E° = -0.771 V

43. Conditions for Corrosion Conditions for Iron Oxidation: Iron will oxidize in acidic medium SO2g H2SO4g H+ + HSO4+ Anions improve conductivity for oxidation. Cl- from seawater or NaCl (snow melting) enhances rusting Conditions for Prevention: Iron will not rust in dry air; moisture must be present Iron will not rust in air-free water; oxygen must be present Iron rusts most rapidly in ionic solution and low pH (high H+) The loss of iron and deposit of rust occur at different placm on object Iron rust faster in contact with a less active metal (Cu) Iron rust slower in contact with a more active metal (Zn)

44. Corrosion Prevention

45. Impressed current cathodic protection system in seawater Sacrificial anode system in seawater from, Richard Baxter, Jim Britton, www.stoprust.com

46. Cathodic Protection of an Iron Storage Tank

47. 전기분해 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

48. Electrolysis of Water • 2H2O(l) → 2 H2(g) + O2(g)

49. Electrolysis of Water • 2H2O(l) → 2 H2(g) + O2(g) DG0 = 474.4 kJ/mol • Cathode: • 2 H+(aq) + 2 e- → H2(g) Eored = 0.00 V • Anode: • 2 H2O(l) → O2 + 4 H+(aq) + 4 e-Eoox = -1.23 V • 2 SO42-→ S2O82- + 2 e-Eoox = -2.05 V 39 kWh of electricity and 8.9 liters of water are required to produce 1 kg of hydrogen at 25°C and 1 atm. In reality, 50.3-70.1 kWh of electricity is required.

50. Electrolysis of NaCl(aq) • 2H2O(l) + 2 Cl-(aq) → H2(g) + Cl2(g) + 2 OH-(aq)