1. The Nucleus: A Chemist’s View

2. Nuclear Symbols Element symbol Mass number, A (p + + n o ) Atomic number, Z (number of p + )

3. Balancing Nuclear Equations  A reactants =  A products  Z reactants =  Z products 235 + 1 = 142 + 91 + 3(1) 92 + 0 = 56 + 36 + 3(0)

4. Balancing Nuclear Equations #2 226 = 4 + ____ 222 88 = 2 + ___ 86 Atomic number 86 is radon, Rn Rn

5. Balancing Nuclear Equations #3 235 + 1 = 139 + 2(1) + ____ 95 39 92 + 0 = 53 + 2(0) + ____ 39 95 Atomic number 39 is yttrium, Y Y

6. Alpha Decay Alpha production (  ): an alpha particle is a helium nucleus Alpha decay is limited to heavy, radioactive nuclei

7. Alpha Radiation Limited to VERY large nucleii.

8. Beta Decay Beta production (  ): A beta particle is an electron ejected from the nucleus Beta emission converts a neutron to a proton

9. Beta Radiation Converts a neutron into a proton.

10. Gamma Ray Production Gamma ray production (  ): Gamma rays are high energy photons produced in association with other forms of decay. Gamma rays are massless and do not, by themselves, change the nucleus

11. Deflection of Decay Particles Opposite charges_________ each other. Like charges_________ each other. attract repel

12. Positron Production Positron emission: Positrons are the anti- particle of the electron Positron emission converts a proton to a neutron

13. Electron Capture Electron capture: (inner-orbital electron is captured by the nucleus) Electron capture converts a proton to a neutron

14. Types of Radiation

15. Nuclear Stability Decay will occur in such a way as to return a nucleus to the band (line) of stability. The most stable nuclide is Iron-56 If Z > 83, the nuclide is radioactive

16. A Decay Series A radioactive nucleus reaches a stable state by a series of steps

17. Half-life Concept

18. Decay Kinetics Decay occurs by first order kinetics (the rate of decay is proportional to the number of nuclides present) N = number of nuclides remaining at time t N 0 = number of nuclides present initially k = rate constant t = elapsed time

19. Calculating Half-life t 1/2 = Half-life (units dependent on rate constant, k )

20. Sample Half-Lives

21. Nuclear Fission and Fusion Fusion: Combining two light nuclei to form a heavier, more stable nucleus. Fission: Splitting a heavy nucleus into two nuclei with smaller mass numbers.

22. Energy and Mass Nuclear changes occur with small but measurable losses of mass. The lost mass is called the mass defect, and is converted to energy according to Einstein’s equation:  E =  mc 2  m = mass defect  E = change in energy c = speed of light Because c 2 is so large, even small amounts of mass are converted to enormous amount of energy.

23. Fission

24. Fission Processes A self-sustaining fission process is called a chain reaction.

25. A Fission Reactor

26. Fusion

27. Review l Oxidation reduction reactions involve a transfer of electrons. l OIL- RIG l Oxidation Involves Loss l Reduction Involves Gain l LEO-GER l Lose Electrons Oxidation l Gain Electrons Reduction

28. Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)? A.PbS, reductant; O 2, oxidant B.PbS, reductant; SO 2, oxidant C.Pb 2+, reductant; S 2- oxidant D.PbS, reductant; no oxidant E.PbS, oxidant; SO 2, reductant

29. Applications l Moving electrons is electric current. 8H + +MnO 4 - + 5Fe +2 +5e -  Mn +2 + 5Fe +3 +4H 2 O l Helps to break the reactions into half reactions. 8H + +MnO 4 - +5e -  Mn +2 +4H 2 O 5(Fe +2  Fe +3 + e - ) l In the same mixture it happens without doing useful work, but if separate

30. H + MnO 4 - Fe +2 l Connected this way the reaction starts l Stops immediately because charge builds up. e-e- e-e- e-e- e-e- e-e-

31. H + MnO 4 - Fe +2 Galvanic Cell Salt Bridge allows current to flow

32. H + MnO 4 - Fe +2 e-e- l Electricity travels in a complete circuit

33. H + MnO 4 - Fe +2 Porous Disk l Instead of a salt bridge

34. Reducing Agent Oxidizing Agent e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode

35. Cell Potential l Oxidizing agent pulls the electron. l Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential E cell l Also called the electromotive force (emf) l Unit is the volt(V) l = 1 joule of work/coulomb of charge l Measured with a voltmeter

36. Zn +2 SO 4 -2 1 M HCl Anode 0.76 1 M ZnSO 4 H + Cl - H 2 in Cathode

37. 1 M HCl H + Cl - H 2 in Standard Hydrogen Electrode l This is the reference all other oxidations are compared to E º = 0 l º indicates standard states of 25ºC, 1 atm, 1 M solutions.

38. Cell Potential Zn(s) + Cu +2 (aq)  Zn +2 (aq) + Cu(s) l The total cell potential is the sum of the potential at each electrode. E º cell = E º Zn  Zn +2 + E º Cu +2  Cu l We can look up reduction potentials in a table. l One of the reactions must be reversed, so change it sign.

39. Cell Potential l Determine the cell potential for a galvanic cell based on the redox reaction. Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) Fe +3 (aq) + e -  Fe +2 (aq) E º = 0.77 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V Cu(s)  Cu +2 (aq)+2e - E º = -0.34 V 2Fe +3 (aq) + 2e -  2Fe +2 (aq) E º = 0.77 V

40. Reduction potential More negative E º – more easily electron is added – More easily reduced – Better oxidizing agent More positive E º – more easily electron is lost – More easily oxidized – Better reducing agent

41. Line Notation solid  Aqueous  Aqueous  solid Anode on the left  Cathode on the right l Single line different phases. l Double line porous disk or salt bridge. l If all the substances on one side are aqueous, a platinum electrode is indicated.

42. Cu 2+ Fe +2 l For the last reaction Cu(s)  Cu +2 (aq)  Fe +2 (aq),Fe +3 (aq)  Pt(s)

43. In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________. a. cathode, oxidation b. anode, reduction c. anode, oxidation d. cathode, reduction

44. Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd|Cd 2+ || Cu 2+ |Cu a. Cu 2+ + Cd → Cu + Cd 2+ b. Cu + Cd → Cu 2+ + Cd 2+ c. Cu 2+ + Cd 2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu 2+

45. Galvanic Cell l The reaction always runs spontaneously in the direction that produced a positive cell potential. l Four things for a complete description. 1)Cell Potential 2)Direction of flow 3)Designation of anode and cathode 4)Nature of all the components- electrodes and ions

46. Practice l Completely describe the galvanic cell based on the following half-reactions under standard conditions. MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º=1.51 V Fe +3 +3e -  Fe(s) E º=0.036V

47. Potential, Work and  G l emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q l Charge is measured in coulombs. -w = q E l Faraday = 96,485 C/mol e - l q = nF = moles of e - x charge/mole e - w = -q E = -nF E =  G

48. Potential, Work and  G  Gº = -nF E º if E º > 0, then  Gº < 0 spontaneous if E º 0 nonspontaneous l In fact, reverse is spontaneous. Calculate  Gº for the following reaction: Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq) Fe +2 (aq) + e -  Fe(s) E º = 0.44 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V

49. Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M l if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M l if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M

50. The Nernst Equation  G =  Gº +RTln(Q) -nF E = -nF E º + RTln(Q) E = E º - RT ln(Q) nF 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E º = 0.48 V l Always have to figure out n by balancing. l If concentration can gives voltage, then from voltage we can tell concentration.

51. The Nernst Equation l As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and E cell = 0 0 = E º - RTln(K) nF E º = RT ln(K) nF nF Eº = ln(K) RT

52. Batteries are Galvanic Cells l Car batteries are lead storage batteries. Pb +PbO 2 +H 2 SO 4  PbSO 4 (s) +H 2 O

53. Batteries are Galvanic Cells Dry Cell Zn + NH 4 + +MnO 2  Zn +2 + NH 3 + H 2 O + Mn 2 O 3

54. Batteries are Galvanic Cells Alkaline Zn +MnO 2  ZnO+ Mn 2 O 3 (in base)

55. Batteries are Galvanic Cells l NiCad NiO 2 + Cd + 2H 2 O  Cd(OH) 2 +Ni(OH) 2

56. Corrosion l Rusting - spontaneous oxidation. l Most structural metals have reduction potentials that are less positive than O 2. Fe  Fe +2 +2e - E º= 0.44 V O 2 + 2H 2 O + 4e -  4OH - E º= 0.40 V Fe +2 + O 2 + H 2 O  Fe 2 O 3 + H + l Reactions happens in two places.

57. Water Rust Iron Dissolves- Fe  Fe +2 e-e- Salt speeds up process by increasing conductivity O 2 + 2H 2 O +4e -  4OH - Fe 2+ + O 2 + 2H 2 O  Fe 2 O 3 + 8 H + Fe 2+

58. Preventing Corrosion l Coating to keep out air and water. l Galvanizing - Putting on a zinc coat l Has a lower reduction potential, so it is more easily oxidized. l Alloying with metals that form oxide coats. l Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

59. l Running a galvanic cell backwards. l Put a voltage bigger than the potential and reverse the direction of the redox reaction. l Used for electroplating. Electrolysis

60. 1.0 M Zn +2 e-e- e-e- Anode Cathode 1.10 Zn Cu 1.0 M Cu +2

61. 1.0 M Zn +2 e-e- e-e- Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2

62. Calculating plating l Have to count charge. Measure current I (in amperes) l 1 amp = 1 coulomb of charge per second q = I x t l q/nF = moles of metal l Mass of plated metal l How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag +

63. Calculating plating 1.Current x time = charge 2.Charge ∕Faraday = mole of e - 3.Mol of e - to mole of element or compound 4.Mole to grams of compound Or the reverse if you want time to plate

64. Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

65. How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO 3 ) 3 at a current of 2.00 A?

66. Other uses l Electrolysis of water. l Separating mixtures of ions. l More positive reduction potential means the reaction proceeds forward. l We want the reverse. l Most negative reduction potential is easiest to plate out of solution.

67. Redox Know the table 2. Recognized by change in oxidation state. 3. “Added acid” 4. Use the reduction potential table on the front cover. 5. Redox can replace. (single replacement)

68. 6. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element. I - + IO 3 - + H +  I 2 + H 2 O 7. Decomposition. a) peroxides to oxides b) Chlorates to chlorides c) Electrolysis into elements. d) carbonates to oxides

69. 69 Examples 1. A piece of solid bismuth is heated strongly in oxygen. 2. A strip or copper metal is added to a concentrated solution of sulfuric acid. 3. Dilute hydrochloric acid is added to a solution of potassium carbonate.

70. 70 23. Hydrogen peroxide solution is added to a solution of iron (II) sulfate. 24. Propanol is burned completely in air. 25. A piece of lithium metal is dropped into a container of nitrogen gas. 26. Chlorine gas is bubbled into a solution of potassium iodide.

71. 71 Examples 5. A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide. 6. A solution of tin ( II ) chloride is added to an acidified solution of potassium permanganate 7. A solution of potassium iodide is added to an acidified solution of potassium dichromate.

72. 72 70. Magnesium metal is burned in nitrogen gas. 71. Lead foil is immersed in silver nitrate solution. 72. Magnesium turnings are added to a solution of iron (III) chloride. 73. Pellets of lead are dropped into hot sulfuric acid 74. Powdered Iron is added to a solution of iron(III) sulfate.

73. A way to remember l An Ox – anode is where oxidation occurs l Red Cat – Reduction occurs at cathode l Galvanic cell- spontaneous- anode is negative l Electrolytic cell- voltage applied to make anode positive

74. A student places a copper electrode in a 1 M solution of CuSO 4 and in another beaker places a silver electrode in a 1 M solution of AgNO 3. A salt bridge composed of Na 2 SO 4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. l (a) Draw a diagram of this cell. l (b) Describe what is happening at the cathode (Include any equations that may be useful.)

75. A student places a copper electrode in a 1 M solution of CuSO 4 and in another beaker places a silver electrode in a 1 M solution of AgNO 3. A salt bridge composed of Na 2 SO 4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. l (c) Describe what is happening at the anode. (Include any equations that may be useful.)

76. A student places a copper electrode in a 1 M solution of CuSO 4 and in another beaker places a silver electrode in a 1 M solution of AgNO 3. A salt bridge composed of Na 2 SO 4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. l (d) Write the balanced overall cell equation. l (e) Write the standard cell notation.

77. A student places a copper electrode in a 1 M solution of CuSO 4 and in another beaker places a silver electrode in a 1 M solution of AgNO 3. A salt bridge composed of Na 2 SO 4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH 3 ) + (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu 2+ (aq) concentration in the cell after the ammonia has been added?