Redox reaction( 氧化還原反應) , Oxidation number and Electrolysis( 電解)

1. Redox reaction(氧化還原反應), Oxidation number and Electrolysis(電解) International Junior Science Olympiad (IJSO) Dr. Yu-San Cheung yscheung@cuhk.edu.hk Department of Chemistry The Chinese University of Hong Kong

2. Oxidation(氧化) and Reduction(還原) Oxidation: Reaction of an element or a compound with O2 to give an oxide e.g., 4 Na(s) + O2(g)  2 Na2O(s) (sodium oxide) 2 H2(g) + O2(g)  2 H2O(l) (“hydrogen oxide”?) Reduction: Reverse of oxidation e.g., 2 CuO(s)  2 Cu(s) + O2 (g)

3. Oxidation(氧化) and Reduction(還原) Reduction: Reverse of oxidation e.g., 2 CuO(s)  2 Cu(s) + O2 (g) CuO(s) can also be converted to Cu(s) with hydrogen: e.g., CuO(s) + H2(g)  Cu(s) + H2O(l) Therefore, we may also say that: Reduction is a reaction of an element or a compound with H2. e.g., H2(g) + Cl2(g)  2 HCl(g) [Cl2(g) is reduced] Here, we consider that H2 and O2 have opposite properties.

4. Oxidation (氧化) and Electron Transfer(電子轉移) In the following reaction, copper is oxidized and loses electrons to have a positive charge: e.g., 2 Cu(s) + O2(g)  2 CuO(s) Therefore, we may also say that: In an oxidation, an element or a compound loses electron(s) to have a positive charge. Similarly: in an reduction, an element or a compound gains electron(s) to have a positive charge. e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s) Which ion is oxidized? __________ Which ion is reduced? __________

5. Oxidation Number(氧化數) Consider: H+(aq) + OH–(aq)  H2O(l) Is OH–(aq) oxidized or reduced? (Gaining “hydrogen” but with charge increased) Oxidation Number/Oxidation State: Numbers assigned to elements, ions, and compounds to help us to tell whether they are oxidized or reduced in a reaction. It is also assigned to individual atoms in ions and molecules.

6. Rules of Assigning Oxidation Number (O.N.) From high to low priority: • O.N. for atoms in elements: 0 • Overall O.N. for neutral molecules and compounds: 0 e.g., H2(g), Na(s), NaCl(s), CO2(g) • Overall O.N. for ions: equal to the charges e.g., Na+ & NH4+(O.N.= +1), SO4 2–(O.N.= –2)

7. Rules of Assigning Oxidation Number (con’t) For individual atoms in neutral molecules and compounds: • F: –1 • Group 1A metals (Li, Na, K, …): +1 • Group 2A metals (Be, Mg, Ca, …): +2 • H: +1 (except in metal hydride, see below) • O: –2 • Cl: –1 • Br, I: –1; N, P: –3; S: –2

8. Examples: SO4 2–: overall: –2O.N. for O: –2  O.N. for S = +6 SO2: overall: 0 O.N. for O: –2  O.N. for S = +4 SF2: overall: 0 O.N. for F: –1  O.N. for S = +2 H2S: overall: 0 O.N. for H: +1 O.N. for S = –2 PCl5: overall: 0O.N. for Cl: –1  O.N. for P = +5 PO4 3–: overall: –3O.N. for O: –2  O.N. for P = +5 PO3 3–: overall: –3O.N. for O: –2  O.N. for P = +3 PH3: overall: 0 O.N. for H: +1 O.N. for S = –3 H2O: overall: 0 O.N. for O: –2  O.N. for H = +1 CaH2: containing Ca2+ and H–  O.N. for H = –1

9. Examples: ClO4–: overall: –1O.N. for O: –2  O.N. for Cl = +7 ClO3–: overall: –1O.N. for O: –2  O.N. for Cl = +5 ClO2–: overall: –1O.N. for O: –2  O.N. for Cl = +3 ClO–: overall: –1O.N. for O: –2  O.N. for Cl = +1 MnO4–: overall: –1O.N. for O: –2  O.N. for Mn = +7 Note: Mn has bonding with oxygen atoms. It does not exist as Mn7+ ion.

10. Oxidation Number(氧化數)for Oxygen O.N. for O is usually –2, except (including but not limited to): • In fluorine-oxygen compounds • e.g.,: OF2 • Neutral molecule: overall O.N. = 0 • Then assign O.N. = –1 for F • So, O.N. = +2 for O • How about FIO3? • O.N. = –1 for F, O.N. = –2 for O, finally we have • +7 for I • In some ions: • Peroxide: O2 2– O.N. for O atom: –1 • Superoxide: O2– O.N. for O atom: –1/2

11. Redox reaction(氧化還原反應) (Oxidation-Reduction) Oxidation of an atom: O.N. increases Reduction of an atom: O.N. decreases In a reaction, O.N. increase of an atom must be accompanied with O.N. decrease of another atom e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s) O.N. change: Fe: +2  +3 Ag: +1  0 We say that: Fe2+(aq) is oxidized to Fe3+(aq) Ag+(aq) is reduced to Ag(s)

12. O.N. and Electron Transfer(電子轉移) e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s) Ag+: gaining electron, being reduced Fe3+: losing electron, being oxidized In general: - gaining electron, being reduced - losing electron, being oxidized

13. Oxidizing Agent(氧化劑) and Reducing Agent(還原劑) Oxidizing agent (oxidant): • oxidizing another species • being reduced (O.N. decreased) Reducing agent (reductant) : • reducing another species • being oxidized (O.N. increased) e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s) Oxidizing Agent: Ag+(aq) • Reducing Agent: Fe2+(aq)

14. Common Oxidizing Agents (氧化劑) Ions of metal at low position in the reactivity series (e.g., Ag+ Ag) MnO4– & Cr2O7– in acidic medium: MnO4– Mn2+; Cr2O7– Cr3+ Conc. HNO3(aq): NO3– NO2(g) Conc. H2SO4(aq): SO42– SO2(g) Cl2(g) 2 Cl–;Br2(l) 2 Br– O2(g) O2–; H2O2(aq)  H2O(l)

15. Common Reducing Agents(還原劑) Metal at high position in the reactivity series (e.g., Na  Na+) C(s)  CO(g) or CO2(g) CO(g)  CO2(g) SO32– SO42– Fe2+ Fe3+ 2I– I2 H2(g) 2 H+

16. Remarks: Some chemical can act as both oxidizing agent and reducing agent. e.g., SO2(g)  S(s) oxidation / reduction SO2(g)  SO42–(aq) oxidation / reduction Species with high O.N. atom has higher chance to be an oxidizing agent. Similarly, species with low O.N. atom has higher chance to be a reducing agent.

17. Acid-Base Reaction & Redox Reaction H+(aq) + OH–(aq)  H2O(l) O.N.: H +1 +1 +1 O–2–2 No atom has its O.N. change. In general, acid-base reaction is NOT a redox reaction. Exercise Verify the conclusion for the following: HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l)

18. Concentration(濃度)Effect on Oxidizing/Reducing Power • Example • Dilute nitric acid (HNO3) reacts with magnesium but not copper. Concentrated nitric acid reacts with copper. • Dil. HNO3(aq) with Mg: • 2 H+(aq) + Mg(s)  H2(g) + Mg2+(aq) • Dil. HNO3(aq) with Cu: no reaction • Conc. HNO3(aq): (unbalanced equation) • NO3–(aq) + Cu(s)  NO2(g)+ Cu2+(aq) • Dil. HNO3(aq) acts as an acid, conc. HNO3(aq) can act as an • oxidizing agent. Similarly for H2SO4(aq).

19. Oxidation Number (氧化數) in Chemical Naming Roman number is sometimes used for O.N./O.S. e.g., “The Oxidation State of Mn in MnO4– is +VII.” The Romanic number system is also used in the “Stock system” to distinguish different compounds. e.g., Cu2O, containing Cu+ ion: copper(I) oxide CuO, containing Cu2+ ion: copper(II) oxide SO42–, sulphate(VI) ion (more common: sulphate) SO32–, sulphate(IV) ion (more common: sulphite)

20. O.N. of Atoms(原子) in Various Species Exercise: verify the O.N.

21. Electrolysis(電解) • Charging and discharge of rechargeable battery • Charging: applying voltage  new substances • (electrical energy  chemical energy) • Discharge: giving out electrical energy • (chemical energy  electrical energy) • Electrolysis: chemical reaction by applying voltage

22. Electrolysis(電解) of molten PbBr2 Anode (“+”, attracting anions): Br -Br + e- Cathode (“-”, attracting cations): Pb2+ + 2e- Pb “An Ox”: anode - oxidation “Red Cat”: reduction - cathode

23. Electrolysis(電解) of CuSO4 solution Ions attracted to anode: OH-(aq) & SO42-(aq) Oxidation: 4OH-(aq)  O2(g) + 2H2O(l) + 4e- SO42-(aq): no reaction Ions attracted to cathode: Cu2+(aq) & H+(aq) Reduction: Cu2+(aq) + 2e- Cu(s) H+(aq): no reaction

24. Preference of ion discharge • Electrochemical series • Na  Na+ + e- (More easily) • Ag  Ag+ + e- (More difficultly) • Therefore, we can infer that sodium ions gain electrons to form • atoms more difficultly than silver ions.

25. Different species have different abilities to gain electrons to be reduced. These abilities are summarized in the electrochemical series. Example: http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29

26. Characteristics • The half-reactions are reductions (i.e., gaining electrons on the left- • hand-side). • The half-reactions are equilibriums. • The abilities of gaining electrons are quantified by “standard • electrode potentials”. The smaller the potentials (near the top) are, • the more difficultly the reductions occur. • Compare • Na(s)  Na+(aq) + e-Eo = +2.71 V • Ag(s)  Ag+(aq) + e-Eo = -0.80 V • (Note that when the reactions are reversed, the signs of Eo are changed.)

27. Characteristics • Higher concentration or pressure: more favorable to go to the • opposite side. “Standard” values: 1 M for concentration and 1 atm • pressure for pressure. • e.g., Ca2+(aq) + 2e- Ca(s) Eo = -2.87 V • If increasing Ca2+ concentration, more favorable to the right-hand- • side, larger E (less negative or even positive). • e.g., NO3-(aq) + 2H+(aq) + e- NO2(g) + H2O(l) Eo = +0.78 V • If increasing NO2(g) pressure, more favorable to the left-hand-side, • smaller E.

28. Characteristics • A complete reaction consists of a reduction and an oxidation. • e.g., anode: 4OH-(aq)  O2(g) + 2H2O(l) + 4e- (1) • cathode: Cu2+(aq) + 2e- Cu(s) (2) • They are combined to form a complete reaction, in which no • electron shows up.

29. Characteristics • Voltage of an electrochemical cell: • (1): 4OH-(aq)  O2(g) + 2H2O(l) + 4e- -0.40 V • (2)x2: 2Cu2+(aq) + 4e- 4Cu(s) +0.34 V • IMPORTANT: Eo does not change when the equation is “doubled”. • Adding: Eo = -0.40 + 0.34 = -0.06 V

30. Metal wire H2(g) at 1 atm Glass tube Pt electrode H+(aq, 1 M) http://www.yorku.ca/skrylov/Teaching/Chemistry1001/ electrochemistry.pdf Characteristics • An electrochemical cell consists of two • half-reactions. A single half-reaction • does not exist alone and the absolute • values of Eo for half-reactions cannot • be measured. Therefore, the Eo of • one of the half-reactions, • 2H+(aq) + 2e- H2(g), • is set to zero. The electrode for this • half-reaction is shown on the right • and is called “standard hydrogen • electrode (SHE)”. The Eo of all the • others can be determined as values • relative to this “standard”.

31. Characteristics e.g., 2H+(aq) + 2e- H2(g) E1o = 0 V Zn2+(aq) + 2e- Zn(s) E2o = ? Consider Zn(s)  Zn2+(aq) + 2e- -E2o Adding: 2H+(aq) + Zn(s)  H2(g) + Zn2+(aq) Ecello = E1o + (-E2o) = -E2o Ecello of the last electrochemical cell is measured as +0.76 V. Therefore, E2o = -0.76 V.

32. Characteristics • If the voltage is positive, the reaction occurs spontaneously. If • the voltage is negative, the reaction does not occur • spontaneously and an external voltage must be applied. The • external voltage must be larger than the magnitude of the cell • voltage. • e.g., H2(g) + Zn2+(aq)  2H+(aq) + Zn(s) Eo = –0.76 V • The external voltage applied must be at least 0.76 V.

33. Characteristics • In electrolysis, the preference of ion discharge depends on Eo of • the relevant half-reaction potential. For example, • 4OH-(aq)  O2(g) + 2H2O(l) + 4e- -0.40 V • 2SO42-(aq)  S2O82-(aq) + 2e- -2.01 V • The first half-reaction is preferred because its Eo is larger.

34. Concentrations of species Increasing concentration increases the discharge tendency of an ion. For example, 4OH-(aq)  O2(g) + 2H2O(l) + 4e- -0.40 V 2Cl-(aq)  Cl2(g) + 2e- -1.36 V For dilute NaCl solution, OH- is discharged because the Eo value of the first half-reaction is preferred. But for concentrated NaCl solution, Cl- concentration is high enough for Cl- to be discharged.

35. Electrodes (電極) • Commonly used graphite and platinum electrodes are inert and have no effect on the preference of ion discharge. But some may. • Mercury electrode(汞電極) • If graphite or platinum electrodes are used in the electrolysis of • concentrated NaCl solution, only H+ is discharged at the cathode. • But if mercury electrode is used for the cathode, Na+ is • discharged because sodium metal forms an alloy with mercury. • (This method is used in industry for the production of sodium.)

36. Electrodes (電極) • Metal electrode • When an anion discharges at anode, it gives out electrons. If a • metal electrode is used as the anode, the metal atoms may • also give out electrons to form metal ions, i.e., the metal • electrode may compete with the anion in giving out electrons. • For example, if copper electrode is used as the anode in the • electrolysis of copper sulfate solution, the copper electrode • becomes thinner and thinner.

37. Electrodes(電極) cf. Cu(s)  Cu2+(aq) + 2e- -0.34 V 4OH-(aq)  O2(g) + 2H2O(l) + 4e- -0.40 V Copper metal of the anode completes with OH-. The potential of the first half-reaction is larger. Copper metal, rather than OH-, gives out electrons. In principle, if platinum electrode is used, platinum may also give out electrons to form platinum ion. But in practice, it seldom happens due to the very negative value of Eo for this process: Pt(s)  Pt2+(aq) + 2e- -1.20 V

38. Summary of common cases:

39. Salt Bridge The container in which a half-reaction occurs is called a “half-cell”. In the diagram shown, the two half-cells are in the same beaker. But in some cases they must be separated physically, because the species of the two half-cells react directly without electrons going through the external circuit.

40. Salt Bridge For example, an electrochemical cell can be constructed for the following reaction: Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) We can break it into two half-reactions: Cu(s)  Cu2+(aq) + 2e– –0.34 V 2Ag+(aq) + 2e– 2Ag(s) +0.80 V If we put everything into the same beaker, it does not work as expected.

41. Salt Bridge Reason: when Ag+ is in contact with Cu, they react on the surface of the copper plate. Electronsare given out by Cu to Ag+ directly and they do not go through the external circuit. Voltmeter Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)

42. Salt Bridge Therefore, the Ag+ solution and Cu2+ solution must be separated in two beakers. Voltmeter Salt-bridge: • “connecting” the two half-cells. • providing ions to keep the half-cells electrically neutral • simplest version: a strip of filter-paper soaked in KNO3 or NH4NO3 Electron flow Electrical current NO3– K+ Cu Ag Salt bridge Cu2+ NO3– Ag+ Cu2+ 1 M Cu(NO3)2(aq) 1 M AgNO3(aq) Modified from: http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

43. Fuel Cell(燃料電池) • Burning fuel  …  electrical energy: some energy is lost in heating • Fuel cell: chemical energy  electrical energy • Example: H2 fuel cell • H2 fuel cell is reversible and it can act as a rechargeable battery Charging (storing up electrical energy): 2H2O(l)  2H2(g) + O2(g) Discharging (releasing electrical energy): 2H2(g) + O2(g)  2H2O(l)

44. H2 fuel cell Half-reactions: (discharging) H2(g)  2H+(aq) + 2e– 0.00 V O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V The cell gives an electrical potential of 1.23 V. Fuel Cell – Car & Experiment Kit Lab Manual, Thames & Kosmos (2000) H2 O2 H2O Anode PEM Cathode PEM: proton exchange membrane

45. Fuel Cell(燃料電池) Other than H2, some other compounds can also be used for fuel cell. For example: • Methanol (CH3OH) • Ethanol (CH3CH2OH) These kinds of fuel cell may not be reversible, but easier to handle and more energy-rich.

46. Silver-Zinc(鋅)Battery(1.8V) Half reactions of discharge: http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

47. The Nickel-Cadmium Rechargeable Battery (1.4 V) Half reactions of discharge: http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

48. Eo and K (Equilibrium Constant(平衡常數)) nFEo = RT·ln(K) e.g, Calculate K for the equilibrium: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) Solution: Cu(s)  Cu2+(aq) + 2e––0.34 V 2Ag+(aq) + 2e– 2Ag(s) +0.80 V Eo = –0.34 V+0.80 V = +0.46 V

49. Eo and K (Equilibrium Constant(平衡常數)) n = no. of electrons in the half-reactions = 2 F = 96485 C mol–1 (Faraday constant) Put into the equation, nFEo = RT·ln(K), (2) (96485 C mol–1) (0.46 V) = (8.314 J mol–1 K–1) (298 K) ln(K) K = 3.6 x 1015 dm3 mol–1 Note: the unit of K is determined by the expression: K = [Cu2+(aq)]/[Ag+(aq)]2 (mol dm–3 for concentration, atm for pressure)