《無機化學簡明教程》課件-Chapter 6 Redo

Chapter6OxidationandReduction(RedoxReaction)

Fourissues：1.Basicknowledge2.StandardReductionPotentials3.IntroductionofVoltaicCells4.Electrolysisand

ElectrolyticCells1Basicknowledge1-1

Concepts1-2

BalancingRedoxEquations1-3

Voltaiccells1-3-1Whatkindsofredoxreactionscanbeharnessedtoproduceelectricalenergy?1-3-2ElectrodeTypesandCellNotation1-3-3CellPotentials

1-1ConceptsZn+Cu2+=Zn2++CuC6H12O6+6O2=6CO2+6H2O(1)OxidationNumber假設(shè)分子中成鍵的電子都歸電負性較大的原子，而得到的某元素的一個原子所帶的形式電荷數(shù)（表觀電荷數(shù)）。Rulesforassigningoxidationnumber：u

Theoxidationnumberofanelementinanelementarysubstanceis0.u

Generally,incompounds,theoxidationnumberofHis+1，theoxidationnumberofOis-2,theoxidationnumberofhalideFis-1,theoxidationnumberofGroup1metals(IA)is+1,theoxidationnumberofGroup2elements(IIA)is+2.u

Thesumoftheoxidationnumbersofalltheatomsinaneutralspeciesis0;inanion,itisequaltothechargeofthation.Eg.S4O62-,oxid.no.S

4+2

6=-2,oxid.no.S=2.5CrO5oxid.no.Cr+2

5=0;oxid.no.Cr=-10Thus,theoxidationnumberofcanbeintegersorfractions.(2)DefinitionofOxidationandReduction：Oxidationisdefinedasanincreaseinoxidationnumber,andreductionasadecreaseinoxidationnumber.Aredoxreactioncanbedefinedasareactionthathasoxidationnumberchanges.Inaredoxreaction,thereareusuallyatleasttworeactants.Onewithoxidationnumberdecreaseisreferredtoastheoxidizingagent,theotherwithoxidationnumberincreaseisreferredtoasthereducingagent.Zn–2e

Zn2+

（Oxidation,Znisreducingagent）Cu2++2e

Cu(reduction,Cuisoxidizingagent)Thesetworeactionequationsarecalledhalfequations.Obviously,aredoxreactionequationsiscomposedoftwohalfequations.

lSelfRedoxreactions:theoxidizingandreducingagentisthesamecompound.

歧化反應(yīng):theoxidizingandreducingagentisthesameelementinasubstance.2KClO3=2KCl+3O2Cl2+H2O=HCl+HClORedoxelectronicpair：同一元素的兩種不同的氧化態(tài)，構(gòu)成氧化還原電對。Ox/Redeg.Cu2+/Cu,Zn2+/Zn,H+/H2,Cl2/Cl-

Ox1+Red2

Ox2+Red1

1-2BalancingRedoxEquationshalfequationsmethodorion-electronmethod.Procedures：u

Writethenetionicequationsinsteadofcompleteionicequations.Splittheequationintotwohalf-equations,oneoxidationandonereduction.u

Balancetheatomsofthetwohalf-equations,firstwithrespecttoatomsthathaveoxidationnumberchangesandthenwithrespecttoHandO.Inacidicsolution,addH+(YoucannotaddOH-!)

andaddH2OonthesideoflessOatoms.Inbasicsolution,addOH-andH2O(YoucannotaddH+!).AddOH-onthesideoflessOatoms,andaddOH-onthesideofmoreHatomswhenOatomsareequal.u

Balancethechargeofthetwohalf-equationsFindtheleastcommonmultipleofthechargesofthetwohalf-equations,combinethemsoastomakethenumberofelectronsarinedinreductionequaltothenumberlostinoxidation.eg1.MnO4-+C2O42-

Mn2++CO2(Acidicsolution)MnO4-+8H++5e

Mn2++4H2O+)C2O42-

2CO2+2e2MnO4-+5C2O42-+16H+

2Mn2++10CO2+8H2Oeg2.FeS2+HNO3

Fe2(SO4)2+NO2FeS2+8H2O

Fe3++SO42-+16H++15e+)2H++NO3-+e

NO2+H2OFeS2+14H++15NO3-

Fe3++2SO42-+15NO2+7H2Oeg3.ClO-+CrO2-

Cl-+CrO42-

（Basicsolution）

ClO-+H2O+2e

Cl-+2OH-+)CrO2-+4OH--3e

2H2O+CrO42-3ClO-+2CrO2-+2OH-

H2O+3Cl-+CrO42-1-3Voltaiccells

1-3-1Whatkindsofredoxreactionscanbeharnessedtoproduceelectricalenergy?Inprnciple,thespontaneousredoxreactionswithenoughreactionratecanbeharnessedtoproduceelectricalenergybycarryingthemoutinavoltaiccell.u

G<0u

enoughhighreactionrate

1-3-2

ElectrodeTypesandCellNotation（1）Fourtypesofelectrode：l

Metal-metalionselectrodeElectrodereaction：Zn2++2e

ZnElectrodenotation：Zn(s)

Zn2+

l

Gas-ionelectrodeStandardhydrogenelectrodeElectrodereaction：2H++2e

H2Electrodenotation：Pt

H2(g)

H+Thesymbol(Pt)isusedtoindicatethepresenceofaninertplatinumcathode.

l

Metal-insolublemetalsaltselectrode/Metal-oxide-anionelectrode,eg.CalomelelectrodeandAgClelectrodearesuchkindofelectrodes.Calomelelectrode：OnthesurfaceofHg,thereisalayerofHg2Cl2,whichisinsaturatedKClsolution.Electrodereaction：2Hg2Cl2+2e

2Hg+2Cl-Electrodenotation：Hg-Hg2Cl2(s)

Cl-AgClelectrodeElectrodereaction：AgCl+e

Ag+Cl-Electrodenotation：Ag-AgCl(s)

Cl-

l

Redoxelectrode（ion-ionelectrode）Electrodereaction：Fe3++e

Fe2+Electrodenotation：Pt

Fe3+,Fe2+

(2)Cellnotation(-)Zn(s)

Zn2+

Cu2+

Cu(s)(+)Zn+2H+

Zn2++H2(g)

(-)Zn(s)

Zn2+

H+

H2(g)

Pt(+)(-)Pt

H+

H2(g)

Cu2+

Cu(s)(+)Redoxequations

Cellnotations

1-3-3

CellPotentials

=

+-

-2.StandardReductionPotentials2-1Origin2-2Measurement2-3Affectingfactors——NernstEquation2-4Applications2-5電勢圖解及其應(yīng)用2-1OriginM(s)

Mn+(aq)+neThepotentialexistingbetweensurfaceofthemetalanditssaltsolutioniscalledelectrodepotential.Itissymbolizedas

M+/M.2-2Measurement(1)StandardHydrogenelectrodePt

H2(g)

H+2H++2e

H2aH+=1mol·kg-1,PH2=1.013

105PaSettingthepotentialofstandardhydrogenelectrodeequaltozero,

°H+/H2=0(2)Standardelectrodepotential(StandardReductionPotential)referstothepotentialwhenanelectrodewithallitscomponentsinstandardstateconstitutesacellwithastandardhydrogenelectrode.

°Zn2+/Zn=-0.76v

°Cu2+/Cu=0.34vExplanations：l

Thevalueof

°hasnothingtodowiththenumberofcharges,nomattergainandloss；

Zn2++2e

Zn

°Zn2+/Zn=-0.76v1/2Zn2++e

1/2Zn

°Zn2+/Zn=-0.76vl

Thebiggerthe

°Ox/Redvalue，thestrongercapacityofOxgainingcharges,whichmeansthatOxisastrongoxidizingagent;consequently,itsRed.specieshasaweakpowertolosecharge,itisaweakreducingagent.Correspondingly，thesmallerthe

°Ox/Redvalue,thestrongercapacityofRedlosingcharges,whichmeansthatRed.isastrongreducingagent;consequently,itsOxspecieshasaweakpowertogaincharge,itisaweakoxidizingagent.l

當同一元素有多種氧化態(tài)時，同一物質(zhì)在一個電對中為Ox，在另一電對中為Red，使用時應(yīng)注意選取正確的值。Eg.

°Fe3+/Fe2+=0.771vFe3++e

Fe2+

°Fe2+/Fe=0.44vFe2++2e

Fel

Inacidictable：[H+]=1mol/LInbasictable：[OH-]=1mol/Leg.A:

°O2/H2O=1.229vO2+4H++4e

2H2OB:

°O2/OH-=0.401vO2+2H2O+4e

4OH-

2-3Affectingfactors——NernstEquationConcentration,pressure,andtemperature2-3-1DeducingofNernstEquation

-c,P

E~Q

rGm~Q

(1)Relationshipbetween

rGmandE：-

rGm

W-

rGm=Wmax=nFE電功=電動勢

電量J/molvc/molSince1joule(J)equalsto1volt(V)

1Coulomb(C)1molelectrons=1.6

10-19

6.02

1023

=96485C/mol

96500C/mol=Faradayconstant-

rGm

=nFE

(2)E~Q

rGm=

rGm

+RTlnQ-nFE=-nFE

+RTlnQE=E

-RT/nFlnQ

T=298K,F=96500c/mol,R=8.314J·mol/KE=E

-0.059/nlgQNernstEquationZn+Cu2+=Zn2++Cu

Cu2+/Cu-

Zn2+/Zn=(

°Cu2+/Cu-

°Zn2+/Zn)-

0.0591(lg[Zn2+]/[Cu2+])/n=(

°Cu2+/Cu+0.0591(lg[Cu2+])/n)–(

°Zn2+/Zn+0.0591(lg[Zn2+])/n)

Cu2+/Cu=

°Cu2+/Cu+0.0591(lg[Cu2+])/n

Zn2+/Zn=

°Zn2+/Zn+0.0591(lg[Zn

])/nOx+ne

RedNernstEquation

(3)ExplanationsonNernstEquationl

[Ox]/[Red]表示電極反應(yīng)式中，氧化型物質(zhì)一邊的所有物質(zhì)的濃度以系數(shù)為指數(shù)的冪的乘積/還原型物質(zhì)一邊的所有物質(zhì)的濃度以系數(shù)為指數(shù)的冪的乘積。

Eg.MnO4-+8H++5e

Mn2++4H2O

MnO4-/Mn2+=

°MnO4-/Mn2+

+(0.0591/5)?(lg[MnO4-][H+]8/[Mn2+])

lAccordingtoNernstEquation,

valueincreaseswithincreasingof[Ox]anddecreasingof[Red].2-3-2ApplicationsofNernstEquation

eg1.At298K,putAginitssaltsolutionwhoseconcentrationis0.10mol/L.Calculate

Ag+/Ag(

Ag+/Ag=0.799v)Solution：

Ag+/Ag=

Ag+/Ag+0.0591lg[Ag+]=0.799-0.0591=0.7399(v)eg2.AddNaCltoaAg

AgNO3solution，setting[Cl-]=1.0mol/L，Calculate

AgCl/Ag.(

Ag+/Ag=0.799v,Ksp

AgCl=1.6

10–10)Solution：：

Ag+/Ag=

Ag+/Ag+0.0591lg[Ag+]=0.799–0.0591lg1.6

10–10=0eg3.At298K，addNaOHtoasolutioncontainingFe2+andFe3+，whentheequilibriawerereached，[OH-]=1.0mol/L.Pleasecalculate：

°Fe(OH)3/Fe(OH)2

(

°Fe3+/Fe2+=0.771v,Ksp

Fe(OH)3=4.0

10–38,Ksp

Fe(OH)2=8.0

10–16)Solution：

Fe3+/Fe2+=

°Fe3+/Fe2++0.0591lg[Fe3+]/[Fe2+]=0.771+0.0591lg(KspFe(OH)3/KspFe(OH)2)=0.771+0.0591lg(4.0

10–38/8.0

10–16)=-0.55(v)=

°Fe(OH)3/Fe(OH)2eg4.NO3-+4H++3e

NO+2H2O,

°NO3-/NO=0.96v,[NO3-]=1.0mol/L,PNO=P

.Calculatethe

when（1）[H+]=1

10–7mol/Land（2）[H+]=10mol/L.Solution：

=

°+0.0591（lg[NO3-][H+]4P

/PNO

）/3=0.96+0.0591(lg10-28)/3=0.41v(2)

=

°+0.0591（lg[NO3-][H+]4P

/PNO

）/3=1.039vItisobviousthattheoxidizingabilityofNO3-increasewiththeincreasingof[H+].2-4Applicationsofstandardreductionpotentials：Thestandardreductionpotentialscanbeusedinthreeaspects:(1)Judgewhethertheoxidizingandreducingagentarestrongorweak；(2)Predictthespontaneousdirectionofaredoxreaction;P235,Example5-13(3)CalculateKofaredoxreaction,whichisthefifthmethodofcalculatingequilibriumconstantinourcourse.∵

rG

m=-RTlnK

,

rG

m=-nE

F∴l(xiāng)gK

=nE

F/(2.303RT)=nE

/0.0591

eg1.PutZn(s)in0.1MCuSO4,Calculate[Cu2+]atequilibrium.Solution:Zn+Cu2+=Zn2++Cu(-)Zn(s)

Zn2+

Cu2+

Cu(s)(+)E

=

°Cu2+/Cu-

°Zn2+/Zn=0.34-(-0.76)=1.10VlgK

=nE

/0.0591=2

1.10/0.0591=37.2∴K

=1.6

1037

Zn+Cu2+=Zn2++Cut00.10te[Cu2+]0.1-[Cu2+]

0.1K

=[Zn2+]/[Cu2+]=1.0/[Cu2+]=1.6

1037[Cu2+]=6.3

10-39mol/Leg2.Ag++e

Ag

Ag+/Ag=0.799vAgCl(s)+e

Ag+Cl-

AgCl/Ag=0.222vCalculateKspAgCl

Solution:(-)Ag-AgCl(s)

Cl-(1.0M)

Ag+(1.0M)

Ag(s)(+)(-)Ag+Cl--e

AgCl(s)(+)Ag++e

AgAg++Cl-

AgCl(s)lgK=lg1/KspAgCl=nE°/0.0591

=1

0.577/0.0591=9.75∴KspAgCl=1.8

10–10Eg3.KspCuCl=1.2

10–6Cu++e

Cu

°Cu+/Cu=0.515vCu2++e

Cu+

°Cu+/Cu=0.159vCalculatetheK1ofCu+Cu2+

2Cu+(2)CalculatetheK2Cu+Cu2++2Cl-

2CuCl(s)Solution:(1)(-)Pt

Cu2+,Cu+

Cu+

Cu(+)(-)Cu+-e

Cu2+(+)Cu++e

Cu2Cu+

Cu+Cu2+lgK=nE°/0.0591=1

(0.515-0.159)/0.0591=6.02

∴K=106.02,K1=9.5

10–7Eg3.KspCuCl=1.2

10–6Cu++e

Cu

°Cu+/Cu=0.515vCu2++e

Cu+

°Cu+/Cu=0.159vCalculatetheK1ofCu+Cu2+

2Cu+(2)CalculatetheK2Cu+Cu2++2Cl-

2CuCl(s)Solution:(2)Cu+Cu2+

2Cu+2Cu++2Cl-

CuCl(s)Cu+Cu2++2Cl-

2CuClK2=K1/Ksp2CuCl=(9.5

10–7)/(1.2

10–6)2

=6.6

105Exercise

：Hg2++2e

Hg

°Hg2+/Hg=0.85vHg22++2e

2Hg

°Hg22+/Hg=0.80vCalculate:(1)TheequilibriumconstantofHg2++Hg

Hg22+;(2)The[Hg2+]in0.10MHg22+(NO3)2solution.

2-5

Diagramofelementalelectrodepotentials2-5-1Diagramofelementalelectrodepotentials

（1）

Definition：當元素具有三種或三種以上的氧化態(tài)時，將元素的不同氧化態(tài)按氧化數(shù)由大到小，自左向右排列，用橫線相連，線上標出相應(yīng)的標準電極電勢值

。Eg.

A:MnO4--0.56MnO42-

2.26MnO20.95Mn3+

1.51Mn2+-1.18Mn

1.511.6931.23(2)Applications：l

判斷歧化反應(yīng)

左

右ABC

左

右

,B

A+C

0.1590.515eg.Cu2+Cu+Cu

0.920.79Hg2+Hg22+Hg

l

Calculate

。Eg.Fe3+0.77Fe2+-0.44Fe?Solution:(1)

Fe3++e

Fe2+

1

rG

m1=-

1F(2)

Fe2++2e

Fe

2

rG

m

2=-2

2F(3)

Fe3++3e

3Fe

3

rG

m

3=-3

3F

rG

m3=

rG

m1+

rG

m23

3F=

1F+2

2F∴

3=(

1+2

2)/3Ingeneral,

1z1

2z2ABC

3z3zstandsforthechangevalueoftheoxidationnumberoftheelement.3PatternsofVoltaiccells3-1PrimarycellsDrycells3-2Secondarybatteries/storagebattery3-3Fuelbatteries/Continouscells4.Electrolysisand

ElectrolyticCellsCathodeAnodeH2O/H+

Cathode：2H2O+2e

H2+2OH-

（Reductionreaction）Anode：H2O–2e

1/2O2+2H+（Oxidationreaction）Overallreaction：H2O

H2+1/2O2(Electrolysisreaction)Ecalculated=

°O2/H2O-

°H+/H2=1.229vEactually=1.7vAnodereaction:2Cl--2e=Cl2(g)Cathodereaction:2H2O+2e=H2(g)+2OH-Overallreaction:2Cl-+2H2O=Cl2(g)+H2(g)+2OH-

overpotentialTheminimumvoltagenecessarytobringaboutelectrolysisisthecalculatedpotentialforthecell.However,thevoltageactuallyneededisalwaysgreaterthanthisminimumvoltage.Thisextravoltageisrequiredtoovercometheresistanceofthecell.Inaddition,someelectrodereactionstakeplaceslowly;energyatleastequaltotheactivationenergymustbesuppliedinorderforthereactiontotakeplace.Theextravoltageneededtomakeslowreactionstakeplaceatapracticalrateiscalledoverpotential.overpotential=Eactually–Ecalculated(2)QuantitativeAspectsofElectrolysis--ElectrolysisLawIn1934,Faradaydiscoveredthattheamountofasubstanceproducedbyelectrolysismustdirectlyproportionaltothecurrentpassedandinde