#所以可利用反應(yīng)CuCl2+Cu=2CuCl制備CuCl。9.15答案：若(1)為正極，c(H+)=0.054mol/L若(2)為正極，c(H+)=0.187mol/L9.160.059V90(O/OH-)=1.229V+xlg(1.0x10-14)4=0.403V24(所求實(shí)際為c(H+)=10-14mol/L時(shí)，O2/H2O電極的電極電勢(shì))9.17=0.281V-0.48V=-0.20V9e(Hg2Cl2/Hg)=0.281V所以：9(H+/H2)=9e(Hg2Cl2/Hg)=0.281V-0.48V=-0.20V9(H+/H)=90(H+/H)+222.303RTPlg{c(H+)/c0}2p(H)/p02???0.20V=0.059Vxlg{c(H+)/c0}c(H+)=4.1x10-4mol/Lc(H+)/c0=.、'：K0xc(HB)/c叭中a?K0=1.7x10-6a9.18若亞汞離子為Hg+，則電極反應(yīng)為：Hg++e-=Hg根據(jù)此電極反應(yīng)，可寫(xiě)出二電極的Nernst方程并可計(jì)算電池電動(dòng)勢(shì)2.303RT9(+)=9?+lg{c(Hg+)/c0}F9(-)=90+lg{c(Hg+)/10c0}Fe=9(+)-9(-)=2.303RTlg10=0.059VF計(jì)算結(jié)果與實(shí)驗(yàn)數(shù)據(jù)不符。若亞汞離子為Hg22+，則電極反應(yīng)為：Hg22++2e-=2Hg根據(jù)此電極反應(yīng)，可寫(xiě)出二電極的Nernst方程并可計(jì)算電池電動(dòng)勢(shì)2.303RT9(+)=90+lg{c(Hg2+)/c0}2F22.303RT9(-)=90+lg{c(Hg2+)/10c0}2F22.303RTe=9(+)-9(-)=lg10=0.03V2F計(jì)算結(jié)果與實(shí)驗(yàn)數(shù)據(jù)相符，故可知亞汞離子為Hg22+。9.19Hg22+=Hg2++HglgK0=F{?0(Hg2+/Hg)-00(Hg2+/Hg2+)}lgK0=22=_20342.303RT?K0=9.2x10_3K0=C(Hg2+"c0c(Hg2+)/c0c(Hg2+)=1c(Hg2+)=帀29.20根據(jù)標(biāo)準(zhǔn)電極電勢(shì)圖0一(PbO2/Pb2+)=1.455V0一(O2/H2O2)=°.695V0一(H2O2/H2O)=1.76V下列反應(yīng)可自發(fā)進(jìn)行：PbO2+H2O2+2H+=Pb2++O2+2H2OPb2++H2O2=PbO2+2H++(2):2H2O2=2H2O+O故PbO2（MnO2等）可催化過(guò)氧化氫歧化分解。9.22KClO3+6HCl=3Cl2+KCl+3H2O3I2+6KOH=KIO3+5KI+3H2O2H2S+SO2=3S+2H2OBi(OH)3+Cl2+3NaOH=NaBiO3+2NaCl+3H2OSn+4HNO3=SnO2+4NO2+2H2OPb+4HNO3=Pb(NO3)2+2NO2+2H2OMnSO4+O2+4NaOH=Na2MnO4+Na2SO4+2H2O2Pb2++Cr2O72－+H2O=2PbCrO4+2H+2Cr(OH)4－+3HO2－=2CrO42－+5H2O+OH－2MnO4－+3Mn2++2H2O=5MnO2+4H+Hg(NO3)2=Hg+2NO2+O2Cu(NO3)2=2CuO+4NO2+O2KNO3=2KNO2+O2Cr2O72－+3H2O2+8H+=2Cr3++3O2+7H2OCr2O72－+4H2O2+2H+=2CrO5+5H2O第十章配位化合物10.2A[Co(SO4)(NH3)5]BrB[CoBr(NH3)5]SO410.3(2)10.4(4)10.5(1)Zn2+:3d104s04p0，sp3雜化，正四面體2)Hg2+:5d104s04p0,sp3雜化，正四面體3)Mn2+:3d54s04p04d0sp3d2雜化，正八面體4)Co3+:3d64s04p0d2sp3雜化，正八面體10.6(4)(3)(3)(1)(4)(3)(1)10.13Ag++2NH3=Ag(NH3)2+c0/mol?L-10.100.400ceq/moLL-1x0.40-2(0.10-x)0.10-xK0{Ag(NH)+}=0.10=1.1x107f32xx0.202x=2.3x10-710.14Ceq/moLL-1c(Ag+)=2.3x10-710.14Ceq/moLL-1Ag++2NH3=Ag(NH3)2+0.0010x0.099K0{Ag(NH)+K0{Ag(NH)+}=

f320.0990.0010x2=1.1x107x=3.0x10-3c(NH)=3.0x10-3mol-L-13c(NH+)=0.10mol-L-1+0.0020mol-L-1一3.0x10-3mol-L-1=0.099mol-L-14c(NH)pOH=pK0(NH)-lg上=4.76-1.52=6.24b3c(NH+)4pH=7.7610.15開(kāi)始生成沉淀時(shí)：c(Ag+)/c0=1.77x10c(Ag+)/c0=1.77x10-9―sp=c(Cl-)/c00.10Ag++2NH3=Ag(NH3)2+ceq/moLL-i1.77X10-9x0.1010.1610.16K?{Ag(NH)K?{Ag(NH)+}=f320101.77x10-9xx2=1?1x107x=x=2.3:、c(NH)=2?3mol-L-13c(NH+)=5.0mol-L-1一2?3mol-L-1=2.7mol-L-14c(NH)pOH=pKe(NH)一lg/=4.82pH=9.18b3c(NH+)4Ag++2S2O32-=Ag(S2O3)23-ceq/mol?L-1X0.100.050Ke=山050=2.9x1013x=1?7x10-13fxx0.102即：c(Ag+)=1.7X10-i3mol?L-i欲生成沉淀所需KC1和KI的濃度分別為：2cexKe(AgCl)c(KCl)=sp=2.0x103mol-L-1c(Ag+)/ce2cexKe(AgI)c(KI)=sp=1.0x10-3mol-L-1c(Ag+)/ce10.17Cu(OH)2+4NH3=Cu(NH3)42++2OH-K?=Kspe{Cu(OH)2}Kf一{Cu(NH3)42+}=2.2x10-7反應(yīng)難于自發(fā)進(jìn)行Cu(OH)2+2NH3+2NH4+=Cu(NH3)42++2H2OK?=Kspe{Cu(OH)2}Kfe{Cu(NH3)42+}{1/Kbe(NH3)}2=1.4x103反應(yīng)易于自發(fā)進(jìn)行，原因?yàn)镹H4+與生成的OH-結(jié)合為NH3，化學(xué)平衡正向移動(dòng)10.18不能，Co3+，+3，小于，大于10.19Zn2++4NH3=Zn(NH3)42+ceq/mol?L-1x0.100.10Ke{Zn(NH)2+}=0,10=2.9x109f34xx0.104c(Zn2+)/c?=x=3.4x10-7q(Zn2+/Zn)=q(Zn2+/Zn)+2.303RTlg{c(Zn2+)/ce}=-0.95V

Cu2++4NH3=Cu(NHCu2++4NH3=Cu(NH3)42+ceq/mol?L-1x0.100.10K?{Cu(NH)2+}=-0.10=2?1X1013c(Cu2+)/c?=x=4.8x10-ii

2.303RT9(Cu2+/Cu)=9(Cu2+/Cu)+lg{c(Cu2+)/c叭=0.04V2F￡=0.04V-(-0.95V)=0.99V10.20CuCl+=Cu2++Cl-15—xK6=-=1.0x=15—xK6=-=1.0x=0.50fx(1.5+x)c(CuCl+)=1.0mol?L-1c(Cu2+)=0.50mol?L-110.21AgCl+Cl-=AgCl2-ceq/mol?L-1x0.10K?=K?(AgCl)K?(AgCl-)=0.105310=5.3x10-5spf2xx=1.9X103m(NaCl)=c(NaCl)VM(NaCl)=1.1X104g加入沉淀劑過(guò)多，可能由于配合物的生成反而使沉淀溶解度增大。10.22(1)Zn2++4NH3=Zn(NH3)42+ceq/mol?L-11.0X10-4x0.20K?=-0.20=2.9x109f1.0X10—4Xx4x=0.029即：c(NH3)=0.029mol?L-1c(OH-)/c6=,-K&(NH)c(NH)/c?=7.2x10-4b33{c(Zn2+)/c?}{c(OH-)/c?}2=5.2x10-11>K?{Zn(OH)}sp2.有沉淀生成(2)同理可計(jì)算得，當(dāng)c(Z