Chapter 20 Electrochemistry：20章电化学

1、ElectrochemistryChapter 20ElectrochemistryChemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. BurstenJohn D. BookstaverSt. Charles Community CollegeSt. Peters, MO 2006, Prentice Hall, Inc.ElectrochemistryAPRIL 8 REDOX REACTIONS HOW TO ASSIGN OXIDATION NU

2、MBERS HOW TO BALANCE REACTIONS IN ACID AND BASIC MEDIUM USING THE HALF REACTION METHOD HW : 13,15, 17,19 ElectrochemistryElectrochemical ReactionsIn electrochemical reactions, electrons are transferred from one species to another. Metals tend to lose electrons and are oxidized, non metals tend to ga

3、in electrons and are reduced.ElectrochemistryElectrochemistryElectrochemistryElectrochemistryOxidation NumbersIn order to keep track of what loses electrons and what gains them, we assign oxidation numbers.ElectrochemistryAssigning Oxidation NumbersElements in their elemental form have an ON= 0.The

4、oxidation number of a monatomic ion is its charge.Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Oxygen has an oxidation number of 2, except in the peroxide ion in which it has an oxidation number of 1. Hydrogen is +1 except in metal hydri

5、des when is 1 . Fluorine always has an oxidation number of 1.ElectrochemistryAssigning Oxidation Numbers The other halogens have an oxidation number of 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanionsThe sum of the oxidation numbers in a polyatom

6、ic ion is the charge on the ion.The sum of the oxidation numbers in a neutral compound is 0.ElectrochemistryOxidation and Reduction A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.ElectrochemistryOxidation and Reduction A spe

7、cies is reduced when it gains electrons.Here, each of the H+ gains an electron and they combine to form H2.ElectrochemistryOxidation and Reduction What is reduced is the oxidizing agent.H+ oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent.Zn reduces H+ by giving it elec

8、trons.Electrochemistry Zn added to HCl yields the spontaneous reactionZn(s) + 2H+(aq) Zn2+(aq) + H2(g). The oxidation number of Zn has increased from 0 to 2+. The oxidation number of H has reduced from 1+ to 0. Zn is oxidized to Zn2+ while H+ is reduced to H2. H+ causes Zn to be oxidized and is the

9、oxidizing agent. Zn causes H+ to be reduced and is the reducing agent. Note that the reducing agent is oxidized and the oxidizing agent is reduced.Electrochemistry Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. Conservatio

10、n of charge: electrons are not lost in a chemical reaction.Half Reactions Half-reactions are a convenient way of separating oxidation and reduction reactions.ElectrochemistryHalf Reactions The half-reactions forSn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)areSn2+(aq) Sn4+(aq) +2e-2Fe3+(aq) + 2e- 2Fe2+(a

11、q) Oxidation: electrons are products. Reduction: electrons are reactants. Loss of Gain of Electrons isElectrons is OxidationReductionElectrochemistryBalancing Equations by the Method of Half Reactions Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (dee

12、p purple). MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. The equivalence point is given by the presence of a pale pink color. If more KMnO4 is added, the solution turns purple due to the excess KMnO4.ElectrochemistryWhat is the balanced chemical equation?1. Write down the

13、 two half reactions.2. Balance each half reaction:a. First with elements other than H and O.b. Then balance O by adding water.c. Then balance H by adding H+.d. If it is in basic solution, remove H+ by adding OH-e. Finish by balancing charge by adding electrons. Multiply each half reaction to make th

14、e number of electrons equal. Add the reactions and simplify. Check!ElectrochemistryHalf-Reaction MethodConsider the reaction between MnO4 and C2O42 :MnO4(aq) + C2O42(aq) Mn2+(aq) + CO2(aq)ElectrochemistryBalancing Equations by the Method of Half Reactions Consider the titration of an acidic solution

15、 of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. The equivalence point is given by the presence of a pale pink color. If more KMnO4 is added, the solution turns purple due to the excess KMnO4.ElectrochemistryH

16、alf-Reaction MethodFirst, we assign oxidation numbers.MnO4 + C2O42- Mn2+ + CO2+7+3+4+2Since the manganese goes from +7 to +2, it is reduced.Since the carbon goes from +3 to +4, it is oxidized.ElectrochemistryFor KMnO4 + Na2C2O4:The two incomplete half reactions areMnO4-(aq) Mn2+(aq)C2O42-(aq) 2CO2(g

17、)2. Adding water and H+ yields 8H+ + MnO4-(aq) Mn2+(aq) + 4H2OThere is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left:5e- + 8H+ + MnO4-(aq) Mn2+(aq) + 4H2OElectrochemistryIn the oxalate reaction, there is a 2- charge on the left and a 0 charge on

18、the right, so we need to add two electrons:C2O42-(aq) 2CO2(g) + 2e-3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives:10e- + 16H+ + 2MnO4-(aq) 2Mn2+(aq) + 8H2O5C2O42-(aq) 10CO2(g) + 10e-Electrochemistry4. Adding gives:16H+(aq)

19、 + 2MnO4-(aq) + 5C2O42-(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g) 5. Which is balanced!ElectrochemistryBalancing in Basic Solution If a reaction occurs in basic solution, one can balance it as if it occurred in acid. Once the equation is balanced, add OH to each side to “neutralize” the H+ in the equation

20、and create water in its place. If this produces water on both sides, you might have to subtract water from each side.ElectrochemistryExamples Balance the following oxidation-reduction reactions:Cr (s) + NO3- (aq) Cr3+ (aq) + NO (g) (acidic)Al (s) + MnO4- (aq) Al3+ (aq) + Mn2+ (aq) (acidic)PO33- (aq)

21、 + Mn4- (aq) PO43- (aq) + MnO2 (s) (basic)1. H2CO (aq) + Ag(NH3)2+ (aq) HCO3- (aq) + Ag (s) + NH3 (aq)(basic)ElectrochemistryApril 9Section 20-3 20-4 Voltaic Cells Spontaneous reactions ELECTRODES POLARITIES SALT BRIDGE DRIVING FORCE- EMF STANDARD REDUCTION POTENTIALS Hw: 23, 25, 27, 29, 31, 33 (a,b

22、), 35ElectrochemistryVoltaic CellsIn spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.ElectrochemistryVoltaic Cells We can use that energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell.Elect

23、rochemistryVoltaic Cells A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode.ElectrochemistryVoltaic CellsOnce even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would st

24、op.ElectrochemistryVoltaic Cells Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.Cations move toward the cathode.Anions move toward the anode.ElectrochemistryVoltaic Cells In the cell, then, electrons leave the anode and flow throu

25、gh the wire to the cathode. As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.ElectrochemistryVoltaic Cells As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. The electrons are taken by the cati

26、on, and the neutral metal is deposited on the cathode.ElectrochemistryApril 10 Free energy and redox rx Cell EMF under nonstandard conditions Nerst Equation- Concentration cellsElectrochemistryElectromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only s

27、pontaneously flow one way in a redox reactionfrom higher to lower potential energy.ElectrochemistryElectromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential, and is designated Ecell.Electro

28、chemistryCell PotentialCell potential is measured in volts (V).One volt is the potential difference required to impart 1J of energy to a charge of 1 coulomb. (1 electron has a charge of 1.6 x 10-19 Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the

29、 electron through the external circuit.ElectrochemistryCell Potential or Electromotive Force (emf) The “pull” or driving force on the electrons. ELECTROMOTIVE FORCE EMF CAUSES THE ELECTRON MOTION!1 V = 1 JCElectrochemistryStandard Reduction PotentialsReduction potentials for many electrodes have bee

30、n measured and tabulated.ElectrochemistryStandard Hydrogen Electrode Their values are referenced to a standard hydrogen electrode (SHE). By definition, the reduction potential for hydrogen is 0 V:2 H+ (aq, 1M) + 2 e H2 (g, 1 atm)ElectrochemistryStandard Cell PotentialsThe cell potential at standard

31、conditions can be found through this equation:Ecell= Ered (cathode) Ered (anode)Because cell potential is based on the potential energy per unit of charge, it is an intensive property.Electrochemistry For the reaction to be SPONTANEOUSE has to be positiveElectrochemistryCell Potentials For the oxida

32、tion in this cell, For the reduction, Ered = 0.76 VEred = +0.34 VElectrochemistryCell PotentialsEcell= Ered(cathode) Ered(anode)= +0.34 V (0.76 V)= +1.10 VElectrochemistry Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. The oxidation of Zn w

33、ith the SHE is spontaneous. Changing the stoichiometric coefficient does not affect Ered. Therefore, 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V. Reactions with Ered 0 are spontaneous reductions relative to the SHE.Electrochemistry Reactions with Ered Cu2+(aq) + Fe2+(aq)Fe3+(aq) + e- - Fe2+(aq) Eo = 0.77

34、 VCu2+(aq) + 2 e- - Cu(s) Eo = 0.34 VReaction # 2 must be reversed.ElectrochemistryCell Potential Calculations Continued2 (Fe3+(aq) + e- - Fe2+(aq) Eo = 0.77 VCu(s) - Cu2+(aq) + 2 e- Eo = - 0.34 V2Fe3+(aq) + Cu(s) - Cu2+(aq) + 2Fe2+(aq)Eo = 0.43 VElectrochemistryOxidizing and Reducing AgentsThe grea

35、ter the difference between the two, the greater the voltage of the cell.REMEMBER TO REVERSE THE SIGN OF THE SPECIE THAT GETS OXIDIZED (THE ONE BELOW!)ElectrochemistryOxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. O.A. PULL electrons The strongest r

36、educers have the most negative reduction potentials. R.A. PUSH their electronsElectrochemistryExamples Sketch the cells containing the following reactions. Include Ecell, the direction of electron flow, direction of ion migration through salt bridge, and identify the anode and cathode:Cr3+ + Cl2 (g)

37、 Cr2O72- + Cl-Cu2+ + Mg (s) Mg2+ + Cu (s)IO3- + Fe2+ Fe3+ + I2Zn (s) + Ag+ Zn2+ + AgElectrochemistryFree EnergyG for a redox reaction can be found by using the equationG = nFEwhere n is the number of moles of electrons transferred, and F is a constant, the Faraday.1 F = 96,485 C/mol = 96,485 J/V-mol

38、ElectrochemistryFree Energy and Cell PotentialUnder standard conditions G = nFEn = number of moles of electronsF = Faraday = 96,485 coulombs per mole of electrons = 96,485 J/VmolE = V Units C= J/vElectrochemistryFind the free energy for the reaction of Zn + Cu2+ - Zn2+ + CuElectrochemistryAPRIL 11 E

39、ffect of concentration on cell EMF Nerst equation Chemical equilibrium and cell EMFElectrochemistryThe Nernst Equation A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the red

40、ox reaction. The Nernst equation relates emf to concentration usingand noting thatQRTGGlnQRTnFEnFElnElectrochemistryThe Nernst Equation This rearranges to give the Nernst equation: The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K: (Note that

41、 change from natural logarithm to base-10 log.) Remember that n is number of moles of electrons.QnFRTEElnQnEElog0592. 0ElectrochemistryCell EMF and Chemical Equilibrium A system is at equilibrium when G = 0. From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = Keq):0592. 0loglog0592.

42、00nEKKnEeqeqElectrochemistryCalculation of Equilibrium Constants for Redox Reactions At equilibrium, Ecell = 0 and Q = K.log( ).KnE00591at25 CElectrochemistryExamplesCalculate Ecell for the following reaction:Zn (s) + Cu2+ Zn2+ + Cu (s)Calculate Ecell if Zn2+ = 1.88 M and Cu2+ = 0.020 MCalculate Ece

43、ll for the following reaction: IO3- (aq) + Fe2+ (aq) Fe3+ (aq) + I2 (aq)1. Calculate Ecell if IO3- = 0.20 M, Fe2+ = 0.65 M, Fe3+ = 1.0 M, and I2 = 0.75 MElectrochemistryConcentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electr

44、odes. For such a cell,would be 0, but Q would not.Ecell Therefore, as long as the concentrations are different, E will not be 0.ElectrochemistryConcentration Cells We can use the Nernst equation to generate a cell that has an emf based solely on difference in concentration. One compartment will cons

45、ist of a concentrated solution, while the other has a dilute solution. Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq). The cell tends to equalize the concentrations of Ni2+(aq) in each compartment. The concentrated solution has to reduce the amount of Ni2+(aq) (to Ni(s), so must be the cathode.El

46、ectrochemistryConcentration CellsElectrochemistryThe Nernst EquationWe can calculate the potential of a cell in which some or all of the components are not in their standard states. EE0.0591n log QElectrochemistryPROBLEM Use the Nernst equation to determine the (emf) at 25oC of the cell. Zn(s)Zn2+(0

47、.00100M Cu2+(10.0M)Cu(s). The standard emf of this cell is 1.10 V.ElectrochemistryQuantitative Aspects of ElectrolysisWe want to know how much material we obtain with electrolysis.1 Ampere is 1 Coulomb per second (A= C/s)1 mole of electrons = 96,485 C = 1 FaradayUse balanced half-equation to equate

48、moles of substance to moles of electronsMolar mass (in g) = 1 mole of substanceElectrochemistryStoichiometry of Electrolysis-How much chemical change occurs with the flow of a given current for a specified time? current and time quantity of charge moles of electrons moles of analyte grams of analyte

49、1 amp = 1 C/s C=amp x sec (convert time to sec)F = 96,485 C/ mol of e-ElectrochemistryElectrolytic CalculationsHow many grams of copper can be plated out when a current of 10.0 amps is passed through a Cu2+solution for 30.0 minutes?Electrochemistry(10.0 C/s)(30.0 min)(60 s/1 min)(1 mol e-/96,485 C)(

50、1 mol Cu/2 mol e-)(63.5 g/1 mol) = 5.94 g Cu ElectrochemistryElectrolytic CalculationsHow long must a current of 5.00 A be applied to a solution of Ag1+ to produce 10.5 g of silver?Electrochemistry (10.5 g Ag)(1 mol/107.86 g)(1 mol e-/1 mol Ag) (96,485 C/1 mole e-)(1 s/5.00 C)(1 min/60.0s) = 31.3 mi

51、n ElectrochemistryPRACTICE PROBLEMSA car bumper is to be electroplated with Cr from a solution of Cr3+. What mass of Cr will be applied to the bumper if a current of 0.50 amperes is allowed to run through the solution for 4.20 hours?1. What volume of H2 gas (at STP) will be produced from the SHE aft

52、er 2.56 minutes at a current of 0.98 amperes?Electrochemistry What volume of F2 gas, at 25C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of potassium metal is produced? At which electrode does each reaction occur?ElectrochemistryElectroche

53、mical Electrolytic Electrochemical Electrolytic Spontaneous Nonspontaneous Energy released Energy absorbed Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq) Electrochemical cell - chemical energy to electrical energy. Electrolytic cell - electrical energy to chemical energy.ElectrochemistryAPRIL 12 Applications of

54、Oxidation-Reduction ReactionsElectrochemistry Redox Applications - Batteries and fuel cells Corrosion prevention HW 47, 51 a, 59, 63ElectrochemistryBatteriesA battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.ElectrochemistryBatteriesElectrochemistry17_370H2

55、SO4electrolytesolutionAnode (leadgrid filled withspongy lead)Cathode (leadgrid filled withspongy PbO2)+A lead storage battery consists of a lead anode, lead dioxide cathode, and an electrolyte of 38% sulfuric acid.ElectrochemistryLead Storage Battery Anode reaction: Pb(s) +H2SO4(aq) - PbSO4(aq) + 2H

56、+(aq) + 2e- Cathode reaction: PbO2(s)+H2SO4(aq+ 2e-+2H+(aq)-PbSO4(aq)+2H2O(l) Overall reaction: Pb(s)+ PbO2(s) + 2H2SO4(aq)-PbSO4(aq)+ 2H2O (l) Electrochemistry17_371Cathode(graphite rod)Anode(zinc inner case)Paste of MnO2,NH4CL, andcarbonCommon dry cell and its components.ElectrochemistryAlkaline B

57、attery Anode: Zn cap:Zn(s) Zn2+(aq) + 2e- Cathode: MnO2, NH4Cl and C paste:2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + 2H2O(l) The graphite rod in the center is an inert cathode. For an alkaline battery, NH4Cl is replaced with KOH.Electrochemistry Anode: Zn powder mixed in a gel:Zn(s) Zn2+(aq)

58、+ 2e- Cathode: reduction of MnO2.ElectrochemistryElectrochemistry17_372Cathode (steel)Anode (zinc container)Solution of HgO (oxidizingagent) in a basic medium (KOHand Zn(OH)2)InsulationMercury battery used in calculators.ElectrochemistryFuel Cells Direct production of electricity from fuels occurs i

59、n a fuel cell. On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen:2H2O(l) + O2(g) + 4e- 4OH-(aq) Anode:2H2(g) + 4OH-(aq) 4H2O(l) + 4e- ElectrochemistryFuel Cells. . .galvanic cells for which the reactants are continuously supplied.2H2(g) +

60、 O2(g) 2H2O(l)anode: 2H2 + 4OH 4H2O + 4ecathode: 4e + O2 + 2H2O 4OHElectrochemistryHydrogen Fuel CellsElectrochemistry17_369Reference solution ofdilute hydrochloric acidSilver wire coated withsilver chlorideThin-walled membraneIon selective electrodes are glass electrodes that measures a change in p