1. Chapter 13, Preparation to work with cells reaction prediction is a necessary skill 1 st step: Make a list Guidelines: most ionics get listed as (aq) ions e.g. list NaNO 3 (aq) as Na + (aq), NO 3 ‾(aq) list low solubility ionics as formula(s) e.g. MnO 2 (s), AgBr(s), AgI(s), PbSO 4 (s) list molecular compounds as formula(SATP state) e.g. H 2 O(l), NO(g), H 2 O 2 (l), N 2 O 4 (g), N 2 O(g)

2. Chapter 13, Preparation list pure elements as element formula(SATP state) e.g. F 2 (g), Cu(s), O 2 (g), etc list strong acids as H + (aq), @#‾(aq) list weak acids as formula(aq) e.g. HNO 2 (aq), H 2 SO 3 (aq) if question says “acidified” list H + (aq) if question says “basic” list OH‾(aq) list H 2 O(l)

3. Chapter 13, Preparation 2 nd step: identify each entity as OA, RA, or \ Use your table and the following generalizations: Metal elements are always RA’s e.g. Ag(s), Zn(s), etc Metal ions are always OA’s but a few, on your table are both OA and RA e.g. Sn 2+ (aq), Fe 2+ (aq), Cr 2+ (aq) Water is both OA and RA

4. Chapter 13, Preparation 3 rd step: pick out SOA and SRA 4 th step: write half-reactions for SOA and SRA and balance electrons to get net balanced equation 5 th step: determine if predicted reaction is spontaneous or non- spontaneous

5. Chapter 13, Preparation Example 1: Nitric acid is painted onto a copper sheet to etch a design List: H + (aq), NO 3 ‾, Cu(s), H 2 O(l) OA OA RA OA/RA 2 NO 3 ‾(aq) + 4 H + (aq) + 2 e‾ N 2 O 4 (g) + 2 H 2 O(l) Cu(s) Cu 2+ (aq) + 2 e‾ S S 2 NO 3 ‾(aq) + 4 H + (aq) + Cu(s) Cu 2+ (aq) + N 2 O 4 (g) + 2 H 2 O(l) spont Note: concentrated HCl(aq) would have no effect on Cu(s)

6. Chapter 13, Preparation Example 2: Engineers removed O 2 (g) from a solution by adding basic sodium sulfite solution List: O 2 (g), OH‾(aq), Na + (aq), SO 3 2 ‾ (aq), H 2 O(l) OA RA OA RA OA/RA O 2 (g) + 2 H 2 O(l) + 4 e‾ 4 OH ‾ (aq) SS 2 x (SO 3 2 ‾ (aq) + 2 OH ‾ (aq) SO 4 2 ‾ (aq) + H 2 O(l) + 2 e‾) 2 SO 3 2 ‾(aq) + O 2 (g) 2 SO 4 2 ‾(aq) 2 SO 3 2 ‾(aq) + 4 OH ‾ (aq) + O 2 (g) + 2 H 2 O(l) 2 SO 4 2 ‾(aq) + 2 H 2 O(l) + 4 OH‾(aq) spont Do WS 54

7. Chemistry 30 – Unit 2 Part 2 Cells and Batteries To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca

8. Chapter 13, Section 13.1 Redox reactions – transfer of electrons In past, OA and RA were in contact and electrons flowed between the two In cells, OA and RA are kept away from each other so that electrons must flow through a wire to be transferred – this is an electric current Now the electrons can do work

9. Chapter 13, Section 13.1 cell – transforms chemical potential energy to electrical energy via redox reaction cell (studied later in chapter) – transforms electrical energy into chemical potential energy

10. Chapter 13, Section 13.1 Cell Definitions and Principles: electrode where oxidation occurs (where SRA reacts) electrode where reduction occurs (where SOA reacts) (salt bridge or porous cup) device to allow flow of ions (current flow): from anode to cathode anions to anode; cations to cathode memory tool: “An ox cared”

11. Chapter 13, Section 13.1 Daniell Cell components: Zn(s) in ZnSO 4 (aq) solution; Cu(s) in CuSO 4 (aq) solution To find out what will happen, list everything present in the cell and use your chart to identify the OA’s and RA’s List List: Zn(s), Zn 2+ (aq), Cu(s), Cu 2+ (aq), SO 4 2- (aq), H 2 O(l) RA OA RA OA OA* OA/RA

12. Chapter 13, Section 13.1 Zn(s), Zn 2+ (aq), Cu(s), Cu 2+ (aq), SO 4 2- (aq), H 2 O(l) RA OA RA OA OA* OA/RA Identify the strongest oxidizing agent, SOA, strongest reducing agent, SRA, and write the half-reactions S S Cu 2+ (aq) + 2 e – Cu(s) Zn(s) Zn 2+ (aq) + 2 e – By definition @ cathode @ anode

13. Chapter 13, Section 13.1 anode – cathode + electron flow anion flow cation flow Zn(s) Cu(s) Zn 2+ (aq) Cu 2+ (aq) KNO 3 (aq) V SO 4 2- (aq) Always use red meter lead for (+) and black for ( – )

14. Chapter 13, Section 13.1 Why do ions flow? Look at the half- reactions Zn(s) Cu(s) Zn 2+ (aq) Cu 2+ (aq) KNO 3 (aq) V SO 4 2- (aq) Cu 2+ (aq) changes to Cu(s) SO 4 2- (aq) concentration unchanged – build up of negative charge around cathode. Cations flow to neutralize charge Zn(s) changes to Zn 2+ (aq) build up of positive charge around anode. Anions flow to neutralize charge anode cathode Cu 2+ (aq) + 2 e − Cu(s) Zn(s) Zn 2+ (aq) + 2 e −

15. Chapter 13, Section 13.1 Daniell Cell in action!Daniell Cell in action! Look at the cell half-reactions again. What would you expect to happen to copper electrode? zinc electrode? Discuss

16. Chapter 13, Section 13.1 The voltmeter reading in the animation was +1.10 V or -1.10 V Do you see any connection to electrical potential numbers from your redox chart in Data Booklet? more to come on this Any idea about the meaning of the (+) or (-)?

17. Chapter 13, Section 13.1 Inert electrodes: many of the half- reactions on your chart do not include a metal element that can function as an electrode Examples: 2 H + (aq) + 2 e – H 2 (g) Fe 3+ (aq) + e – Fe 2+ (aq) Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 e – 2 Cr 3+ (aq) + 7 H 2 O(l) To use these half-reactions in a cell, inert electrodes are used. Common inert electrodes: C(s), Pt(s)

18. Chapter 13, Section 13.1 Cell Notation: For the Daniell Cell: Zn(s) | Zn 2+ (aq), SO 4 2- (aq) || Cu 2+ (aq), SO 4 2- (aq) | Cu(s) Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) common to leave out spectators: anode | anode solution || cathode solution | cathode oxidation half-cell reduction half-cell porous barrier

19. Chapter 13, Section 13.1 Oxidation (@ anode) produces electrons making anode negative Reduction (@cathode) consumes electrons making cathode positive Electrons at anode have lots of potential energy (they are repelled through the wire away from the anode) Electrons at cathode have zero potential energy

20. Chapter 13, Section 13.1 Potential difference (commonly called voltage) = difference in E p per unit charge between anode and cathode E º cell = E º net = ΔE º = E º cathode – E º anode Standard cell potential* Cell potential Cell voltage (V) Standard reduction potential for anode ½ -reaction (V) Standard reduction potential for cathode ½ -reaction (V) **Important note: never multiply the Eº values by balancing coefficients!

21. Chapter 13, Section 13.1 For Daniell cell Eº cell = + 0.34 V – – 0.76 V = + 1.10 V same as on animation! Recall the spontaneity rule from chapter 12: OA RA Note that spontaneous reactions will have a + E º cell

22. Chapter 13, Section 13.1 Reference half-cell: …….|| H + (aq), H 2 (g) | Pt(s) or Pt(s) | H + (aq), H 2 (g) || ….. E o r = 0.00 V

23. Chapter 13, Section 13.1 Cell analysis: When you look at cell notation it’s good idea to check that left hand side of cell notation really is anode ½ cell (where SRA reacts) Example: Ni(s) | Ni 2+ (aq) || Cr 2 O 7 2- (aq), H + (aq) | C(s) RA OA OA OA \ S S inert electrode

24. Chapter 13, Section 13.1 Half-reactions: 3 x (Ni(s) Ni 2+ (aq) + 2 e – ) (anode) Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 e – 2 Cr 3+ (aq) + 7 H 2 O(l) (cathode) Cr 2 O 7 2- (aq) + 14 H + (aq) + 3 Ni(s) 2 Cr 3+ (aq) + 7 H 2 O(l) + 3 Ni 2+ (aq) Spontaneous: why are standard voltaic cells always spontaneous?

25. Chapter 13, Section 13.1 Interruption: Another half-cell separator (instead of salt bridge) Porous cup: unglazed pottery – holds water but ions are small enough to flow through

26. Chapter 13, Section 13.1 Back to the cell: electron flow anode – cathode + V Ni(s) C(s) Ni 2+ (aq) Cr 2 O 7 2- (aq) H + (aq) anions cations

27. Chapter 13, Section 13.1 Other solutions can be used, but KNO 3 (aq) is particularly good for use in a salt bridge Why? Both K + (aq) and NO 3 – (aq) have high solubility with all other ions – no precipitates in salt bridge K + (aq) and NO 3 – (aq) without H + (aq) are weak OA’s and unlikely to interfere with the desired cell reaction

28. Chapter 13, Section 13.1 Investigation 13.A page 488 Prelab: Make predictions for the following cells: AAg(s) in Ag + (aq) with Cu(s) in Cu 2+ (aq) B Zn(s) in Zn 2+ (aq) C Mg(s) in Mg 2+ (aq) D Mg(s) in Mg 2+ (aq) with Zn(s) in Zn 2+ (aq) E Cu(s) in Cu 2+ (aq) FZn(s) in Zn 2+ (aq) with Cu(s) in Cu 2+ (aq) " " " " " " " " " """ We’ll do the first together

29. Chapter 13, Section 13.2 Write cell notation (in proper order) Draw cell diagrams showing electrodes labeled with identity and also as + or -, anode or cathode Label electrolyte solution in each half- cell Show direction of electron flow, direction of ion flow, predicted E° cell, anode and cathode half-reactions

30. Chapter 13, Section 13.2 Commercial voltaic cells: Dry Cells, also called carbon-zinc cells or zinc chloride cells (Eveready “cat”) zinc anode (case), carbon cathode, electrolyte paste made up of MnO 2 (s), starch, and a very small amount of water Half-reactions: Zn(s) Zn 2+ (aq) + 2 e – 2 MnO 2 (s) + H 2 O(l) + 2 e – Mn 2 O 3 (s) + 2 OH – (aq) 1.5 V, primary cell (non-rechargeable)

31. Chapter 13, Section 13.2 alkaline cells (Duracell, Energizer, etc) more expensive version of dry cell Significant differences between these and regular dry cells

32. Chapter 13, Section 13.2 batteries – multiple cells e.g. a 9 V alkaline is made up of six 1.5 V alkaline cells Button batteries – read page 492-3

33. Chapter 13, Section 13.2 Fuel Cells: batteries where a “fuel” in the anode half-reaction is consumed and products are released from the cell Battery is renewed by replacing the fuel Common fuel cell fuels: hydrogen, methanol, hydrocarbons Conversion of fuels to energy in fuel cells more efficient than combustion (also much cleaner than combustion)

34. Chapter 13, Section 13.2 The hydrogen fuel cell used in the space shuttle provides electricity and water for the astronauts as net reaction is 2 H 2 (g) + O 2 (g) 2 H 2 O(l) Your text has interesting pictures and reading on pages 493-6 on fuel cells

35. Chapter 13, Section 13.2 Discuss questions 2, 6, 7, 10 page 501

36. Chapter 13, Section 13.2 Rusting (oxidation/corrosion of iron) Many metals such as copper, aluminium, and zinc, when they react with O 2 in air, get covered with a passive oxide coat that protects them from further oxidation Not true for iron and steel Rust forms on the surface and continues to form below the surface Once it starts.......

37. Chapter 13, Section 13.2 in corrosion of iron, iron acts as anode and is oxidized the OA is oxygen in presence of water cathode, where OA reacts, is inert conductor on surface of iron – soot in some cases Half-reactions: Fe(s) Fe 2+ (aq) + 2 e – O 2 (g) + 2 H 2 O(l) + 4 e – 4 OH – (aq) Products: Fe 2+ (aq) and OH – (aq) react to form Fe(OH) 2 (s), a precursor of rust

38. Chapter 13, Section 13.4 sootless soot

39. Chapter 13, Section 13.2 Fe(OH) 2 (s) further reacts with air to form Fe(OH) 3 (s) and Fe 2 O 3 x H 2 O(s) This is rust Rust prevention: Physical - cover the iron with paint, tar, or grease, plate it with chromium, alloy it with chromium (stainless steel)

40. Chapter 13, Section 13.2 Rust prevention continued Chemical – attach, by some means, a metal that is a stronger RA than Fe to the Fe Under oxidizing conditions the other metal (sacrificial anode) will be oxidized instead of the Fe

41. Chapter 13, Section 13.2 iron becomes inert cathode where O 2 (g) + 2 H 2 O(l) + 4 e − 4 OH − (aq) takes place Process called cathodic protection since iron becomes inert cathode Examples of cathodic protection: galvanizing, sheradizing, contact with blocks of Zn or Mg Discuss questions 13, 14, 16 page 500

42. Chapter 13, Section 13.2

43. Review sample diploma exam questions on cathodic protection

44. Chapter 13, Section 13.3 Electrolytic Cells: Cells that convert electrical energy to chemical potential energy

45. Chapter 13, Section 13.3 Voltaic Cell + – oxidation electrons produced reduction electrons consumed e-e- e-e- Electrolytic Cell cathode anode reduction electrons consumed oxidation electrons produced spontaneous – anode cathode e.g. C(s)|Cu 2+ (aq)|C(s) SOA Cu 2+ (aq) Cu 2+ (aq) + 2 e – Cu(s) SRA H 2 O(l) 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e – Non-spontaneous Eº net = + 0.34 V – + 1.23 V = – 0.89 V

46. Chapter 13, Section 13.3 Observations: Anode (+) in electrolysis; cathode (–) in electrolysis Electrons in electrolysis circuit flow from (+) to (-) but still from anode to cathode! Anions in electrolytic cell flow towards (+); cations flow towards (-) but this is still ………. Anions to anode; cations to cathode Voltage in electrolytic cells is (-): non- spontaneous Text has good chart on page 503 – turn to it

47. Chapter 13, Section 13.3 What do you think the meaning of the negative E º cell is? Predict anode and cathode half- reactions for following electrolysis reactions: Na 2 SO 4 (aq) with Pt(s) electrodes NaCl(aq) with Pt(s) electrodes KI(aq) with C(s) electrodes CuSO 4 (aq) with C(s) electrodes Use the handout provided Be sure to list H 2 O(l) when finding SOA & SRA

48. Chapter 13, Section 13.3 All of these electrolysis reactions will be demonstrated for you next day

49. Chapter 13, Section 13.3 When you calculate Eº net for electrolysis of H 2 O(l): 2 H 2 O(l) + 2 e – H 2 (g) + 2 OH – (aq) Eº r = – 0.83 V 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e – Eº r = + 1.23 V 2 x ( ) 4 H 2 O(l) 6 H 2 O(l) 2 H 2 (g) + O 2 (g) + 4 H + (aq) + 4 OH – (aq) 2 Eº net = − 0.83 V − + 1.23 V= − 2.06 V : predicted minimum voltage 2.06 V But this is way too high! Why? Nowhere near standard conditions!! Under these conditions Eº r ’s much different E º net = − 0.42 V − + 0.82 V= − 1.24 V − 0.42 V + 0.82 V overpotential

50. Chapter 13, Section 13.3 If you did this electrolysis and applied a voltage of 1.24 V (new predicted minimum voltage) you would find that the reaction would not occur E º r ’s are measured with respect to the reference half-cell (H 2 (g), H + (aq)|Pt(s)) Particularly in half-reactions where gases are formed, a slightly higher voltage called the overpotential is required For formation of H 2 (g) and O 2 (g) overpotential is approximately 0.6 V

51. Chapter 13, Section 13.3 Overpotential needed to understand NaCl(aq) electrolysis In electrolysis of NaCl(aq), predicted reaction didn’t occur at anode Prediction: H 2 O(l) both SRA and SOA anode: 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e – Eº r = +1.23 V cathode: 2 H 2 O(l) + 4 e – H 2 (g) + 2 OH – (aq) Eº r = - 0.83 V Predicted Products:O 2 (g) & H 2 (g) Eº net = – 0.83 V – + 1.23 V = – 2.06 V

52. Chapter 13, Section 13.3 Actual: Products: anode: 2 Cl – (aq) Cl 2 (g & aq) + 2 e – Eº r = +1.36 V cathode: 2 H 2 O(l) + 4 e – H 2 (g) + 2 OH – (aq) Eº r = - 0.83 V Cl 2 (g), H 2 (g), OH – (aq) Eº net = – 0.83 V – + 1.36 V = – 2.19 V Even though production of O 2 (g) seems to be favoured, the overpotential for O 2 (g) is greater than that for Cl 2 making production of Cl 2 favoured called “the chloride anomaly”

53. Chapter 13, Section 13.3 Chlor-Alkali Process – industrial electrolysis of NaCl(aq) Figure 13.25, page 506

54. Chapter 13, Section 13.3 Electrolysis of aqueous ionics to produce elements has several problems: Some ionics have low solubility in water H 2 O(l) is a stronger OA than some metal ions Answer: Use molten ionics (molten salts)

55. Chapter 13, Section 13.3 Down’s Cell anode & cathode both C(s) Figure 13.26, page 510

56. Chapter 13, Section 13.3 Rechargeable cells and batteries (secondary cells and batteries) Example: lead-acid battery Figure 13.28, page 511 while battery discharges it’s voltaic; while it recharges it’s electrolytic

57. Chapter 13, Section 13.3 You should know the half-reactions – they are in your Data Booklet Example: nicad – discuss Know ecological concerns Read pages 511-2

58. Chapter 13, Section 13.4 Cell Stoichiometry Necessary definitions: rate of flow of electric charge; 1 ampere of current = 1 Coulomb (unit of charge) per second Charge of 1 mol of electrons called 9.65 x 10 4 C/mol No need to memorize; page 3 Data Booklet

59. Chapter 13, Section 13.4 = # mol of electrons (mol) I = current (A) (must be amps) t = time (s) (must be seconds)

60. Chapter 13, Section 13.4 Example: question 9, page 516 Zn 2+ (aq) + 2 e − Zn(s) n 1 750 mA 3.25 h n 2 m=? current, time info always related to

61. Chapter 13, Section 13.4 Example: question 11, page 516 Ni(s) Ni 2+ (aq) + 2 e − n 1 1.20 g n 2 35.5 min I=?

62. Chapter 13, Section 13.4 Do worksheet 60 Answers: 1.51 min, also indicate electrode 2.6.35 kg, note this is Cl 2 not Cl - also indicate electrode 3.5.96 h 4.274 A 5.9.35 g

63. Chapter 13, Section 13.4 Electroplating – The process of plating a metal coating on top of another metal by electrolysis Sounds easy, but there are numerous problems What are some?

64. Chapter 13, Section 13.4 Extraction and refining of metals Extraction – Extracting the metal from its ore; often done by electrolysis of solutions or moltens Refining – Purification of metals after extraction − + ? ? anode: impure Cu cathode: purified Cu

65. Chapter 13, Section 13.4