1. Chapter 17 Electrochemistry Study Of Interchange Of Chemical And Electrical Energy Using RedOx chemistry to generate an electrical current – moving electrons

2. Review ReDox reactions = transfer electrons 2

3. Review Rules for Assigning Oxidation States p. 156 Atom in an element = 0 Monoatomic ion = same as charge Fluorine = -1 in compound Oxygen = -2 in compound Hydrogen = +1 in covalent compounds  of oxidation numbers = overall charge Apply what you just relearned: What are the oxidation states for each element? O 2, Cl -1, Mg +2, NH 3, NO 2, NO 3 -, CrO 4 -2, MnO 4 -1 3

4. Oxidation – Reduction Reactions Identify the atoms that are oxidized and reduced, and specify the oxidizing and reducing agents in the following reactions. 2Al (s) + 3I 2(s) → 2AlI 3(s) 2PbS (s) + 3O 2(g) → 2PbO (s) + 2SO 2(g) 4

5. Method of Balancing Redox Reactions in Acidic Solutions 1.Write separate equations for the oxidation and the reduction half- reactions 2.For each half-reaction a.Balance the elements except hydrogen and oxygen b.Balance the oxygen using H 2 O c.Balance the hydrogen using H + d.Balance the charge using electrons 3.If necessary, multiply one of both balanced half-reaction by an integer to equalize the number of electrons transferred in the half-reactions 4.Add the half-reactions and cancel identical species 5.Check that the elements and charges are balanced 5

6. Method of Balancing Redox Reactions in Basic Solutions 1.Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H + were present. 2.To both sides of the equation, add a number of (OH) - that is equal to the number of H +. 3.Form H 2 O on the side containing both H + and (OH) -, eliminate the number of H 2 O that appear on both sides of the equation. 4.Balance this equation: Al (s) + MnO 4 - (aq) → MnO 2(s) +Al(OH) 4 - (aq) 6

7. POP Quiz Complete and balance this equation for a redox reaction that takes place in basic solution. 7

8. Applications Is the following a redox reaction? (i.e. do oxidation states change from reactant to product) Moving electrons is electric current. 8H + +MnO 4 - + 5Fe +2 +5e -  Mn +2 + 5Fe +3 +4H 2 O Break the reactions into half reactions. Reduction: MnO 4 -  Mn +2 Oxidation: Fe +2  Fe +3 In the same mixture electrons transfer without doing useful work, but if you separate oxidation rxn. from reduction rxn. we can force the electrons to flow through a wire. 8

9. Connected this way the reaction starts Stops immediately because charge builds up. 9

10. Galvanic Cell Galvanic Cell (Voltaic Cell) can contain a salt bridge or a porous disk which allows the electrons to flow in a complete circuit. 10

11. Galvanic Cell An electrochemical process involves electron transfer. The species that acting as a reducing agent supplies electrons to the anode. The species acting as the oxidizing agent receives electrons from the cathode. Galvanic Cell AnimationGalvanic Cell Animation 11

12. Constructing a Galvanic Cell H 2 O + PO 3 -3 + MnO 4 - → 3PO 4 -3 + MnO 2(s) + 2OH - Draw a Galvanic Cell and Identify the components What is being reduced? What is being oxidized? Which direction does the electrons flow? Which electrode is the cathode? Which electrode is the anode? 12

13. Anode - e-e- e-e- e-e- e-e- e-e- e-e- Cathode + Oxidation Occurs at Anode (Anode Oxidation is neg.) Reduction Occurs at Cathode (Red Cat is pos.) Electrons flow An. Ox. to Red. Cat. PO 3 -3 MnO 4 - 13

14. POP Quiz 3Zn 2+ + 2Al → 3Zn + 2Al 3+ Draw a Galvanic Cell and Identify the components What is being reduced? What is being oxidized? Which direction does the electrons flow? Which electrode is the cathode? Which electrode is the anode? 14

15. Cell Potential Def. Pull (“driving force”) causes electron flow = E cell Called electromotive force (emf) or cell potential Oxidizing agent pulls the electron. Reducing agent pushes the electron. Unit is the volt(V) V = 1 joule of work/coulomb of charge 1V = 1J/C 15

16. Digital Voltmeters Draw only a Negligible Current and are Convenient to Use 16

17. Reaction in a Galvanic Cell with Standard Hydrogen Electrode 17

18. Standard Hydrogen Electrode This is the reference all other oxidations are compared to E o = 0 This is the set up that is used to measure Reduction Cell Potential and calculate the cell potential for all other metals. e -  18

19. A Galvanic Cell involving the Half-Reactions Mnemonic: e - flow An. Ox. to Red. Cat. 19

20. Standard Reduction Potential Potential of a given species to become reduced when all species are in their standard state. The Standard Hydrogen Potential is the reference potential against which all half-reactions potentials are assigned. Standard States: [H + ] = 1.0 M P H2 = 1 atm T = 298K IUPAC has universally accepted the half-reaction potentials based on the assignment of zero volts to this rxn. 2H+ + 2e- → H2 Hydrogen Reduction Cell Potential is E o = 0 20

21. Cell Potential = E o cell Zn(s) + Cu +2 (aq)  Zn +2 (aq) + Cu(s) The total cell potential is the sum of the potential at each electrode. We can look up reduction potentials in a table (A25) & combine using this formula: 1.One of the reactions must be reversed, so some signs must be reversed. 2.Electrons lost = Electrons gain, so don’t need to multiply value of E o by integers when balancing equations 21

22. Cell Potential E o Determine the cell potential for a galvanic cell Cu (s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) Step 1: Write the Half Reactions 1.Fe +3 (aq) + e -  Fe +2 (aq) E o = 0.77 V 2.Cu +2 (aq)+2e -  Cu(s) E o = 0.34 V Step 2: Reverse equation #2 & change the sign Cu(s)  Cu +2 (aq)+2e - E o = -0.34 V Step 3: Multiply equation #1 by integer 2 2Fe +3 (aq) + 2e -  2Fe +2 (aq) E o = 0.77 V Step 4: E o cell = E o reduction + E o oxidation 22

23. Important Facts 1.The higher the reduction potential the greater the tendency to be reduced. 2.Between 2 species, the species with the largest E o will be written as reduction and the other will be oxidization reaction, therefore it must be flipped. 3.The standard potential for an Galvanic Cell ( E o cell ) comes from adding E o for each half-reaction. 4.When multiplying a half-reaction by an interger to equalize e- number, DO NOT MULTIPLY the E o by that number. E o is an intensive property which does not depend on quantity. 23

24. Time To Get Ready For Carnegie Hall For the following cell & data, identify the cathode & anode, write the balanced reaction and calculate E o cell Al 3+ + 3e - → Al E o = -1.66 V Mg 2+ + 2e - → Mg E o = -2.37 V 24

25. Line Notation solid (anode)  Aqueous  Aqueous  solid (cathode) Anode on the left  Cathode on the right Single line = phases difference (solid | liquid). Double line || porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated. For the reaction on slide 20: Cu (s)   Cu +2 (aq)     aq   Fe +2 (aq),Fe +3 (aq)   Pt (s) 25

26. Complete Galvanic Cell Description Given Just Half Reactions The cell will always runs spontaneously in the direction that produces + E o Four things for a complete description. 1)Cell Potential E o & Balanced Equation 2)Direction of flow gives positive E o 3)Designation of anode and cathode e - flow An. Ox. to Red. Cat. 4)Nature of all the components: electrodes and ions 26

27. Schematic of Galvanic Cell Involving Half-Reactions 27

28. Practice Completely describe the galvanic cell based on the following half-reactions under standard conditions. Include line notation, E o cell and a labeled sketch of the cell. Cu +2 + 2e -  Cu E o = 0.34V Fe +2 + 2e -  Fe E o = -0.44V 28

29. Speaking of Electricity 29

30. Potential, Work and  G q is charge Work is viewed as being done by system Work flow out is minus sign When cell produces a current, E o cell is Positive therefore E o & w have opposite signs Charge is measured in Coulombs(C) Solve the equation for w 30

31. Maximum work in a cell is equal to the maximum cell potential, theoretically The charge on 1 mole of electrons is a constant called a Faraday (F) 1F = 96,500 C/mol e - 31

32. if E o 0 not spontaneous, no current if E o > 0, then  G o < 0 spontaneous Under standard conditions: Calculate  G o for the following reaction: Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq) Fe +2 (aq) + 2e -  Fe(s) E o = - 0.44 V Cu +2 (aq)+2e -  Cu(s) E o = 0.34 V Can also use the equation to predict spontaneity, if  G o > 0 the process will not occur, Study Ex. 17.4 p. 803 32

33. Cell Potential and Concentration Qualitatively - Can predict direction of change in E o from LeChâtelier. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E o cell = 0.48V Predict if E o original cell will be > or < E o cell for the following conditions: 1.if [Al +3 ] = 2.0 M and [Mn +2 ] = 1.0 M 2.if [Al +3 ] = 1.0 M and [Mn +2 ] = 3.0M 33

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35. The Nernst Equation  G =  G o +RTln(Q) -nF E = -nF E o + RTln(Q) (RT/F = 0.02567 @ 25 o C) Calculate E for the following reaction: 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E o = 0.48 V [Mn 2+ ] = 0.50M & [Al 3+ ] = 1.50M Always have to figure out n by balancing. If concentration can gives voltage, then from voltage we can tell concentration. 35

36. The Nernst Equation As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and E o cell = 0 which is a “dead battery” Therefore 36

37. Batteries are Galvanic Cells Car batteries are lead storage batteries. – Pb + PbO 2 + H 2 SO 4  PbSO 4 (s) + H 2 O 37

38. Common Dry Cell Zn + NH 4 + + MnO 2  Zn +2 + Mn 2 O 3 + NH 3 + H 2 O 38 Alkaline version uses KOH and NaOH in paste instead of NH 4 Cl

39. NiCad NiO 2 + Cd + 2H 2 O  Cd(OH) 2 +Ni(OH) 2 39 NiCad batteries can be recharged multiple times before the battery is no good.

40. Corrosion Rusting - spontaneous oxidation. Most structural metals have reduction potentials that are less positive than O 2 Reduction Potentials: – Fe +2 +2e -  Fe E o = -0.44 V – O 2 + 2H 2 O + 4e -  4OH - E o = 0.40 V – Al +3 + 3e -  Al E o = 0-1.66 V 40

41. Corrosion of Iron Oxid.Fe  Fe +2 +2e - E o = 0.44 V Red.O 2 + 2H 2 O + 4e -  4OH - E o = 0.40 V Fe +2 + O 2 + H 2 O  Fe 2 O 3 + H + Reaction happens in two places. 41

42. Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coat – Has a lower reduction potential, so it is more. easily oxidized. Alloying with metals that form oxide coats. – Stainless Steel – Cr + Ni Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead. 42

43. Electrolysis Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating, charging a battery, producing aluminum, electrolysis of water 43

44. Electrolysis Calculating Plating 44 Calculate mass of Cu plated out when apply 10.0 amps for 30.0 minutes. 1.Since 1 amp = 1 C / s Solve for C C = 10.0 amps x 1800 s = 18,000 C 2.1F = 96,500C/mole of e - Solve for # of mole e - 18,000 C/96,500 = 0.187 moles e - 3.Each Cu +2 needs 2 e - to become an atom 0.187 moles e - x 1mol Cu/2mol e - = 0.0935 mol Cu 4.Multiple mol of Cu by Cu molar mass 0.0935 mol Cu x 63.55gCu/1 mol Cu = 5.94g Cu It is all Stoichiometry Last Carnegie Time p.834 #77