Presentation on theme: "Chapter 20 Principles of Reactivity: Electron Transfer Reactions"— Presentation transcript:
1Chapter 20 Principles of Reactivity: Electron Transfer Reactions Chem 106, Chapter 20Chapter 20 Principles of Reactivity: Electron Transfer ReactionsCopyright 2007, David R. Anderson
2Oxidation-Reduction Reactions Balancing oxidation-reduction reactions Balancing Redox Equations by the Oxidation Number Change Method (Other Stuff page)Voltaic (Galvanic) CellsCell constructionCell potentialEffect of concentration on cell potentialCommercial voltaic cellsElectrolytic CellsElectrolysisQuantitative aspects of electrolysis
3I. Oxidation-Reduction Reactions Review:2 Na + Cl2 2 NaCl(0) (0)(+1)(–1)oxidation = increase in oxidation number (loss of electrons)reduction = decrease in oxidation number (gain of electrons)e.g., Assign oxidation numbers to the species being oxidized and reduced in the following equation, and label the oxidizing agent and reducing agent.NaI + 3 HOCl NaIO3 + 3 HCl
4I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactionshard way: ion-electron (half-reaction) method (text)easy way: oxidation number change method (Web - Other Stuff)1. Determine the oxidation numbers of the species being oxidized and reduced (and make sure there are the same number of each on each side).2. Balance the changes in oxidation numbers by multiplying each species by the appropriate coefficient (i.e., balance the electrons gained and lost).3. Balance charges with: H+ in acidic solution OH– in basic solution4. Balance H (and O!) with H2O.
7II. Voltaic (Galvanic) Cells Produce electricity:chemical energy electrical energy2Na+ + 2Cl– 2Na + Cl2 DG >> 0requires input of electrical energy (electrolysis)2Ag+ + Ni 2Ag + Ni2+ DG < 0produces energybut with Ag+ and Ni in contact, we can’t generate electricity; electrons just flow from Ni to Ag+have to use a voltaic (galvanic) cell…
8II. Voltaic (Galvanic) Cells electrical deviceA. Cell constructione–anode(–)cathode(+)salt bridgeKCl in gelatinallows electrolytic conduction without mixing–+Cl–K+NiNi(NO3)2(aq)AgAgNO3(aq)Ni2+Ag+Ni Ni2+ + 2e–(oxidation)Ag+ + e– Ag(reduction)net: Ni + 2Ag+ Ni2+ + 2Ag
9II. Voltaic (Galvanic) Cells A. Cell constructionPotentiometer (voltmeter)measures cell potential (voltage or electromotive force)depends on:species in redox equationconcentrationstemperaturecell potential, E= actual cell voltagestandard cell potential, Eº= voltage at standard state25ºC1 atm pressure1 M concentrationNiAgNi2+Ag+salt bridge or porous partitionShorthand notation:Ni | Ni2+ || Ag+ | Ag
10II. Voltaic (Galvanic) Cells B. Cell potential1. standard reduction potential, Eºa. Eº = tendency for a species to be reducedAg+ + e– Ag Eº(Ag+)Ni2+ + 2e– Ni Eº(Ni2+)can’t measure directlyCan only measure difference in a voltaic cell:2Ag+ + 2e– 2Ag Eº(Ag+)Ni Ni2+ + 2e– –Eº(Ni2+)(not 2 x)net: Ni + 2Ag+ Ni2+ + 2Ag Eºcell = Eº(Ag+) – Eº(Ni2+)= 1.05 VGeneral: Eºcell = Eº(species reduced) – Eº(species oxidized)
11II. Voltaic (Galvanic) Cells B. Cell potential1. standard reduction potential, Eºb. standard: hydrogen electrode at standard state1 atm H21 MH+Pt electrodeby definition:at standard state, the reduction2H+ + 2e– H2 has Eº = V (exactly)
12II. Voltaic (Galvanic) Cells B. Cell potential1. standard reduction potential, Eºc. standard reduction potentials- measure others against the standard hydrogen electrode:DVMAg1 MAg+H+1 atmH2Find:cathode: 2Ag+ + 2e– 2Aganode: H2 2H+ + 2e–Eºcell = 0.80 VSince Eºcell = Eº(Ag+) – Eº(H+)0.80 V = Eº(Ag+) – 0.00 V Eº(Ag+) = V(i.e., more easily reduced than H+)
13II. Voltaic (Galvanic) Cells B. Cell potential1. standard reduction potential, Eºc. standard reduction potentialsNi | Ni2+(1 M) || H+(1 M) | H2Find:cathode: 2H+ + 2e– H2anode: Ni Ni2+ + 2e–Eºcell = 0.25 VSince Eºcell = Eº(H+) – Eº(Ni2+)0.25 V = 0.00 V – Eº(Ni2+) Eº(Ni2+) = –0.25 V(i.e., less easily reduced than H+ or Ag+)
14II. Voltaic (Galvanic) Cells B. Cell potential1. standard reduction potential, Eºd. determining cell potentialsNi | Ni2+(1 M) || Ag+(1 M) | AgAg+ + e– Ag Eº = VNi2+ + 2e– Ni Eº = –0.25 Vmore easily reduced 2Ag+ + 2e– 2AgNi Ni2+ + 2e–Ni + 2Ag+ Ni2+ + 2AgEºcell = Eº(Ag+) – Eº(Ni2+)= V – (–0.25 V)= V
15II. Voltaic (Galvanic) Cells B. Cell potential2. spontaneity of redox reactionseasiest toreduce(strongestoxidant)hardest tooxidize(weakestreductant)Eº+2.87 V+0.80 V0.00 V–0.25 V–3.05 VF2 + 2e–Ag+ + e–2H+ + 2e–Ni2+ + 2e–Li+ + e–2F–AgH2NiLihardest toreduce(weakestoxidant)easiest tooxidize(strongestreductant)A species on the left will react spontaneously with a species on the right that is below it in the table.Or: The species with the more positive Eº will be reduced, and the species with the more negative Eº will be oxidized (Eºcell always > 0).
16II. Voltaic (Galvanic) Cells B. Cell potential2. spontaneity of redox reactionsGiven the two half reactions below, what is the net cell reaction? What is Eº? Draw a galvanic cell using these half cells and label the anode and cathode, their charges, and the direction electrons flow in the circuit.Fe3+ + 3e– Fe Eº = –0.04 VZn2+ + 2e– Zn Eº = –0.76 V
17II. Voltaic (Galvanic) Cells B. Cell potential3. cell potential and free energyDGsys = –wsurrw = # e–s energye–= coulombs joulescoulomb(= joules)= coulombs voltscoulombs = nF (n = # of moles of e–s in redox reaction)(F = 96,500 coulombs/mol) w = nFEDG = –nFEDGº = –nFEºEºcell > 0, DGº < 0, spontaneous (voltaic)Eºcell < 0, DGº > 0, nonspontaneous (electrolytic)
18II. Voltaic (Galvanic) Cells B. Cell potential3. cell potential and free energye.g., Zn | Zn2+(1 M) || Fe3+(1 M) | Fe2Fe3+ + 3Zn 2Fe + 3Zn2+ Eºcell = 0.72 VWhat is DGº for the cell?
19II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential1. Nernst equationDG = DGº + RTlnQ–nFE = –nFEº + RTlnQE = Eº –RTlnQnFat 25ºC: E = Eº –0.0592nlogQNernstequation
20II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential1. Nernst equatione.g., Ni | Ni2+ (0.05 M) || Ag+ (0.01 M) | AgNi + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V
21II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential2. applicationsa. measuring Kspe.g., AgCl(s)Ag+ + Cl– Ksp = [Ag+][Cl–]DVMAgNiAgCl(s)0.10 MCl–[Ag+] = ?1.0 MNi2+Find: E = 0.53 V; What is Ksp for AgCl?Ni + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V
22II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential2. applicationsb. measuring pHFind E = 0.94 V; What is pH?Ag1.0 MAg+1 atmH2lemon juice[H+] = ?DVMH2 + 2Ag+ 2H+ + 2Ag Eº = 0.80 V
23II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential2. applicationsA cell was constructed using the standard hydrogen electrode ([H+] = 1.0 M) in one compartment and a lead electrode in a 0.10 M K2CrO4 solution in contact with undissolved PbCrO4 in the other. The potential of the cell was measured to be 0.51 V with the Pb electrode as the anode. Determine the Ksp of PbCrO4 from this data. (Pb2+ + 2e– Pb Eº = –0.13 V)
24II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential2. applicationsA galvanic cell was constructed with a Cu electrode in a solution of 1.0 M Cu2+ in one compartment and a hydrogen electrode immersed in a sample of a soft drink. The cell potential was measured to be V. What was the pH of the soft drink? (Cu2+ + 2e– Cu Eº = V)
28III. Electrolytic cells A. Electrolysiselectrical energy chemical energye.g., NaCl(l) Na (l) + Cl2(g) DG >> 0cell: 2Na+ + 2Cl– 2Na + Cl2Eº = Eº(Na+) - Eº(Cl2)= (-2.71) - (1.36)= VDGº = kJ/mol
29III. Electrolytic cells A. ElectrolysisEe.g., NaCl(aq) ?Eºcathode: 2H2O + 2e– H2 + OH– V(reduction) Na+ + e– Na VH2O more easilyreduced than Na+Eºanode: Cl2 + 2e– 2Cl– V(oxidation) O2 + 4H+ + 4e– 2H2O VH2O more easilyoxidized than Cl–net: (2H2O + 2e– H2 + OH–) x 22H2O O2 + 4H+ + 4e–2H2O 2H2 + O2
30III. Electrolytic cells A. ElectrolysisEe.g., CuCl2(aq) ?Eºcathode: Cu2+ + 2e– Cu V(reduction) 2H2O + 2e– H2 + OH– VCu2+ more easilyreduced than H2OEºanode: Cl2 + 2e– 2Cl– V(oxidation) O2 + 4H+ + 4e– 2H2O VH2O more easilyoxidized than Cl–net: (Cu2+ + 2e– Cu) x 22H2O O2 + 4H+ + 4e–2Cu2+ + 2H2O 2Cu + O2 + 4H+
31IV. Electrolytic cells B. Quantitative aspects of electrolysis Units of charge: 1 faraday (F) = 1 mol e–s1 coulomb = 1 amp ·1 sec (A·s)experimentally: 1 F = 96,500 Ce.g., How many moles of Na and Cl2 are produced in the electrolysis of NaCl(l) when a current of 25 A is applied for 8.0 hours?
32IV. Electrolytic cells B. Quantitative aspects of electrolysis e.g., How long would it take to deposit 21.4 g of Ag from a solution of AgNO3 using a current of 10.0 A?