电化学原理习题答案.pdf

copyright天下downskyer电化学原理第一章习题答案电化学原理第一章习题答案1、解：2266KClKClHOHO0.001141.31.010142.31010001000cKKKKcm11=+=+=+=溶液2、解：EVFii=，FEVii=，10288.0=+scmVH10050.0=+scmVK10051.0=scmVCl3、解：62.5501212=+=eqcmKCloHCloKOHoOHo2-14WOccccK1.008101.004HH+=设故，2811c5.510cm1000oHO=4、（1）121Cltt1t76.33mol(KCloKCloClcm+=+=中）121121121KNaCl73.49mol50.14mol76.31mol(NaClooocmcmcm+=同理：，中）（2）由上述结果可知：121ClNa121ClKmol45.126mol82.142=+=+cmcmoooo，在KCl与NaCl溶液中Clo相等，所以证明离子独立移动定律的正确性；（3）vscmvscmuvscmuFuaoololoio1020.51062.71091.724N24K24CC=+5、解：Cu(OH)2=Cu2+2OH-设=y；2Cuc+OHc=2y则KS=4y3因为u=ui=KH2O+10-3yCu2+2yOH-以o代替（稀溶液）代入上式，求得y=1.3610-4moldm3所以Ks=4y3=1.00610-11(moldm3)36、解：=+，令=y，3AgIO+Ag3IOAgc+3IOc=y，则=ySK2，K=iK=+（yOHK2310+Ag+y3IO）作为无限稀溶液处理，用0代替，=+yOHK23103AgIO则：y=43651074.1104.68101.11030.1=Lmol=ySK2=3.038102)(Lmol7、解：HAco=HClo+NaAco-NaClo=390.7，121eqcmHAco=9.02121eqcm0=0.023，=1.69K_2)1(V5108、解：由欧姆定律IR=iSKSl=KilK=1000cIR=1000ilc=V79.05.01261010105339、解：公式log=-0.5115|+ZZI（设25）C1课后答案网copyright天下downskyer（1）=0.9740，I=212iizmI=212iicz，=（）m+mm1（2）=0.9101，（3）=0.6487，（4）=0.811410、解：=+Ha+Hm，pH=-log=-log（0.209+Ha4.0）=1.08电化学原理第二章习题答案电化学原理第二章习题答案1、解：()+2326623SbOHeSbHO+，()236HH+6e+，电池:2322323SbOHSbHO+解法一：00GEnF=83646F=0.0143V，E=+0E2.36RTF2232323logHSbOSbHOPaaa=0.0143V0E解法二：0602.32.3loglog6SbSbHHRTRTaaFF+=+=+；2.3logHRTaF+=000.0143SbEE+=V2解：，()+22442HOeHO+)224HH+4e+；电池:22222HOHO+2220022.3log4HOHOPPRTEEEFa=+=查表：0+=1.229V，0=0.000V，001.229EV+=视为无限稀释溶液，以浓度代替活度计算()242SnSne+，()，电池:32222FeeFe+23422SnFeSnFe2+23422022.3log2SnFeSnFeCCRTEEFCC+=+=（0.771-0.15）+220.05910.001(0.01)log20.01(0.001)=0.6505V()，(0.1)AgAgme+()(1)AgmeAg+(1)(0.1)AgmAgm+电池：(1)0(0.1)2.3logAgmAgmaRTEEFa+=+，（其中，=0）0E查表：1m中3AgNO0.4V=，0.1m中3AgNO0.72V=，2.310.4log0.0440.10.72RTEVF=3、解：2222|()()|()ClHgHgClsKClmClPPt()2222HgClHgCle+，()222CleCl+，222HgClHgCl2+电池：2课后答案网copyright天下downskyer3222200002.3log2ClHgHgClPaRTEEEFa+=+=O1.35950.26811.0914(25C)EV，=设由于E与无关，故两种溶液中的电动势均为上值Cla其他解法：E+=0，亦得出0E+=按Cla计算+，查表得甘汞，则E+=甘汞4、解法一：23(1)|(1)()HPtHatmHClaAgNOmAg+=()222HHe+222，()AgeAg+g，2222HAgHA+电池：有E+=+，02.3log()AgAgAgRTEmF+=。解法二：223|()()()HgHgClsKClAgNOmAg饱和，02.3log()AgAgAgRTEmF+=+甘汞同上解法:23(1)|(1)Pb()PbHPtHatmHClaNOm+=22Pb2PbHH+电池：220PbPbPb2.3log()2RTEmF+=同上解法:232e(1)|(1)FeCl()FeCl()FHPtHatmHClamm+=32+21FeFe2HH+电池：232302.3log()FeFeFeFeRTEaF+a=5、解：424()()ZnZnSOmHgSOsHg，22402.3log()2SOZnRTEEaaF+=222()()()HPbPbClsHClmHPPt；-20222.3log()2HHClRTEEaaPF+=+6、解：2(1)|(1)(0.1)()HPtHatmHClaKClmAgClsAg+=；()AgCleAgCl+，02.3logClRTaF+=，查表得：0+=0.2224V，0.79V=；0.22240.0591log(0.790.1)0.2876EV+=；7、解：2.312.31loglogaClClRTRTEFaFa=2.3logClClaRTFa=（=0.01m，=0.1m）aa可计算出：，0.10.780.078()Claaa=0.010.9040.00904()Claaa=又2.32log(bRTEtaaF+=)，2.30.078(21)log()(20.3891)0.0591log0.01230.00904baRTEEtaaVF+=若用浓度计算，则，0.05910.046abEVEV=0.01baEEV=8、解：22(1)|(1)|()PtHatmHIaIsPt=课后答案网copyright天下downskyer4+222()222HHe，()IeI+I，222HIH+电池：0202.3log()2HIRTEEaEF=030.53460.1310(3525)0.5333EV=，0E电池表达式中正负极未写错。00ln2965000.5333102.9GnFERTKkJ=0lnRTEKnF=，2965000.5333ln40.198.314298K=，172.8610K=第问不变（即E不变），00151.52GGk=J，85.3510KK=+2222HgCleHgCl，两者均改变。9、解法一:()222HHe()+222222HHgClHgClH+电池：022.3log()22HClRTEEaaF+=查表得，0.790.1Cla=02.32.3loglogHClRTRTaEEaFF+=+得，又lo，则log6.9Ha+glog14HOHaa+=7.18107.9410OHa=解法二:，00.0591logCla+=0.0591log0.0591HapH+=E+=，6.90.0591EpH+=，7.183107.9410()OHamoldm=10、解：，；2222HgHge+22242HgSOHgSO+4242422HgSOHgSOe+；按：222202.3log2HgHgHgRTaF+=+222402.32.3loglog22HgHgSORTRTKsaFF+=+按：22244402.3log2HgHgSOSOSORTaF=2224420072.30.0591log0.7986log(6.510)0.63822HgHgSOSOHgHgRTKsVF+=+=+=11、解：()AgClAgCle+，()2212HgCleHgCl+，2212AgHgClAgClHg+电池：1965000.04554.391GnFEkJm=ol343911965002980.338105.329PEHGnFTkJmolT=+=+=31965002980.3381032.617PESnFTJmolKT=i12、解：()22ZnZn+e，()，22CueCu+22ZnCuCuZn+电池：课后答案网copyright天下downskyer51.1031.1012965001.101296500293189.8298293PEQpHnFEnFTkJmolT=+=+=1.1031.10129650077.2298293PESnFTJmolKT=ilnGnFERTK=，221ZnCuaa+=0021.101log37.20581nFEEEKRTRT=，377.9510K=22GGGFF13、解法一：（），Cu（），22CueCu+010.337V=eCu+020.520V=-，000000312123=+=00031220.33720.5200.154V，=；00330.1549650014861GFJm=ol，所以反应为自发。解法二：由0lnRTKnF=计算由=-，则132KKK=；0002331212ln(lnln)()RTRTRTFFKKKFFFRTRT=544+22442OHeHO+()001220.1V=14、解法一：()，()，224HHe22222HOHOg+电池：1KKp=，|22222lnln44HOHOpRTRTEKFFp=p()222228122247602.312.30.05910.0591logloglog(9.710)log1.226444411HOHOpRTRTEVFKpFpp=解法二：22.30.0591loglog(0.020.544)0.0622HHRTaVF+=，()2220222.30.0591(loglog)1.229log(0.020.544)log247601.1622OOHOHRTapF+=+=+=V()221.160.061.22OHEV=15、解：有2Zn+2Cu+H+三种离子可在阴极还原，22202.30.0591log0.763log(0.010.39)0.8322ZnZnZnRTaVF+=+=+=22202.30.0591log0.337log(0.010.41)0.2722CuCuCuRTaVF+=+=+=2.3log0.0591(5)0.30HHRTaVF+=所以，阴极析出的顺序为：Cu、2HZn课后答案网copyright天下downskyer当Zn析出时电位为-0.83V，（由，故可用平衡电位替代析出电位，即可忽略电极的极化）0i220.05910.83log2CuCua+=+，22log(0.830.337)39.490.0591Cua+=，2403.2410Cua+=240403.24107.9100.41CuaC+=molL，说明铜离子几乎完全极化了16、解：，；22PbPbe+222PbFPbF+222PbFPbFe+；00212.3log2RTKsF=+，002122log()(0.35020.1263)7.582.30.0591FKsRT=+=，得，82.6310Ks=17、()22ZnZn+e，()22222HgCleHgCl+22222HgClZnHgClZn+电池：()()()22300022.312.30.0591loglog2log4222aaaaaaaZnClRTRTEEEmmEmFaaF+=+=()300.0591log42bbbEEm=()()3340.059130.0591logloglog224aaaababbbbmmEEmm=+()()221.00441.100850.25148logloglog0.3130.059130.05910.00500abaabbEEmm=0.49ab=18、解：电池为无迁移浓差电池，电极对负离子（aCl）可逆。电池b为有迁移浓差电池，电极对负离子（Cl）可逆。2221112.32.3loglogaamRTRTEFaFm=，22110.08220.0082logloglog1.6952.320.05910.082aFEmRTm=，212.02=。212lnbaRTEtFa+=，33222157.9100.704482.8102lnbbaEEtmRTEFm+=00000HHHClHClt+=+12000.704483.859.03HClHtc1+=12100(1)24.77HClCltcmeq+=meq，6课后答案网copyright天下downskyer19、解：02.32.3logaAbBaRTRTmpHnFnFb=+22232333220.0591140.05911.33log1.330.1380.00985log66CrOCrOCrCraapHpHaa+=+=022.30.0591log0.2681log0.26810.0591log2aAbClClBaRTaanFb=+=，即lo()222logloglogloglogHOCdOHOHHCdOaaaaK+=2gloglogHCdOOHaaK=由14logOHpHa=+，214loglogHCdOpHaK=+20、解:线与线大约在时相交7.68pH=在8.7，14.5pH平，所以金属表面带正电荷。XC+=C_C+=C_=0=0X00a=0a=0dX=平=平C-C+CC-C+CXX1add8课后答案网copyright天下downskyer5、解：XX=0=0ddXX=1=1ddXX=2=2X00a=0a=01adddd2X00=12lnlnaRTRTCFF=+常数10=6、解：在0的范围内Cd，表明在和接近0q0q=的电极表面有阴离子吸附，故为加入了表面活性阴离子（无机离子）X0a=0dd1X7、解法一：根据dqCdd=，如图所示，321aa+b321aa+ba-ba-b在a点：()23213bbq+=+aa；在()点：ba()12122bbq=aa；9课后答案网copyright天下downskyer213122qqqb=()bb=aa2122qCb3=.解法二：取的平均电位1q()1122aaabb+=处的表面电荷密度；取的平均电位2q()2122aaabb+=+处的表面电荷密度；则1211aqb=，222aqb3=。则有，213121222222aaqqqqqCbbbb2+=。8、解法一：00.11(0.119)0.30V=a，10，且由题给数据知q，较大。c101212ln2lnlnryRTRTRTRTcFCFF=+a，11yaqqC=a；1022ln2lnlnrRTRTRTRTcFqFF=+aa1230.328.8510402988.310.05912log2log0.3log(10)0.1c=+30.2220.0591log(10)c=当0.001cmo=lL时，10.222V=12:0.222:0.33=a，双层结构较分散；当0.1cmol=L时，10.1038V=，11:3=a，分散层较小。解法二：1082rFqcRTShRT=，令12FZRT=，则32.64810qcShZ=；当0.001cmo=lL时，337.7642.648101qShZ=，()()1237.764ln37.76437.76412.3log75.54.325ZSh=+，120.222ZRTVF=同理可求0.1cmol=L时，10.1038V。9、解：从Cd曲线可知0.37V=0；未加时，NaCl0.250.27V=0稳，电极（钢）表面带正电荷，缓蚀剂为阳离子型，难吸附，不起缓蚀作用；加以后，ClNaCl吸附，缓蚀剂离子在Cl层吸附，故有缓蚀作用。10课后答案网copyright天下downskyer11电化学原理第四章习题答案电化学原理第四章习题答案1、解：，设，22CueCu+25oTC=2072.32.3log0.337100.1322CuRTRTaVFF+=+=+212HeH+，设，查表25oT=C0.130=，a=平，c=平则：()2.3log0.0591log20.1300.230.28cHRTaVF+=为使氢气不析出，阴极电位不可负于0.28V。2、解：222HeH+，已知，log6.5Ha+=()0.0591log1.230.846cHHaV+=平3、解：()22ZneZn+22，()ZnZne+()202.30.0591log0.763log1722ZnaZnRTaVF+=+=+=平若阴极上析氢：222HeH+c，()0.0591log0.0591log0.20.2651.061.1354cHHHaV+=平不使氢析出，则需c，设1.135cV=，则：()0.8171.1350.318ZncZnV=平最高槽压：V()0.8171.1350.50.818acIRV=+=+=4、解：因为在B液中极化度小于A液，所以铁在B溶液中易于腐蚀溶解0.00.81.00.00.40.5-VABjc(Acm2)-VABjc(Acm2)5、解：设初始速度为，改编后速度为；1j2j111expwjnFvnfKcRT=，222expwjnFvnFKcRT=321211410expexpexp0.1938.31293jwwwjRTRT=，2210.19318.6jj，比原来低81%mAcm=课后答案网copyright天下downskyer126、解：32296500110193jnFvAm=扩散扩散，22965000.2548250jnFvAm=电子电子与2193jA=m比较：该电极过程的控制步骤是扩散步骤。2193jjAm=扩散，24825019348057jjAm=电子jjj电子电子，电子转移步骤为非控制步骤，处于准平衡态。7、解：设原扩散速度为，活化能为；搅拌后速度为，活化能为。1v1W2v2W11expWvKcRT=，22expWvKcRT=，1221exp1000WWvvRT=，()31221lnln108.3129817106WWWRTvvJmol=扩散活化能应降低17106Jmol。8、解：()()202.30.0591log0.763log0.3310.7722cZnZnRTmVF+=+=+=，00aH=。0.777acEV=V=1.24V，而，VE，是电解池。作为电解池，负极为阴极，故Zn电极发生阴极极化，()VacEIR=+()()V1.240.7770.1640.10.199caEIRV=+=+=电化学原理第五章习题答案电化学原理第五章习题答案1、解：出现铜红色，表明锡离子扩散速度较小，在凹处离子浓度降低较多，采用间隙电流有助于消除凹处锡离子浓度与体浓度之差，而避免镀层中Cu含量过高。2、解：设的极限扩散电流为为为；1dj2dj3dj0dcjnFD=，322ddjj又由于正离子在阴极还原时，反应粒子的电迁移将使稳态电流密度增大，所以中有大量高外电解质，可忽略电迁移影响，其反应不同于故1ddjj：dGcGaXdXaaccdXdGcGaXX00a=0a=00=：G平dGcGaX2、()0.7j()05810.0581logloglog210log0.040.42411cRTRTjjFF=+=+=V5、牺牲阳极反应：22ZnZn+e0.05aV=，估计在弱极化区，且1.1n=，0420.9965000.051.1965000.05expexp1.13108.312988.31298ajjAcm=，240.05442.51.1310pRcmj=（本题可用TAFEL公式近似计算()0:50:ajj1，不可以用线性极化公式计算。）6、22CueCu+202log2.3CuFaRT+=+平，()()2020.310.3372log0.942.30.0577CuFaRT+=平，20.116Cuam+=olL，（若查得，则00.3448V=20.06Cuam+olL=）02.32.3loglogccRTRTjjFF=+平，()()()0910.310.23logloglog1.3100.4732.30.0577cFjjRT+=+=+=平，22.97cjAc=m7、02.3logcRTjjF=，()900.0581log1log102.3log0.940.556cRTjFj=090.9420.556logloglog108.992.30.0581ccFjjRT=+=+=，829.810cjA=cm课后答案网copyright天下downskyer16比原来速度（cj）增加了近九个数量级，可见对j影响之大，但实际中将受dj所限，不会达到dj值8、由于对阴极反应：，故可用TAFEL关系计算0cjjc：()902.3log0.0591log1log1100.532cRTjVjF=，0V0.832jEV=，()()V1.7650.8320.5320.40.001acEIRV=+=+=由于a很小，故应按线性极化或弱极化区进行计算028.31298112.832965000.001aaRTjjAnFcm=上述计算结果表明：阳极反应过程的可逆性很好（0j很大）；且阳极过程之0j比阴极反应过程的0j大得多。所以在题给条件下，槽电压的变化主要取决于阴极极化的大小；阳极极化很小，呈线性极化规律。9、电极反应：22244HOOeOH+用作图法求解：lo时g0j=0.36cV=，0.36aV=。设TAFEL区斜率为tg，0.093logbtgj=，00.36log3.870.093ajb=，041.35102jAcm=10、由作图法求解：lo时g0j=1.185cV=，1.185aV=。设TAFEL区斜率为tg，0.289logbtgj=，2.30.05910.200.289RTbF=01.185