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Chapter 19 Electrochemistry

Chapter 19 Electrochemistry. HW 19 is due before 11:50 pm, Wednesday, 6/5/2013. Final Exam Monday 6/10/2013; 10:30-12:30 SL 150. End-of-Chapter Problems pp 817 - 826, Ch 19. 1 2 3 4 6 7 8 14 15 17 18 19 20 21 22 25

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Chapter 19 Electrochemistry

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  1. Chapter 19Electrochemistry HW 19 is due before 11:50 pm, Wednesday, 6/5/2013. Final Exam Monday 6/10/2013; 10:30-12:30 SL 150

  2. End-of-Chapter Problems pp 817 - 826, Ch 19 1 2 3 4 6 7 8 14 15 17 18 19 20 21 22 25 31 33 39 43 49 51 57 61 65 67 69 75 81 85 93 95

  3. I. Introduction A. Terms Electrochemistry - deals with the relationship between redox chemical reactions and electricity. Redox Reactions - Chemical reactions which involve a transfer of electrons. Electrochemical Cell - a system consisting of electrodes that dip into an electrolyte in which a reaction either generates oruses an electric current. Cell 1: Voltaic Cell (battery) - An electrochemical cell in which a spontaneous reaction generates an electric current. Cell 2: Electrolytic Cell - An electrochemical cell in which an electric current forces a reaction to occur.

  4. I. IntroductionB. review - Redox Reactions - Two General Classes of Chemical Reactions: Redox & Non-Redox - Redox reactions = ones in which electrons are transferred & two processes are associated with redox reactions: 1) Electrolysis: Electricity forces a chemical reaction to occur. Example: 2 H2O + electricity 2 H2(g) + O2(g) 2) Battery or a Voltaic Cell: spontaneous redox reaction generates electrical potential. Example (1 M Cu+2 & Zn) Zno + Cu+2 Zn+2 + Cuo + electricity(1.1 volts)

  5. B. Review - Redox Reactions • Oxidation = loss of electrons. Know • Reduction = gain of electrons. Know • Both must occur together, but we frequently break them up into two half reactions. Example: Zno + Cu+2 -----) Zn+2 + Cuo Oxidn: Zno -----) Zn+2+ 2 e- Redn: Cu+2+ 2 e- -----) Cuo a) Oxidizing Agent causes oxidn. & is reduced (Cu2+) b) Reducing Agent causes redn. &is oxidized (Zno) - Classes of Redox Reactions: Combination, Decomposition, Displacement, Combustion

  6. B. Review - Oxidation Numbers (ON) • Definition: The charge that an atom (or group of atoms) would have if it were ionic • Usefulness: 1) Naming variably charged compounds. Example: FeCl3 = Iron(III)chloride 2) Determining if a reaction is redox. 3) Balancing a redox reaction. • Rules for Determining ON: Note - priority of rules are important; lowest # rule takes precedent when a conflict occurs. Know Rules

  7. B. Review - Oxidation Numbers (ON). Rules 1) The sum of ONs must add up to give the charge 2) The ON of a neutral element by itself is 0 3) Group 1, 2, 13 elements in ionic form are +1, +2 & +3 respectively (H -1 when combined with 1, 2, 13) 4) F is -1 5) O is -2 (O is -1 when found as a peroxide, O2-2) 6) Cl, Br, I are -1 (when together, most electronegative rules)

  8. B. Review of ON Examples: Determine the ON of singleN in each: (Treat as a Quiz & Complete in < 5 minutes) N2 N3-1 NH3 N2O NO NF3 N2O4 NO3-1 0 -1/3 -3 +1 +2 +3 +4+5 - 1) Which one would be most reactive and why? - 2) Methanol/Nitric Acid experience in the lab. - 3) HClO4 - What is ON of Cl? Comments on perchloric acid.

  9. B. Review – ON Examples - The top two are redox: 1) Mg + Cl2 MgCl2 2) CH4 + 2 O2 CO2 + 2 H2O 3) HCl + NaOH H2O + NaCl - In any redox reaction you should be able to determine: The ON for each element Which reagent has been oxidized Which reagent has been reduced Which reagent is the oxidizing agent Which reagent is the reducing agent What are the two half reactions How many moles of e- are transferred in each half rxn

  10. II. Balancing Redox Rxns A. acidic media - Many are difficult and require a systematic method: Note that you must first determine if the reaction occurs in acid or base. - Steps: 1-3 for acidic media &Steps 1-6 for basic media 1.Assign ON’s & split into two half reactions 2. Complete & balance each half reaction: a. balance atom undergoing redox b. balance O by adding H2O’s c. balance H by adding H+’s d. balance ionic charge (not ON) by adding e-’s 3. Combine half reactions and finish (if in acidic media): a. 1st - multiply ½ Rxns by # to balance e-’s lost & gained. b. 2nd - add ½ Rxns & cancel same species on both sides.

  11. II. Balancing Redox Rxns B. Basic media 4.If in basic media add OH- equal to # H+to both sides. 5. Convert H+ & OH-on same side to H2O& cancel H2O’s if needed. 6. Both elements & charge should balance

  12. II. Balancing Redox Rxns C. Example 1 in acidZn + NO3- + H+ -----) Zn+2 + NH4+ + H2O 1)Zn0 -----) Zn2+ONs:0 ---) +2 for Zn (Oxidn.) NO3- -----) NH4+ONs: +5 ---) -3 for N (Redn.) 2) Zn0 -----) Zn2+ + 2 e-(balance redox atom & NO3- -----) NH4+ + 3 H2O balance O with H2O) 10 H+ + NO3- -----) NH4+ + 3 H2O (balance H with H+) 10 H+ + NO3- + 8 e------) NH4+ + 3 H2O (bal. Charge w e-) 3) 4[Zn -----) Zn2+ + 2 e-](8e- lost) 1 [10 H+ + NO3- + 8 e------) NH4+ + 3 H2O](8e- gained) 4 Zn + 10 H+ + NO3- + 8 e----) 4 Zn2+ + 8 e- + NH4+ + 3 H2O 4 Zn + 10 H+ + NO3- -----) 4 Zn2+ + NH4+ + 3 H2O

  13. II. Balancing Redox Rxns C. Example 2 (basic)Pb(OH)3- + ClO- -----) PbO2 + Cl- 1) ClO- -----) Cl- +1 ---) -1 on Cl (rdn) Pb(OH)3- -----) PbO2 +2 ---) +4 on Pb (oxn) 2) 2 H+ + ClO- + 2 e- -----) Cl- + H2O Pb(OH)3- -----) PbO2 + H2O + H+ + 2e- 3) 2H+ + ClO- + Pb(OH)3- + 2e- -----) Cl- + 2H2O + H+ + PbO2 + 2e- H+ + ClO- + Pb(OH)3- -----) Cl- + 2H2O + PbO2 4) OH-+ H+ + ClO- + Pb(OH)3- ---) Cl- + 2H2O + PbO2 + OH- 5) H2O + ClO- + Pb(OH)3- ---) Cl- + 2 H2O + PbO2 + OH- 6)ClO- + Pb(OH)3- Cl- + H2O + PbO2 + OH-

  14. II. Balancing Redox Rxns C. Example 3MnO4- + S2O32- -----) Mn2+ + SO42- (acidic) 1)a)MnO4- ----) Mn2+ [+7 ---) +2 Mn Redn] b)S2O32- ----) SO42- [+4 ---) +6 per S Oxn] 2)a)5e- + 8H+ + MnO4- ----) Mn2+ + 4H2O (x 8) b)1S2O32- -----) 2SO42- 5H2O + S2O32- -----) 2SO42- + 10H+ 5H2O + S2O32- -----) 2SO42- + 10H+ + 8e-(x 5) 3) 40e- + 64H+ + 8MnO4- + 25H2O + 5S2O32- ---) 8Mn2+ + 32H2O +10SO42- + 50H+ + 40e- 14 H+ + 8 MnO4- + 5 S2O32-----) 8 Mn2+ + 7 H2O + 10 SO42-

  15. II. Balancing Redox Rxns C. Example 4MnO4- + I- -----) MnO2 + IO3- (basic) 1) MnO4- -----) MnO2 +7 ---) +4 Mn I- -----) IO3- -1 ----) +5 I 2) 4H+ + MnO4- + 3e- -----) MnO2 + 2H2O 3H2O + I- -----) IO3- + 6H+ + 6e- 3) 2 [4H+ + MnO4- + 3e- -----) MnO2 + 2H2O] 1 [3H2O + I- -----) IO3- + 6H+ + 6e-] 8H+ + 2MnO4- + 6e- + 3H2O + I- ----) 2MnO2 + 4H2O + IO3- + 6H+ + 6e- 2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3- 4) 2OH- + 2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3-+ 2OH- 5) 2H2O + 2MnO4- + I- ----) 2MnO2 + H2O + IO3-+ 2OH- H2O + 2MnO4- + I- ----) 2MnO2 + IO3-+ 2OH- 6) Double check balancing of atoms & charge.

  16. III. Voltaic Cells A. Introduction - A voltaic cell (battery) is an electrochemical cell in which a spontaneous redox reaction generates an electric current. - It consists of two half cells which are connected: Oxidation half cell consists of electrode (anode = - ) & electrolyte. Reduction half cell consists of electrode (cathode = + ) & electrolyte. Internal connection of a salt bridge that allows the flow of ions but prevents mixing of electrolytes from two half cells. External connection of anode to the cathode. Electrons travel from anode to cathode.

  17. B. Example Zn + Cu2+ -----) Zn2+ + Cu Zn Anode - Zn ---) Zn2+ + 2e- Cu Cathode +Cu2+ + 2e- ---) Cu Zn | Zn2+ (1.0 M) || Cu2+(1.0 M) | Cu

  18. C. Voltaic Cell Shorthand Notationsame example: Zn+2 & Cu2+ at 1.0 M Zn + Cu2+ -----) Zn2+ + Cu Zn | Zn2+ (1.0 M) || Cu2+(1.0 M) | Cu Anode (-) (oxidation) Zn ---) Zn2+ + 2e- Cathode(+) (reduction) Cu2+ + 2e- ---) Cu Know the following Conventions: Anode written on left Cathode written on right | = phase boundary || = salt bridge Zn = anode electrode Cu = cathode electrode , = separates different ions in the same half cell M = Molar concentration of ions This battery develops 1.1 volts at 25 oC

  19. C. Voltaic Cell Notation (ions at 0.1 M) Cd + 2Ag+ -----) Cd2+ + 2Ag Cd Anode - Cd ---) Cd2+ + 2e- Ag Cathode + 2Ag+ + 2e- ---) 2Ag Cd | Cd2+ (0.1 M) || Ag+ (0.1M) | Ag (this cell develops 1.2v)

  20. C. Voltaic Cell NotationAdditional Examples Pt | H2 | H+|| (Half Cell) Hydrogen electrode as anode(electrode on left) H2 -----) 2H+ + 2 e- Three phases: Pt solid, H2 gas, H+ solution ||Fe3+, Fe2+ | PtCathode Reaction (electrode on right) Fe3+ + e- -----) Fe2+ ||Cd2+ | CdCathode Reaction (electrode on right) Cd2+ + 2e- ------) Cd Tl | Tl+|| Sn2+ | Sn 2 Tl + Sn2+ -----) 2 Tl+ + Sn 2 Tl ----) 2 Tl+ + 2 e- (anode) Sn2+ + 2e- ---) Sn (cathode)

  21. D. Work & Voltage of a Cell • Much of the early work on cell physics & chemistry came from Michael Faraday - 1791 to 1867 • He was responsible for uncovering ONs, terminology (anode, cathode, etc), benzene, electromagnetism, Faraday’s Laws & Faraday’s constant. • 1 Faraday = Charge on 1 mol of electrons = 96,485 c

  22. D. Work & Voltage of a Cell • Work done in a redox rxn: ∆G = - n x F x Ecell • n= moles of e- transferred • Ecell = voltage of the cell • F = Charge of 1 mol of e- = 1 Faraday = 96,485 coulombs/mol (c/m) • Notes: ∆G = - n x F x Ecell • Units: Joules = mol x (coul/mol) x volt = coul x volt = J • 1 coulomb x 1 volt = 1 joule • Since batteries are spontaneousredox rxns the sign is -:∆G = - nFEcell • Free Energy, ΔG = maximum useful work done • ΔG = - nFEcell • At 1M concentrations, 1 atm pressure & 25 oC: ΔG = ΔGo • ΔGo = - nFEocell

  23. E. Standard Cell Voltages - some notes - Can measure the cell [voltage/potential/EMF] (EMF = electromotive force) using a voltmeter; and can calculate theoretical voltages from equations & tables. - The voltage (E) developed by a cell is the sum of the oxidation and reduction potentials: Ecell = Eoxidation + Ereduction(note: you supply signs) - Note: can’t separate the two half reactions in the lab; so, how calculate E r & O? By setting an arbitrary standard and comparing all other half cell reactions to this standard. - Arbitrary Standard is the hydrogen half cell reduction under standard conditions (1 M, 1 atm, 25oC) ; a value of 0.0000 volts was assigned.

  24. E. Standard Cell Voltages H2 Electrode || H+ (1.0 M) | H2 (1.0 atm) | PtEo = 0.0000 volts at 25 oC 2H+ + 2e- H2(1 MHCl & 1 atm H2)

  25. E. Standard Cell Voltages - If we couple the hydrogen electrode under standard conditions with another half reaction and measure the Ecell , then we can determine the E for the unknown half reaction. Eocell = Eo1/2 rxn + Eohydrogen electrode = Eo1/2 rxn+ 0.000 = Eo1/2 rxn - Example:Zn | Zn2+ (1.00M) || H+ (1.00M) | H2 (1.00atm) | Pt Zn + 2H+ Zn2+ + H2Eocell measured as 0.76 volts Eocell = EoZn + EoH2 = EoZn + 0.000; EoZn = Eocell = 0.76 v Notes: 1) Convention: ½ rxns are tabulated as reductions. See Table 19.1, pg 792; if need half rxn as oxidn, then reverse the sign. Example: Zn+2 + 2 e- (---) Zn Eo = - 0.76 v; reverse sign if an oxidn. So, you need to supply the correct sign in problems. 2) Eo1/2 value = constant; defined under specific conditions.

  26. E. Standard Cell Voltages – Calculate EoLi/Iodine battery • Eocell = Eoreduction+ Eooxidation - Use Table 19.1 & must change sign for oxidation. - Battery Rxn: I2 + 2Li -----) 2Li+ + 2I-Eocell = ? - Table values: 2Li+ + 2e- ---) 2Li Eo = -3.04 v I2 + 2e- -----) 2I-Eo = 0.54 v Eocell = 0.54 + 3.04 = 3.58 volts (had to reverse Li ½ rxn) Li Iodine battery – can reverse the rxn by adding electricity to recharge battery - salt bridge

  27. E. Standard Cell Voltages • Use of Standard Cell Voltages (table 19.1, pg 792) 1) Can tell strength of reducing/oxidizing agents: Reaction with largest + volts wants to go most 2) Can determine the Eocell from Standard ½ Cell Voltages: Eocell = Eo ½ reduction+ Eo ½ oxidation 3) Can tell which way a redox rxn wants to go: When mix chemicals, will go so that Ecell is largest 4) Can be used to calculate ΔG: ΔGo = - nFEocell 5)Can be used to calculate Keq: Eo = (2.30 RT / nF) Log K 6) Used to calculate Ecellnot under std conditions.

  28. F. Calculations Involving Eo 1) Can tell strength of reducing/oxidizing agents: Reaction with largest positive Eo1/2 value wants to go most Mg2+ + 2e- ---) Mg Eo1/2 = -2.4 F2 + 2e- ---) 2F- Eo1/2 = 2.9 F2 wants to be reduced more than Mg+2 2) Can determine the Eocell from Standard ½ Cell Voltages: Given: Al | Al3+ || Cr3+ | Cr (Direction mandated by battery rxn) Al ---) Al3+ + 3e- Eo = 1.66 vCr3+ + 3e- ----) Cr Eo = - 0.74 V Eocell = Eoox + Eored = 1.66 + (-0.74) = 0.92 V 3) Can tell which way a redox reaction will spontaneously go. Goes such that Eocell is the largest positive value. Zn2+ + Cu ---) Zn + Cu2+ Eo = - 1.10 v = Not Spontaneous Zn + Cu2+ ---) Zn2+ + Cu Eo = + 1.10 v = Spontaneous

  29. F. Calculations Involving Eo 4) Free Energy from Eo & Eo from ΔGo ΔGo = - nFEocellEo cell = -ΔGo/nF Example:Calculate the ΔGo in J for: Zn | Zn2+ (1.0M) || Ag+(1.0M) | Ag Zn + 2Ag+ ----) 2Ag + Zn2+ Zno -----)Zn+2 + 2e- Eo1/2 = +0.76 v 2Ag+1 + 2e- -----) 2Ago Eo1/2 = +0.80 v Eo = 0.76 + 0.80 = 1.56 v Note: 2 m of e- transferred, but Eo1/2 = 0.80v for Ag (Eocell is a constant) ΔGo = - nFEocell = - 2.00m x 9.65x104c/m x 1.56v = - 3.01x105 J Example: Calculate Eo for a rxn in which ΔGo = -409 kJ & n=2 Eo = -ΔGo/nF = - (-409x103J) / (2m x 9.65x104c/m) = 2.12 J/c = 2.12 v (1 coulomb x 1 volt = 1 joule; 1 volt = 1 joule/coulomb)

  30. F. Calculations Involving Eo5) Get Keq from Eo ΔGo = - nFEo& ΔGo = - RTLn K (from last chapter) - nFEo = - RT Ln K -nFEo = - 2.30 RT Log K Eo = (2.30 RT) Log K @ 298 K (25 oC) 2.30RT/F = 0.0592 V nF Eo = (0.0592) Log Keq(true at 25 oC) n Example: Calc K at 25oC forZn + Cu2+ Zn2+ + Cu Eo = 1.10 V Eo = (0.0592) Log K1.10 = (0.0592) Log K n 2 Log K = 37.2 K = 1037.2K = 1.58 x 1037

  31. F. Calculations Involving Eo 6) Cell EMF for Nonstandard Conditions (not 1 M or not 25 0C) From: ΔG = - nFE & ΔG = ΔGo + RT Ln Q (last chapter) Get: E = Eo - RT/nF Ln Q = Eo - 2.30RT/nF Log Q E = Eo - 0.0592/n X Log Qat 25 oC = NERST EQUATION (know) - Can now calculate E at various temperatures & concentrations. - n = # moles of electrons in the balanced redox reaction - Q = same form as Keq except initial M used - not equilibrium concentrations. Example:Calculate E at 25 oC for Zn | Zn2+(1.00x10-5) || Cu2+(0.100) | Cu Zn (s) + Cu2+ (------) Zn2+ + Cu (s) Eo = 1.10 V Q = [Zn2+]/[Cu2+] = 1.00x10-5M / 0.100 M = 1.00x10-4 E = Eo - 0.0592/n X Log Q = 1.10 - 0.0592/2 X Log (1.00x10-4) E = 1.10 - (-0.12) = 1.22 V

  32. IV. Electrolytic Cells, Examples - Place inert electrodes in a solution and force a non-spontaneous redox rxn to occur by running electricity through the solution with a battery = electrolysis. 1. Molten NaCl - at cathode: 2Na+ + 2e- ---) 2Na (Eo = -2.7 v) - at anode: 2Cl- ---) Cl2 + 2e-(Eo = -1.4 v) 2NaCl(s) + elec. -----) 2Na(s) + Cl2 (g)(add > 4.1 v) 2. Aqueous NaCl - at cathode: 2H2O + 2e- ---) H2 + 2OH- - at anode: 2Cl- ---) Cl2 + 2e- 2H2O + 2Cl- + elec. -----) H2 + 2OH- + Cl2 (add > 2.2 v) 3. Water + non reacting Electrolyte(like Na2SO4) - at cathode: 4H2O + 4e- -----) 2H2(g) + 4OH- - at anode: 2H2O -----) 4e- + 4H+ + 1O2(g) 2H2O + elec. ----) 2H2(g) + 1O2(g)(add > 2.1 v)

  33. IV. Electrolytic Cells Electroplating - process of depositing a metal on an electrode. Can be used for analysis, beauty, or for protective coating. - Can be used for quantitative analysis (one with largest + Eo will plate 1st) : Example: Analysis of Cu in brass (mix of Sn, Pb, Cu) Dissolve brass in HNO3; add two Pt electrodes to the HNO3 solution; apply > +0.34 volts. Cu forms at the cathode: Cu2+ + 2e ----) Cuo Sn+2 Eo = -0.14 v Pb+2 Eo = -0.13 v Cu+2 Eo = +0.34 v Weigh the cathode before and after the process to quantitatively determine the amount of Cu in the brass. - Note: 1. The metal ion with the largest + reduction potential will plate out on the cathode first. 2. Can also determine amount of Cu by measuring current (amps) and time necessary to plate out all of the Cu (see next section).

  34. IV. Electrolytic Cells - Stoichiometry - Note that electrolysis can be used for stoichiometric calculations. - Equations and values to know: coulombs = amperes x seconds c = a x sec a= c/sec 1F = Charge on 1 mole of e- = 96500 c Example: How many g of Cu are present in the previous example if a current of 0.852 amp is needed for 600 seconds in order to remove all of the blue color? 1Cu+2 + 2e- ------) 1Cu(s) c = amps x sec = 0.852 a x 600 s = 511 coulombs 511 c x 1 mole e- x 1 mole Cu x 63.5 g Cu= 0.168 g Cu 96500 c 2 mole e- 1 mole Cu - Note: Can also weigh a Pt cathode before & after; would gain 0.168 g.

  35. V. More Applications of Electrochemistry • pH Electrode: Ecell = Ereference electrode + Eindicator electrode

  36. V. Few More Applications of Electrochemistry • pH electrode: AgCl(s) + H+Inside (---) Ag(s) + H+outside + Cl- [Cl-] = 1 M n = 1 [H+Inside ] = constant Ecell = Eo - (0.0592/n) Log Q = Eo - 0.0592 Log {[H+out]x[Cl-] / [H+Ins]} Ecell = K’ - 0.0592 Log [H+out] = K’ + 0.0592 pH Ecell = 0.0592 pH + K’ [1) Eqn a pH meter uses; 2) a plot of voltage vs pH will be a straight line with a slope of 0.0592; 3) when calibrating a pH meter, you are setting the slope = 0.0592 and the Y intercept = K’ ] • Other Ion Selective Electrodes (ISE) (once they understood correct theory for pH electrode in 1952, then started making additional ISE) - F-, Cl-, Ca+2, NO3-, NH4+, etc - Note: very few interferences with H+ electrode; a few interferences with F- electrode; many interferences for the rest.

  37. V. Few More Applications of Electrochemistry IE RE • Example: Fluoride Selective Electrode LaF3 Crystal - Potential developed across Et = Constant - 0.0592 Log F-(at 25oC) F- F- - plot of Et versus Log [F-] = st line (Quant. Anal.) F- F- F- F- Potential interferences: OH-, Al+3, Fe+3, Ionic Strength OH- - Use acidic buffer in stds & unk solutions. Al+3,Fe+3 - add complexing agent to stds & unk solutions. Ionic Strength - add large amount of noninterfering ions to all solns. - Do all three by using high ionic strength buffer (with complx agent)

  38. V. Few More Applications of Electrochemistry • Electrochemical Detectors for Liquid Chromatography - Control the voltage of an electrode system in an LC mobile phase. If compound of interest goes past the cell & can be oxidized at the selected voltage, then will get a current which is proportional to the concentration of the compound. Can detect pg amounts of chemicals.

  39. V. More Applications of Electrochemistry • Batteries Example: Car Battery Pb + HSO4- -----) PbSO4 + H+ + 2e- PbO2 + 3H+ + HSO4- + 2e- -----) PbSO4 + 2H2O PbO2 + Pb + 2H+ + 2HSO4- (-----) 2 PbSO4 + 2H2O One cell develops 2 V; so, a six cell battery connected in series will develop 12 Volts

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