物理化学试卷与练习03-2003级物化答案电池

物理化学答案（可逆电池1.10H2(p½O2(p)(E/T)p8.54×10-4V·K- (5分E2=1.2495电池反应：Cd(s)+Cu2+─→Cd2++ (1分E(Cu2+/Cu)(Cd2+/Cd)0.740 (1分mrG$=-zEF142.8kJ·mol- (1分mmKexp(-rG$/RT (2分m5

H2(g)+ =Hg(l)+Hg(l)+½O2(g)= =H+(aq)+OH- H2(g)½O2(gH+(aqOH (3分rG$=fG$(H+)+fG$(OH-)=-157.5kJ·mol- m求得：fG$=-157.5kJ·mol- (2分m10[答](1T298K(E/T)p=－[40.6+1.9(T/K-293)-0.03(T/K-293)2]×10-6V·K-=－4.935×10-5V·K- (3分 (E/T)p＝－40.6×10-6V·K- (3分在恒压时T有微小变化,(rSm/T)p＝rCprCp,] (4分5

ECl2/ClCu2+/Cu)1.0225 (2分lnKzEF/RT (2分求得K=3.91×10 (1分5

E=E-RT/zF×ln[1/a(ZnCl2)]=E-=1.043mrG$=-zEF=-201.3kJ·mol-m15[答 2AgCl(s)→Ag(s1 （2分2

H$TS⑴E

r

r H$H$(AgCl)1.2703105J 2r2r

（5分得E=1.105⑵Qr=TrSm2.042×104 （2分 S

)pr

VK

(2分 zE$

p⑶lnp

43.04

2)2 22

=4.21033

（4分10[答](1T298K(E/T)p=－[40.6+1.9(T/K-293)-0.03(T/K-293)2]×10-6V·K-=－4.935×10-5V·K- (3分 (E/T)p＝－40.6×10-6V·K- (3分在恒压时T有微小变化,(rSm/T)p＝rCprCp,] (4分9.109.10[答Hg(l)|Hg2+||OH （4分⑵E

0.099V0，能自发进行 （4分⑶E0

[Hg2]2(107[Hg2+]=2.09×10-2mol·kg- Hg2+浓度大于此值时不再反应 （2分5rG$=rH$-TrS$=-318.1kJ·mol- (2分 mErG$/zF1.65 (2分mm(E/T)prS$/zF5.2×10-4V·K- (1分m10电池反应：Ag(s)+(1/2)Hg2Cl2(s)=AgCl(s)+ (3分mrH$=-zF[ET(E/T)p5326J·mol- (3分mmrS$=zF(E/T)p32.61J·K-1·mol- (2分mQrTrSm9718J·mol- (2分2 -24.10kJ·mol-10[答 Pt,H2|H+||I- （4分E=E0.753V0= （2分E<0，电池为非自发，即H2不能还原 （4分5S(槽)=- (1分S(池)=(rHm-rGm)/T=[Qp-(- (2分S(总)=S(槽)+S(池)=Wf/T=3.65J·K- (1分还需rHm或 (1分15.1015.102CuCu2 E2(0.5210.337)0.368 (2分mK'=exp(-rG$/RT)=mK1/K6.03×10- (3分[Cu+]2/(0.01-(1/2)[Cu+])=[Cu7.77×10-5mol·kg- (2分2Cu(NH32NH3 E0.631 (1分2mrG$=-zFE60.9kJ·mol- (2分m16.10电池反应：Ag(s)+(1/2)Hg2Cl2(s)─→AgCl(s)+ (1分rH$=fH$[AgCl(g)]+fH$[Hg(l)]-fH$ m1/2)fH$[Hg2Cl2(s5.44kJ·mol- (2分mrS$=S$[AgCl(g)]+S$[Hg(l)]-S$[Ag(s)]-(1/2)S$ 32.9J·K-1·mol- (2分rG$=rH$-TrS$=-4.37J·K-1·mol- (2分 mErG$/zF0.045 (1分mm(E/T)prS$/zF3.41×10-4V·K- (2分m17.10电池反应：(1/2)H2+AgCl─→Ag+ (1分mrG$=-zEF19314J·mol- (3分mrSmzF(E/T)p8.45J·K-1·mol- (3分Cp,m=(rHm/T)p=174.9J·K-1·mol- (3分18.10 (2分 Zn(s)─→Zn2+(a=1)+ Cu2+(a=12e- (2分mrG$=rGmzFE211.03kJ·mol- (2分mmrS$= 82.8J·K-1·mol- (2分mrH$=rGmTrS$=-234.87kJ·mol- (1分 mQrTrS$=-23.846kJ·mol- (1分m5rSmzF(E/T)p13.51J·K-1·mol- (1分QrTrSm4026J·mol- (1分rHmQp43Qr1.731×105J·mol- (1分=- (1分ErGm/zF1.836 (1分15[答 （3分 （3分⑴-⑵得 （2分 ==(–z1E1F+2z2E2F ==(–z1E1F+2z2E2F–2.372×105)J·mol-=–2.392×105J·mol-（2分（5分21.10rG$(293K)=-zFE(293K)=-25.98kJ·mol-(1分rG$(303K)=-zFE(303K)=-25.67kJ·mol-(1分rG$(298K)25.82kJ·mol-(2分rS$(298K)=-(rG$/T)p=-31.00J·K-1·mol-(3分mmm rH$=rG$+TrS$=-35.06kJ·mol- (3分 22.10Pb(s2AgCl(s2Ag(s (2分E=E=0.49004mrG$=-zFE94.56kJ·mol- (1分mrG$(2)zFE$=-40.74kJ·mol- (1分 rG$=rG$(1)rG$(2)53.87kJ·mol- (1分 rH$(1)=rG$(1)+ 105.3kJ·mol- (2分rH$(2)=rG$(2)+ 48.05kJ·mol- (2分rH$=rH$(1)rH$(2)57.2kJ·mol- (1分 23.10Pb(s2AgCl(s2Ag(s (2分E=E=0.49004mrG$=-zFE94.56kJ·mol- (1分mrG$(2)zFE$=-40.74kJ·mol- (1分 rG$=rG$(1)rG$(2)53.87kJ·mol- (1分 rH$(1)=rG$(1)+ 105.3kJ·mol- (2分rH$(2)=rG$(2)+ 48.05kJ·mol- (2分rH$=rH$(1)rH$(2)57.2kJ·mol- (1分 24.10 (2分mm=2fG$(H+)+2fG$(I-)-fG$(H2)-fG$ 103.34kJ·mol- (2分mE=ErG$/zF0.5354 (2分mmKexp(-rG$/RT (1分mE=E=0.5354mrG$=-51.67kJ·mol-mK (1分25.10电池反应：H2(g,p)+(1/2)O2(g,p)= (2分mrGmfG$(H2O237.19kJ·mol- (2分mErGm/zF1.228 (2分m(E/T)p=E/T+fH$m8.7×10-4V·K- (4分26.10 Hg(l)+Br-(aq)─→(1/2)Hg2Br2(s)+ AgBr(s)+e-─→Ag(s)+Br- Hg(l)+AgBr(s)─→(1/2)Hg2Br2(s)+ (3分E=0.06804rGmzFE6.566kJ·mol- (2分rSmzF(E/T)p30.108J·K-1·mol- (2分rHm=rGm+TrSm=2412J·K-1·mol-1 (2分)通2F电量Wf=zFE=13.13kJ·mol- (1分)27.10AgCl(s)Ag(1分E=(AgCl/Ag)-(Ag+/Ag)=-0.5767(1分K=exp(zFE/RT)=1.76×10-(1分m=m×(K)1/2=1.326×10-5mol·kg-(2分lg=-A│z+z-│I1/2=-==(3分mrG$=-zFE55.64kJ·mol- (2分m28.10mrGmrG$=-zEF474.8kJ·mol- (1分mmrHm2fH$(298K)571.8kJ·mol- (1分mrUmQWrHm-∑RT564.4kJ·mol- (1分rSm=(rHm-rGm)/T=-325.6kJ·mol-1 (1分)因rGm<0 过程(1)为自发过程，又∵W =0-rGm-Wf>0过程为不可逆过 (2分(1)rGm、rHm、rUm、与(1)相 (2分Wf374.78kJ·mol-1-rGmWf100.0kJ·mol- (2分29.10 I2+2e-─→2I-(aq)(- 2Ag(s)+2I-(aq)─→2AgI(s)+电池反应：2Ag(s)+I2(s)= (3分289500C3molQr=zFT(E/T)p=8632.9短路时Wf= Qp=H=-190.26EQr-H)/zF0.687 (4分QzF(E'-E)+zFT(E/T)p10.383kJ·mol- (3分5E=E(298K)-4.06×10-5(T/K-=1.0186V-4.06×10-5(T/K-rGmzFE(298K196.6kJ·mol- (2分rHm=rGm+TrSm=rGm+198.9kJ·mol- (3分10rG$ rH$-TrS$=-2750kJ·mol- m最大电功是-rG$=2750kJ·mol- (2分mrF$=-[rG$- m=-[rG-m2739kJ·mol- (∑=- (3分10Zn(s2AgCl(s2Ag(s (2分rGmzFE195.9kJ·mol- (2分rSmzF(E/T)p94.96J·K-1·mol- (2分rHmrGmTrSm224.2kJ·mol- (2分rUmrHm(pVrHm224.2kJ·mol- (1分rFmrGm195.9kJ·mol- (1分QrTrSm28.30kJ·mol- (1分Qp=rUm+W=rUm=-224.2kJ·mol- (2分Qp=rUm+Wf=rUm+zFE'=-69.8kJ·mol- (2分33.10E1.01823 (2分=- (2分rGmzEF196.5kJ·mol- (2分rSmzF(E/T)p9.65J·K-1·mol- (2分rHmrGmTrSm199.4kJ·mol- (2分2Cu(s)+2AgAc(s)─→Cu(0.1mol·kg-1)+2Ag+2Ac-(0.2mol·kg-(3分rGmzEF71.769kJ·mol- (1分rSmzF(E/T)p38.6J·K-1·mol- (1分rHmrGmTrSm60.293kJ·mol- (1分Ag│Ag+‖Ac-(AgAc,Ag)=0.6413Ksp2.07×10 (4分5rGmzEF4.391kJ·mol- (2分rSmzF(E/T)p32.6J·K-1·mol- (2分rHmrGmTSm5.324kJ·mol- (1分36.104[答](1)(-)Zn(齐)＋SO2＋7H2O－2e- (2分44(+)Hg2SO4(s)＋2e-─→SO2＋2Hg(l) (1分)电池反应:Zn(齐)＋Hg2SO4(s)＋7H2O─→ZnSO4·7H2O(s)＋2Hg(l) (2分)4(2)E298＝1.4202 (1分(E/T)p298＝－0.00133V·K- (2分rHm＝－2EF＋2FT(E/T)p350.59kJ·mol- (2分37.10解设计电池电池反应rG$＝44.92kJ·mol- E＝－rG$/(2F)＝－0.2327 PbSO4,Pb)＝E＋Pb2+,Pb)＝－0.3587 (4分mKsp＝exp[－rG$/(RT)]＝1.3×10- (3分mKsp＝x(x+0.01)＝1.3×10- x1.3×10- (3分38.10(1)mrG$(1)zE1F89.108kJ·mol- (1分mmrS$(1)zF(E/T)p0.02779J·K-1·mol- (1分mrH$(1)=rG$(1)+TrS$ 97.39kJ·mol- (1分(2)mrG$(2)zE2F178.65kJ·mol- (1分mmrS$(2)zF(E2/T)p0.0569J·K-1·mol- (1分mrH$(2)=rG$(2)+TrS$ 195.60kJ·mol- (1分H2(p)+(1/2)O2(p)─→ fG$(HgO)=rG$(3)-rG$(2)58.54kJ·mol- (1分 fH$(HgO)=rH$(3)-rH$(2)90.25kJ·mol- (1分 fG$(Cu2O)=rG$(1)+rG$(3)-rG$ 147.65kJ·mol- (1分fH$(Cu2O)=rH$(1)+rH$(3)-rH$ 187.64kJ·mol- (1分39.10H2(p½O2(p)mrHm=fH$=-zEF+m则：(E/T)p=-8.54×10-4V·K- (5分mzF(E2/T2-E1/T1)=rH$(1/T1-mE21.2495rGmzFEFE22.509kJ·mol- (2分mrGm=G$+ 求 p=16.05 (3分m2(Fe3+/Fe)=[2(Fe2+/Fe)+(Fe3+,Fe2+/Pt)]/3=-0.03635Sn2++2e-─→ G$=-2F Sn4++2e- G$=-2F Sn4++4e-─→ G$=-4F G$=G+G$=-4F 3求得：$=0.005 (3分3 (2分5 3Sn2+─→3Sn4++6e- 2Fe3++6e-─→2Fe电池反应：2Fe3++3Sn2+─→2Fe+ (2分E=E-RT/zF×ln[a3(Sn4+)/a2(Fe3+)a3(Sn2+)]=-0.170 (3分5 E(B)=0.608Na(Hg)+H+(1mol·kg-1)─→Na+(1mol·kg-1)+E(A)(Na+/Na0.05916lga(Na(Hg1.2 (2分(B)Na(Hg)+H2O(l)─→NaOH(0.01mol·kg-1)+E(B)=-(Na+/Na)-0.059160.608 (3分 5电池反应：(1/2)H2(p)+(1/2)Cl2(p)─→ (1分E1ERT/Flna(HCl10mol·kg- (1分E2ERT/Flna(HCl6 kg- (1分E=E2-E1=RT/Fln[a(HCl,10mol·kg-1)]/[a(HCl,6mol·kg-=RT/Fln[p(HCl,10mol·kg-1)]/[p(HCl,6mol·kg-0.0873 (2分5电池反应：Sn(s,白)Sn(s,灰291K达平 G=G= (1分S=(H-G)/T=-6.91J·K-1·mol-1 (1分)在273–298K间视H为常数273K：GHTS124J·mol- (1分EG/zF0.00064 (1分298K：G=49J·mol-E0.00025 (1分5电池反应：Cd(s)+2AgCl(s)=2Ag(s)+Cd2+(aq)+2Cl-(aq) (1分)E=(AgCl(s)/Ag(s))-(Cd2+/Cd(s))=0.625V (1分)E=E-RT/2F×ln[a(Cd2+)a2(Cl-)]=E-RT/2F×ln4.74×10-0.753 (3分242Ag2H+(0.2mol·kg-1SO20.1mol·kg-4Ag2SO4 (1分4E=E-RT/zF×lna(H2)/a2[H+]a[SO244a(H2)/a2[H+]a[SO2]4(EE)298/(EE)308 (2分求得：(E-E)308=-0.072 (2分1010.0% 为负极，4.93%Tl(Hg)为正 (2分E (2分稀释热Qp=Hm=-zEF+zFT(E/T)p=41.76J·mol- (3分E(40℃E(30℃E/T)p×(40300.030953 (3分2 H+(1.0mol·kg-1)─→H+(0.5mol·kg-E=-RT/F×ln(0.5/1.0)=0.01785组成电池： (1分E(Cd2+/Cd)(Fe2+/Fe0.037 (1分Kexp(zFE/RT (1分设平衡时[FeSO4]= = x0.00947mol·dm- (1分[CdSO40.00053mol·dm- (1分10

H2(p1)─→2H++2e- 2H++2e-─→电池反应：H2(p1)─→ (3分ErGm/2FRT/2F×ln(f2/f10.0795 (2分10 (2分) Sn2+(aq)-2e-─→Sn4+(aq)2Fe3+(aq2e- (2分E+)-(-)0.62 (2分lnKzEF/RT K (4分10mU(NaU(ClF0.01067S·m2·mol- (2分kmc1.07S·m- (2分(- 1/2H2(p)-e-─→H+(0.100mol·kg- AgBr(s)+e-─→Ag(s)+Br-(0.100mol·kg-电池反应AgBr(s1/2H2(p)Ag(sHBr(0.100mol·kg-E=E- 求得a= (以上各1分a/(m/m (2分55.10(-)H2(g,p)+2OH-(aq)─→2H2O(l)+(+)HgO(s)+H2O(l)+2e-─→Hg(l)+2OH-电池反应：H2(g,p)+HgO(s)─→Hg(l)+ (2分EE-RT/zF×lna(Hg)a(H2O)/a(H2)a(HgOE0.924 (3分ln(K$/K$)=zF/R(E$/T2-E$ H/R×(1/T2 (4分2求得：E$=E2=0.93 (1分25mrS$=-102.09J·K-1·mol- (1分mrG$=rH$-TrS$=-538.58kJ·mol- (2分 rG$(FrG$-rG$(Na276.704kJ·mol- (2分 (1/2)H2(g1/2)F2(g)H+(aq)FEF2/FmrG$=-zEFFF2/F (2分mrG$=fG$(HfG$(FfG$(F (2分 m (F2/F-)=-fG$(F-)/F=2.867 (1分m2

Cu2+(a2)E=RT/2F×ln(a2/a1)=0.017510Cu2+(0.1mol·kg-1)+2e-→RT/2F×ln0.10.307 (2分Cu(OH)2(s)+2e-→Cu(s)+2OH-(0.1mol·kg-=RT/2F×ln[0.1]20.165 (2分故知 为正极，OH- 为负 (2分 Cu2+(aq)+2OH-(aq)= (1分E(Cu2+/Cu)(Cu(OH)2/Cu)0.472 (3分10

电池反应Zn(s)+Hg2Cl2(s)─→2Hg(l)+ (2分E=E-RT/2F×lna(Zn2+)a2(Cl-E (2分lg=-A│z+z- 求得= (2分代入上式求得E=1.030 (2分mrG$=-zEF198.8kJ·mol- (2分m5 Ag│Ag+(a=1)，Br- AgBr(s)=Ag+(a=1)+Br- (2分E=(+)-(-)=-0.7279VlnK$=zFE/RT=-K$=4.8662×10- (3分10 (1分)电池反应：2Ag(s)+(1/2)O2(0.21p)─→Ag2O(s)E=E-RT/2F×ln=(O2,OH-)-(Ag2O,Ag)-0.05915/2×lg0.0470 (4分E0AgOH-后， ,Ag)=-0.31V<(Ag2O,Ag)，这时负极发生2氧化成Ag(CN)的反应，而不是生成Ag2O的反应 (5分210求得a= (2分aa)2 (1分rGmzEF37.153kJ·mol- (2分101 Cd(s)+2OH-─→Cd(OH)2(s)+ 12 2H2O+ ─→2OH-+ 2 Cd(s)+2H2O─→Cd(OH)2(s)+ (2分rGmzFE (1分rHmzFEzFT(E/T)p115.11kJ·mol- (2分rSmzF(E/T)p386J·K- (1分利用Kw(H2O)=1×10- 求出$=$=-0.828 Cd(s)│Cd2+(a1)‖OH- Cd(OH)2=Cd2++E$=$-(Cd2+/Cd)=-0.425 K$=exp(2FE$/RT4.29×10- (4分 10 (2分E=E-(Sn2+/Sn41.66 (2分Sn4++4e-─→SnSn2++2e-─→相减 Sn4++2e-rG$(Sn2+,Sn4+)=rG$(Sn,Sn4+)-rG$ =-2F求得：(Sn2+,Sn40.154VE1.834 (3分E>0，正向反应能自发进 (1分lgKzFE/2.303RT K10182 (2分10

电池反应：Pb(s2H+(a+0.12Cl-(a-0.1)PbCl2(s)+H2(a(H2)=设计电池：PbCl2(s)─→Pb2E=(PbCl2,Pb)-(Pb2+,Pb)==-0.1071(PbCl2,Pb)=E+(Pb2+,Pb)=-0.2331则：E=0.1445 (各2分10mrG$=-zFE=-zF((PbO2/PbSO4)-m404kJ·mol- (2分EERT/F×ln[a(H2SO41.84 (2分rGmzFE355kJ·mol- (1分5电池反应：1/2H2(p1/2=H+(0.1mol·kg-1)+Cl-(0.1mol·kg-=E-RT/F×ln[a(H+)a(Cl-=1.505电池反应Ag(s)─→ (2分E=- 求 a(Ag)= (2分=a(Ag)/x (1分10 (2分rGm=rHm-rHmQpQv+∑RT255.26kJ·mol- (1分rSmzF(E/T)p97.35J·K-1·mol- (1分代入上式得：rGm=-226.25kJ·mol- (1分ErGm/zF1.1723 (2分E=((Ag,Ag2O/OH-)-(H2/H+)-0.05915/2×lg(10-Ag,Ag2O/OH0.3442 (3分10

(2分E=((+)-(-- (3分=((+)-(-))-E0m，当m≥4mol·kg-1即可 (5分10电池反应：H2+1/2O2+H2O─→2H++ (2分mmrG$=-2EF=-77.2kJ·mol-1 (1分)H2+1/2O2→H2O(l) fG$(H2O)mmm 2H2O(l)2H 2G$(电离mH21/2O2H2O2H (3分m求得rG$(电离)=80kJ·mol- (2分mmrG$(电离)=- Kw=9.5×10- (2分m54Cr2HSO2CrSO4 (1分4E=( -$)-0.05915/2×lg[1/a2(H+)a(SO2 =0.152E $)+0.05915/2×lg(a)3=0.147 (2分 73.10Cd(s)│Cd2+(a=1)‖I- (1分E+)-)0.9384 (2分mrG$=-zEF181.1kJ·mol- (2分mmKexp(-rG$/RT5.56×10 (2分m1E不 (1分mrG$'=-90.55kJ·mol- (1分mK'K)1/2 (1分74.10串联电池总反应：H+(m1Cl-(m1)H+(m2ClE=-RT/F×ln(()2m2)2求得,1/,2= (3分电池反应Cl-(m1)ClE(实测ECt+ (3分=t+×(HCl)=0.00588S·m2·mol-H =t-×=0.0025S·m2·mol- (4分 5(Fe3+/Fe23(Fe3+/Fe2 (2.5分(Sn4+/Sn22(Sn4+/Sn) (2.5分5电池反应Ag(s)+1/2Hg2Cl2(s)─→AgCl(s)+ (1分rG$=fG$(AgCl(s))-1/2fG$ 4.395kJ·mol- (2分mE=ErG$/zF0.04554 (2分m77.10E=RT/zF×lnKp0.1923 (5分EE-RT/4F×ln[p2(H2O)/p4(HCl0.1151 (5分78.10Fe、CdCd(s)│Cd2+(0.1mol·kg-1),Fe2+(0.1mol·kg-Cd(sFe2+(0.1mol/kg)Fe(sCd2+(0.1E=((Fe2+/Fe)-(Cd2+/Cd)-=-0.0383E< 说明Fe(s)首先氧化成 (5分E=E-RT/2F×ln[a(Cd2+)/a(Fe2+)]=0.0054E> 说明Cd(s)首先氧化成 (5分79.10电池反应2H2(p)+O2(p)─→2H2O(l,3.2 (2分O2(p2H2(p)2H2O(l,3.2 2H2O(g, ←──────2H2O(g,3.2∴-rGm=G1+G2+=0+zRTln(p2/p1)-473.6kJ·mol- (4分mErG$/zF1.227 (4分m81.5

Zn2+(a=0.02)ERT/2F×ln(0.004/0.020.0207 (3分 SO2(a=0.01)─→SO2 ERT/2F×ln(0.001/0.010.0296 (2分5 H2(2p)─→E=-RT/2F×ln(1/2)=0.00890 H+(a=0.1)─→(3分E=-RT/F×ln(0.01/0.1)=0.0591(2分5 1/2Cl2(2p)─→ERT/F×ln(1/2)1/20.00890 (3分 Cl-(a=0.1)─→Cl-ERT/F×ln(0.01/0.10.0591 (2分5电池反应：Zn(s2H+(a(H+)=0.02Zn2+(a(Zn2+)=0.01)E(计算)=E(实测)-EJ=0.725 (2分E(计算)=-(Zn2+/Zn)- (2分又a(H2p(H2)/pp(总p(水)]/p故(Zn2+/Zn)=-0.7658 (1分5 Cl-(0.5mol·kg-1)─→Cl-(0.05mol·kg-1)EC=-RT/F×ln[(0.05×0.812)/(0.5×0.649)]=0.0534VEJ=E(实测)-EC=0.0002VEJ=(1-2t-t- (1分10无迁移电池(即浓差电池)反应：Ag+(a+,2)─→ (2分有迁移电池总反应：t-AgNO3(a2)=t- (3分E(浓差)=-RT/F×ln[a1(Ag+)/a2(Ag+)]=0.03478 (2分 (2分E(液接)=E(电池)-E(浓差)=2.5×10- (1分2 2H++2e-─→H2(p=10.1325 H2(101.325kPa)2H (1分电池反应：H2(101.325kPa)─→H2(10.1325 (1分E=-RT/zF×ln(p2/p1)=0.029653对于反应：I2+I-─→I E=-0.0010 (2分3KcIII (1分 zEF z= (1分3-0.0010=0.05916/2×lg[I33[I0.462mol·dm- (1分35电池反应：Ag+(0.10mol·kg-1)+Cl-(0.05mol·kg-1)→ (2分EERT/F×ln[a(Ag+)a(Cl0.5766 (4分lnKa= Ka=9.62×Ksp1/Ka1.0410- (4分5组成电池：电池反应：CuI(s)─→Cu++ (2分E=(Cu/Cu+)-(Cu /CuI)=-0.71V (1分)lnKsp=zFE/RTKsp1.0×10- (2分5 Zn(s)+2AgCl(s)→2Ag(s)+Zn2+(0.01021mol·kg-2Cl-(2×0.01021mol·kg- (