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物理化学课后习题答案第九章

物理化学习题解答(九)习题p109~1161解:(1) Pt︱H2(p H2)︱HCl(a)︱Cl2(p Cl2)︱Pt正极:Cl2(p Cl2)+ 2e-→2Cl-(a)负极:H2(p H2) –2e-→2H+(a)电池反应:H2(p H2) + Cl2(p Cl2)==2HCl(a) (2) Pt︱H2(p H2)︱H+(a H+)‖Ag+(a Ag+)︱Ag(s) 正极:Ag+(a Ag+)+ e-→Ag(s)负极:H2(p H2) –2e-→2H+(a H+)电池反应:H2(p H2) + Ag+(a Ag+)==2H+(a H+)+ Ag(s)(3) Ag(s)︱AgI(s)︱I-(a I-)‖Cl-(a Cl-)︱AgCl(s)︱Ag(s)正极:AgCl(s) + e-→Ag(s) + Cl-(a Cl-)负极:Ag(s) + I-(a I-)– e-→AgI(s)电池反应:AgCl(s) + I-(a I-)==AgI(s) + Cl-(a Cl-)(4) Pb(s)︱PbSO4(s)︱SO42-(a SO42-)‖Cu2+(a Cu2+)︱Cu(s)正极:Cu2+(a Cu2+) + 2e-→Cu(s)负极:Pb(s) + SO42-(a SO42-)–2e-→PbSO4(s)电池反应:Pb(s) + Cu2+(a Cu2+) + SO42-(a SO42-)==PbSO4(s) + Cu(s)(5) Pt︱H2(p H2)︱NaOH(a)︱HgO(s)︱Hg(l)正极:HgO(s) + H2O (l)+ 2e-→Hg(l) + 2OH-(a OH-)负极:H2(p H2)+ 2OH-(a OH-) –2e-→2H2O(l)电池反应:HgO(s) + H2(p H2)==Hg(l)+H2O(l) (6) Pt︱H2(p H2)︱H+(a H+)︱Sb2O3(s)︱Sb(s) 正极:Sb2O3(s) + 6H+(a H+)+ 6e-→2Sb(s) + 3H2O(l)负极:H2(p H2) –2e-→2H+(a H+)电池反应:Sb2O3(s) + 3H2(p H2) ==2Sb(s) + 3H2O(l)(7) Pt︱Fe3+(a1),Fe2+(a2)‖Ag+(a Ag+)︱Ag(s)正极:Ag+(a Ag+)+ e-→Ag(s)负极:Fe2+(a2) – e-→ Fe3+(a1)电池反应:Ag+(a Ag+) + Fe2+( a2)==Fe3+( a1)+Ag(s)(8) Na(Hg)(a am)︱Na+(a Na+)‖OH-(a OH-)︱HgO(s)︱Hg(l)正极:HgO(s) + H2O (l)+ 2e-→Hg(l) +2OH-(a OH-)负极:2Na(Hg)(a am) –2e-→2Na+(a Na+) + 2Hg(l)电池反应:2Na(Hg)(a am) + HgO (s) + H2O(l)==2 Na+(a Na+) + 2OH-(a OH-) + 3Hg(l)2解:(1)AgCl(s)==Ag+(a Ag+) + Cl-(a Cl-)电池:Ag(s)︱Ag+(a Ag+)‖Cl-(a Cl-)︱AgCl(s)︱Ag(s)正极:AgCl(s) + e-→Ag(s) + Cl-(a Cl-)负极:Ag(s)–e-→Ag+(a Ag+)电池反应:AgCl(s)==Ag+(a Ag+) + Cl-(a Cl-) (2)AgCl(s) + I-(a I-) ==AgI(s) + Cl-(a Cl-)电池:Ag(s)︱AgI(s)︱I-(a I-)‖Cl-(a Cl-)︱AgCl(s)︱Ag(s)正极:AgCl(s) + e-→A g(s) + Cl-(a Cl-)负极:Ag(s) + I-(a I-)– e-→AgI(s)电池反应:AgCl(s) + I-(a I-) ==AgI(s) + Cl-(a Cl-)(3) HgO(s) + H2(p H2)==Hg(l)+H2O(l)电池:Pt(s)︱H2(p H2)︱NaOH(a)︱HgO(s)︱Hg(l)正极:HgO(s) + H2O(l) + 2e-→Hg(l) + 2OH-(a)负极:H2(g) + 2OH-(a) –2e-→2H2O(l)电池反应:HgO(s) + H2(p H2)→Hg(l) + H2O(l) (4) Fe2+(a Fe2+) + Ag+(a Ag+)== Fe3+(a Fe3+) + Ag(s)电池:Pt(s)︱Fe3+(a Fe3+),Fe2+( a Fe2+)‖Ag+(a Ag+)︱Ag(s)正极:Ag+(a Ag+)+ e-→Ag(s)负极:Fe2+( a Fe2+) – e-→ Fe3+( a Fe3+)电池反应:Fe2+(a Fe2+)+Ag+(a Ag+)== Fe3+( a Fe3+) + Ag(s)(5) 2 H2(p H2) + O2(p O2)==2H2O(l)电池:Pt︱H2(p H2)︱H+(a H+)︱O2(p O2)︱Pt(s) 正极:O2(p O2) + 4H+(a H+) + 4 e-→2H2O(l)负极:2H2(p H2) – 4e-→4H+(a H+)电池反应:2 H2(p H2) + O2(p O2)==2H2O(l) (6) Cl2(p Cl2) + 2I-(a I-)==I2(s)+2Cl-(a Cl-)电池:Pt︱I2(s))︱I-(a I-)‖Cl-(a Cl-)︱Cl2(p Cl2)︱Pt正极:Cl2(p Cl2) + 2e-→2Cl-(a Cl-)负极:2I-(a I-)–2e-→ I2(s)电池反应:Cl2(p Cl2) + 2I-(a I-)==I2(s)+2Cl-(a Cl-) (7) H2O(l)== H+(a H+) + OH-(a OH-)电池:Pt(s)︱H2(p H2)︱H+(a H+)‖OH-(a OH-)︱H2(p H2)︱Pt(s)正极:2H2O(l) + e-→2H2(p H2) + 2OH-(a OH-) 负极:H2(p H2)–2e-→ 2H+(a H+)电池反应:H2O(l)== H+(a H+) + OH-(a OH-) (8) Mg(s) + 1/2O2(g) + H2O(l)== Mg(OH)2(s) 电池:Mg(s)︱Mg(OH)2(s)︱OH-(a OH-)︱O2(p O2)︱Pt(s)正极:1/2O2(g) + H2O(l) + 2e-→ 2OH-(a OH-) 负极:Mg(s) + 2OH-(a OH-)– 2e-→Mg(OH)2(s) 电池反应:Mg(s) + 1/2O2(g) + H2O(l)== Mg(OH)2(s)(9) Pb(s) + HgO(s)==Hg(l) + PbO(s)电池:Pb(s)︱PbO(s)︱OH-(a OH-)HgO(s)︱Hg(l)正极:HgO(s) + H2O(l) + 2e-→ Hg(l) + 2OH-(a OH-)负极:Pb(s) + 2OH-(a OH-) –2e-→PbO(s) + H2O(l)电池反应:Pb(s) + HgO(s)==Hg(l) + PbO(s) (10) Sn2+(a Sn2+) + Tl3+(a Tl3+) == Sn4+(a Sn4+) + Tl+(a Tl+)电池:Pt(s)︱Sn2+(a Sn2+),Sn4+(a Sn4+)‖Tl3+(a Tl3+),Tl+(a Tl+)︱Pt(s)正极:Tl3+(a Tl3+) + 2e-→ Tl+(a Tl+)负极:Sn2+(a Sn2+) –2e-→Sn4+(a Sn4+)电池反应:Sn2+(a Sn2+) + Tl3+(a Tl3+) == Sn4+(a Sn4+) + Tl+(a Tl+)15解:Fe(s) + Cd2+(aq)==Cd(s)+Fe2+(aq)E=Eө– RT/2F×ln{[ Fe2+]/[Cd2+]}(1) E=φөcd2+/Cd–φөFe2+/Fe- RT/2F×ln{[ Fe2+]/[Cd2+]}=-0.40 +0.44–0.0592/2lg{0.1/0.1}=0.04>0反应能自发向右进行,故金属Fe首先被氧化。

(2) E=φөCd2+/Cd–φөFe2+/Fe–RT/2F×ln{[ Fe2+]/[Cd2+]}= –0.40 +0.44–0.0592/2lg{0.1/0.0036}= –0.003<0反应能自发向左进行,故金属Cd首先被氧化。

23解:Pb(s)︱PbSO4(s)︱H2SO4(1.0mol.kg-1)︱PbO2(s)∣PbSO4(s)︱Pb(s)正极:PbO2(s) + 4H+(m)+ SO42-(a SO42-) +2e-→PbSO4(s)+2H2O(l)φPbO= φөPbO2/ PbSO4+RT/2Fln a SO42- a 2/ PbSO44H+负极:Pb(s)+SO42-(a SO42-) –2e-→ PbSO4(s) φPbSO= φөPb2+/Pb+RT/2Fln{Kөsp/ a SO42-} 4/ Pb电池反应:PbO2(s) + Pb(s)+ 2H2SO4(m)==2PbSO4(s)+2H2O(l)E=φPbO2/ PbSO4–φPbSO4/ Pb= φөPbO2/–φөPb2+/Pb–RT/2FlnKөsp + RT/Fln a H2SO4PbSO4E=Eө+ RT/Fln a H2SO4=2.041+ RT/Fln a H2SO4E/V=1.91737+56.1×10-6(t/℃)+1.08×10-8(t/℃)2 E=1.91737+56.1×10-6×25+1.08×10-8×252=1.91878V1.919=2.041+ RT/Fln a H2SO4,ln a H2SO4= –4.7511ln a H2SO4= ln r±3m H+2m SO42-= – 4.75113lnr±+2ln m H++ln m SO42-= –4.75113lnr±+2ln2.0+ln1.0=–4.7511lnr±=–2.0458,r±=0.12924解:正极:I2(s) +2e-→2I-(a3)φI=φөI2/I-+RT/Fln a32/I-负极:2S2O32-(a1) –2e-→S4O62-(a2)φS=φөS4O62-/S2O32- 4O62-/S2O32-+RT/2Fln( a2/ a2)电池反应:2S2O32-(a1) + I2(s)== S4O62-(a2) + 2I-(a3)△r Hөm=2{(2)–(1)+(3)}=2{28.786-46.735+3.43 1}= –29.04kJ.mol-1△r Sөm=2×105.9+146.0–116.7–2×33.47=1 74.16 J.K-1.mol-1△r Gөm=△r Hөm–T△r Sөm=–29.04×103–298×174.16×10-3= –80.94kJ.mol-1 Eө=–△r Gөm/nF=80.94×103/2/96484.5=0.419VEө= φөI2/I-—φөS4O62-/S2O32-=0.419VφөS=φөI2/I-—4O62-/S2O320.419=0.535–0.419=0.116V25解:Pt(s)︱H2(pө)︱H2SO4(0.010mol.kg-1)︱O2 (pө)︱Pt(s)正极:O2 (pө) + 4H+(a H+) + 4e-→2H2O(l)φO= φөO2/H2O +RT/4Fln{(p O2/pө)a H+4} 2/H2O负极:2H2(pө) – 4e-→4H+(a H+)φH+/H= φөH+/H22–RT/4Fln{(p H2/pө)2/a H+4}电池反应:O2 (pө) + 2H2(pө) == 2H2O(l)E= φO2/H2O –φH+/H2= φөO2/H2O–φөH+/H2 + RT/4Fln{(p O2/pө) (p H2/pө)2}E=φO2/H2O –φH+/H2= φөO2/H2O–φөH+/H2=E ө=1.227V∵2H2O(g) ==O2(g) + 2H2(g) Kөp=9.7×10-81Kөp= (p H2/pө)2(p O2/pө)/(p H2O/pө)2 =9.7×10-81( p H2/pө)2(p O2/pө)/(3.200/100)2 =9.7×10-81( p H2/pө)2(p O2/pө)=(3.200/100)2×9.7×10-81=9.933×10-84∴O2 (pө) + 2H2(pө) == 2H2O(l)Kө=1/{(p H2/pө)2(p O2/pө) }=1/9.933×10-84 =1.0068×1083Eө=RT/nFlnKө=8.314×298/4/96484.5×ln1.0068×1083=6.4196×10-3×(6.7770×10-3+191.1145 6)=1.227V26解:Pb(s)︱PbCl2(s)︱HCl(0.010mol.kg-1)‖H2(10kPa)︱Pt(s)正极:2H+(a H+) + 2e-→H2(10kPa)φH+/H= φөH+/H22–RT/2Fln{(p H2/pө)/a H+2}负极:Pb(s) + 2Cl-( a Cl-) –2 e-→PbCl2(s)φPbCl=φөPb2+/Pb+RT/2Fln(Kөsp/ a Cl-2)2/Pb电池反应:2HCl(a HCl) + Pb(s) == H2(10kPa) + PbCl2(s)E=φH+/H2–φPb2+/Pb=φөH+/H2–φөPb2+/Pb –RT/2FlnKөsp –RT/2Fln{(p H2/pө)+RT/Flna HClm PbCl2=0.039 mol.kg-1I=1/2×{0.039×22+2×0.039×(–1)2}=3×0.039mol .kg -1r ±=0.706K өsp =a Pb 2+a Cl -2=a ±3=( r ±m ±/m ө)3=( r ±3m Pb 2+m Cl-2)= 0.7063×0.039×(2×0.039)2=8.35×10-5 m HCl =0.010 mol .kg -1I =1/2×{0.01×12 +0.01×(–1)2}=0.010mol .kg -1r ±=0.950E =φH +/H 2–φPb 2+/Pb =φөH +/H 2 –φөPb 2+/Pb –RT/2FlnK өsp –RT/2Fln {(p H 2/p ө)+RT/Flna HCl= 0.126–0.012839×ln (8.35×10-5)–0.012839×ln 0.1+0.025678×ln (0.9502×0.012)=0.126+0.12057+0.029562 –0.23914=0.037V 27解:Cu(s)︱CuAc 2(0.10mol .kg -1)∣AgAc(s)︱Ag(s)(1) 正极:AgAc(s) + e -→Ag(s) + Ac -( a Ac -)φAg +/Ag = φөAg +/Ag +RT/FlnK өsp / a Ac -负极:1/2Cu(s) – e -→1/2Cu 2+( a Cu 2+) 3482.0039.0312509.0lg -=⨯-⨯-=-=-+±I z z A r 0509.001.011509.0lg -=-⨯-=-=-+±I z z A rφCu 2+/Cu = φөCu 2+/Cu +RT/2Fln a Cu 2+电池反应:2AgAc(s) + Cu(s)== 2Ag(s) + CuAc 2(aq)E = φAg +/Ag –φCu 2+/cu = φөAg +/Ag – φө Cu 2+/Cu+RT/FlnK өsp – RT/F ln( a Ac - a Cu 2+1/2)(2) △r G m = – nFE= – 2×96484.5×0.372= –71.784kJ .mol -1△r S m =nF =2×96484.5×2×10-4=19.297J .mol-1 △r H m =△r G m +T △r S m = –71.784 +298×19.297×10-3= –66.033 kJ .mol-1 (3) E =φAg +/Ag –φCu 2+/cu = φөAg +/Ag – φө Cu 2+/Cu+RT/FlnK өsp – RT/F ln( a Ac -a Cu 2+1/2)E =φAg +/Ag –φCu 2+/cu = φөAg +/Ag – φө Cu 2+/Cu+RT/FlnK өsp – RT/F ln( m Ac -m Cu 2+1/2)0.372= φөAg +/Ag– φө Cu 2+/Cu +0.025678lnK өsp– 0.025678ln( 0.2×0.11/2)0.372=0.799–0.337+0.025678lnK өsp –0.025678ln0.2–0.01284ln0.1lnK өsp = – 6.2697,K өsp = 1.89×10-3 41212102298308372.0374.0)()()(-⨯=--=--=∂∂T T T E T E TE p p T E )(∂∂28解:Hg22+(a Hg22+) + 2e-→Hg(l)φHg22+/Hg =φө Hg22+/Hg + RT/2F ln a Hg22+Hg2SO4(s) + 2e-→Hg(l) + SO42-( a SO42-)φөHg2SO4/Hg =φөHg22+/Hg+ RT/2F ln(K spө/ a SO42-)φөHg2SO4/Hg =φөHg22+/Hg+ RT/2F ln Kөsp=0.789 + 0.01284ln(8.2×10-7)=0.609V29解:(1)Ag(s) + Fe3+(a Fe3+)== Ag+(a Ag+) + Fe2+( a Fe3+)电池:Ag(s)︱Ag+(a Ag+)‖Fe3+(a Fe3+),Fe2+( a Fe2+)︱Pt(s)正极:Fe3+( a Fe3+) + e-→Fe2+( a Fe2+)负极:Ag(s)– e-→Ag+(a Ag+)电池反应:Ag(s) + Fe3+( a Fe3+)== Ag+( a Ag+) + Fe2+( a Fe3+)Eө=RT/nF ln Kө,Kө=exp(nFEө/RT )Eө= φөFe3+/Fe2+–φөAg+/Ag=0.771–0.799= –0.008ln Kө=nFEө/RT,ln Kө=1×96484.5×( –0.008)/(8.314×298)= –0.3115Kө=0.73(2) Hg2Cl2(s)== Hg22+(a Hg2+) + 2Cl-(a Cl-)电池:Hg(l)︱Hg22+(a Ag+)‖Cl-(a Cl-)︱Hg2Cl2(s)︱Hg(l)正极:Hg2Cl2(s) + 2e-→2Hg(l) + 2Cl-(a Cl-)负极:2Hg(l)–2e-→Hg2+(a Hg22+)电池反应:Hg2Cl2(s)== Hg22+(a Hg2+) + 2Cl-(a Cl-)Eө=RT/nF ln Kөsp,Kөsp=exp(nFEө/RT )Eө= φөHg2Cl2/Hg–φөHg2+/Hg=0.268–0.793= –0.525ln Kө=nFEө/RT,ln Kөsp=2×96484.5×( –0.525)/(8.314×298)= –40.89Kөsp=1.74×10-18(3) HBr(a HBr)==H+(a H+)+Br-(a Br-)电池:Pt(s)︱H2(pө)‖HBr(a HBr)︱AgBr(s)︱Ag(s)正极:AgBr(s) + e-→Ag(s) + Br-(a Br-)负极:1/2H2(pө) – e-→H+(a H+)电池反应:AgBr(s) +1/2 H 2(p ө)== Ag(s) +HBr(a HBr )E=E ө– RT/F×ln a HBr= E ө– RT/F×lna HBr= E ө– 2RT/F×lna ±= E ө– 2RT/F×ln (r ±m ±/m ө)= E ө– 2RT/F×ln (0.01r ±)(4) 2Ag 2O(s)== 4Ag (s) + O 2(p O 2)电池:Pt(s)︱O 2(p O 2)︱OH -(a OH -)︱Ag 2O(s)︱Ag(s)正极:2Ag 2O(s) + 2H 2O(l) + 4e -→4Ag(s) + 4OH -(a OH -)负极:4OH -(a OH -) –4e -→O 2(p O 2) + 2H 2O(l)电池反应: 2Ag 2O(s)== 4Ag (s) + O 2(p O 2)E=E ө –RT/4F×ln ( p O 2/p ө]△r G m = –nFE△r S m = nF △r H m =△r G m +T △r S m = –nFE+nFT△r G m =0T=△r H m /△r S m(5) H 2(p ө) +1/2 O 2(p ө)==H 2O(l)电池:Pt(s)︱H 2(p ө)︱H +(a H +)︱O 2(p ө)︱Pt(s) p TE )(∂∂p T E )(∂∂正极:1/2O2(pө)+2H+(a H+) + 2e-→H2O(l)负极:H2(pө)–2e-→2H+(a H+)电池反应:H2(pө)+1/2 O2(pө)==H2O(l)△Gөm= –nFEөEө= φөO2/H2O –φөH+/H2 =1.229–0= 1.229△Gөm= –nFEө= –2×96484.5×1.229= –237158.9J.mol-1 = –237.16kJ.mol-1(6)HA(a HA)== H+(a H+) + A-(a A-)电池:Pt(s)︱H2(pө)︱HA(a HA),NaA(a NaA ),NaCl(a NaCl )︱AgCl(s)︱Ag(s)正极:Ag+(aq) + e-→Ag(s)AgCl(s) ==Ag+(a Ag+) +Cl-(a Cl-)AgCl(s) + e-→Ag(s) + Cl-(a Cl-)φAg+/Ag =φөAg+/Ag+ RT/F ln a Ag+φAg+/Ag =φөAg+/Ag+ RT/F ln Kөsp/a Cl-φAgCl/Ag =φөAg+/Ag= φөAg+/Ag+RT/F ln Kөsp=0.2223VφAg+/Ag=0.2223–RT/F ln a Cl-负极:1/2H2(p H2) –e-→H+(a H+)HA(a HA)==H+(a H+)+A-( a A-)1/2H2(p H2) + A-( a A-)–e-→HA(a HA)φH+/H2=φөH+/H2+ RT/F ln a H+φөH+/H2=RT/F ln Kөa a HA/a A-电池反应:1/2H2(pө) +AgCl(s) +A-(a A-)== Ag(s) + Cl-(a Cl-) + HA(a HA)E= Eө–RT/Fln(Kөa a Cl-a HA/a A-)Eө= φөAgCl2Ag –φөH+/H2 =0.2223–0=0.2223V E= 0.2223–RT/Fln(Kөa a Cl-a HA/a A-)30、298K时、测得如下电池的电动势E与HBr浓度的关系如表所示:Pt(s)︱H2(pө)︱HBr(m)∣AgBr(s)∣Ag(s)试计算:(1)电极Br-(a Br-)∣AgBr(s)∣Ag(s)∣的标准电极电势ө;(2)0.1 mol.kg-1的HBr溶液的离子平均活度因子r±。

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