物理化學(xué)試一、填空題(共1 2分2分(Fe2+/Fe)=+RT/2F×ln0.01=-0.4992(Cu2+/Cu)=+RT/2F×ln0.02=0.2868(96題8155分電池反應(yīng)AgCl(s)＝AglnK K1.810- (2分I=0.01mol·kg-lg (2分AgClm/mK)1/2/1.48×10-m1.48×10-5mol·kg- (1分5分Hg2SO4溶解的電池：Hg(l)│Hg2a1)‖SO2 電池反應(yīng)：Hg2SO4(s)─→Hg2+SO (2分 ERT/zFlnKsp0.180 (2分2(Hg2SO4/Hg)E(Hg2/Hg)0.609 (1分210分Cu(s)─→Cu2+(0.1mol·kg-1)+2AgAc(s)+2e-─→2Ag(s)+2Ac-(0.2mol·kg-Cu(s)2AgAc(s)─→Cu2+(0.1mol·kg-12Ac-(0.2mol·kg-1 (2分(E/T)p=E/T=0.0002V·K-rGm=-zFE=-71.80kJ·mol- rSm=zF(E/T)p=38.6J·K-1·mol-rHmrGm+TrSm60.29kJ·mol- (4分設(shè)電池反應(yīng) AgAc(s)=Ag++E=(Ac-,AgAc/Ag)-(Ag+/Ag)=RT/F×E=(Ac-,AgAc/Ag)-可得(AgAc/Ag0.638RT/F×lnKsp=(0.6380.8)V Ksp1.89×10- (4分5分(+)H2(p)─→2H+(aH+)+(-)Hg2Cl2(s)+2e-─→2Hg+陽(yáng)=- (2分E陰陽(yáng)0.664pHE-)/0.05916 (3分10分 (1分電池反應(yīng)AgCl(s)AgE(Cl-/AgCl(s)/Ag)(Ag+/Ag0.5767 (1分lnKspzFE/RT Ksp1.8×10- (2分Ksp=a(Ag+)a(Cl-)=mKsp)1/2/×m1.51×10-5mol·kg- (2分GzEF55.65kJ·mol- (2分(1)Ag-AgCl(s)│Cl(2)計(jì)算結(jié)果說(shuō)明反應(yīng)AgCl(s)Ag+(aq)Cl自左進(jìn) (2分10分We=pV=∑iRT=1.24Wf=zEF=14.658rHm=fH$(AgCl,s)-fH$(I-)=-6.44kJ·mol- mrUm=rH -∑iRT=-7.68kJ·mol-mrGm=-rFm=-Wmax=15.898kJ·mol-rSm=(rHm-rGm)/T=27.58J·K-1·mol-QRTrSm8.129 (1分Wf= We=1.24Qp=rHm=-6.44 其余狀態(tài)函數(shù)變量同 (4分10分 Pb(s)-2e- 2Ag+(a=1)+2e-─→(Ag+,Ag)=(Ag+/Ag)-RT/2F×ln[1/a2(Ag+)]=0.7991(Pb2+,Pb)(Pb2+,PbRT/2Fln[1/a(Pb20.126 (5分E右左0.7991V0.126V0.925rGm=-zEF=-178.5kJ·mol-*如果電池反應(yīng)的電子得失數(shù)為1，E不變，而rGm=-89.26kJ·mol- (5分9.10分1/2H2(p)─→H+(a)+(H+/H2)=-2.303RT/F×E(標(biāo))=()-(H+/H2)=()+2.303RT/F×pH(標(biāo) Ex=()+2.303RT/F× (pH)x(pH)s+(Ex- (5分(2)E=()-((醌)-pH= (5分*.10分(-)1/2H2(g,p)─→H+(+,m+)+(+)AgBr(s)+e-─→Ag(s)+Br-(-,m-)Pt│H2(g,p)│HBr(,m)│AgBr(s)│Ag(s)AgBr+1/2H2─→Ag+H++Br-E=E-RT/F×ln(a+·a-=E-RT/F×ln-RT/F×2RT/FlnEE2RT/F 求出 (5分1Ag2O(s)2Ag(s)2 (5分從rG$(T)求p(O2)，再用rH 和rS$，求T(分解) 5分Ag(s)│AlCl3(NaCl中)│Al(s)(Zn中電池反應(yīng)：Al(s)─→ (3分E=-RT/3F×lna(合金 a(Al)= (2分10設(shè)計(jì)電池Cu(sCu2+E0.368V lnKzEF/RT=-14.333 (5分K=(2x)2/(0.1-x)=5.96×10-x2.446×10-4mol·kg- (5分5H2Ag2O2AgfG$(Ag2OfG$(H2O,lzEF11.0kJ·mol- (2分 Ag2O＝2Ag+1/2G$=-fG$(Ag2O)=-RTlnK$=- p(O214.08 (3分5E(甘醌甘RT/Flna(H甘(甘)=RT/F×lna(H+)-0.19V=-0.058×pH-0.19V=-0.364 (3分(甘)-(醌)=0.019 (醌 =-0.383 -0.383V=RT/F×lna(H+)=-0.058× pH= (2分10 (3分E=E-RT/zF×lna2(HCl)=E-0.05916EE0.05916lg(m/m)20.2223 (3分E=RT/2F×ln[a2(HCl) p(H211.59 (4分10電池反應(yīng)：H2(pHg2SO4(s)2Hg(l)H2SO4(7mol·kg-H2(pPbO2(s)H2SO4(7mol·kg-1)PbSO4(s) (1分1E1E$+RT/2Fln[a(H2)]ln[a(H2SO4 (2分12E2E$+RT/2Fln[a(H2)a(H2SO4)/(a(H2O))2 (2分2(1)+(2) E1+E2-(E$+E$)=-RT/F× a(H2O (4分10Pb(s S2O2(a=1)─→PbSO4(s)SO21mol·kg- (1分 E=ERT/zFln[a(SO2a(S2O22.427 (2分 mG$=-zFE Ka1.69 (2分mmQ(可逆)=TS$=-13.979kJ·mol- (2分mQ=UW(膨)+WH+GTS+zFEzFE+zFETS96.390kJ·mol- (3分10 (3分E=((Ag+/Ag)-(Fe3+/Fe2+)=0.0281lnK$= K$= (3分 Fe2++Ag+─→Ag(s)+平衡 (0.05-x)mol·kg-aK$=(0.05- x= [Ag0.0442mol·kg- (4分a51/2H2(pAgCl(s)H+(0.1mol·kg-1Cl-(0.1mol·kg-1 (1分E=E-RT/F×ln[a(H+)a(Cl- a(H+)a(Cl-)=a [a(H+)a(Cl-)]1/2 (2分m=(m+m-)1/2=m=0.1mol·kg-am/m (1分pHlga(H (1分5 HgO+H2─→Hg+ mrG(1)zFE178.7kJ·mol- (2分mH2(g)+1/2O2(g)─→ (1)-(2)式：HgO(s)─→Hg(l)+1/2 G$(3)G$(1)G$(2)58.5kJ·mol- (2分 mG$(3)=- K=5.57×10-11=mp(O23.1410-16 (1分2110-=(甘)-E=0.2067 (2分(Hg2/Hg)RT/2Fln[a(Hg2 (2分 2(Hg2/Hg)RT/2F×lnKspRT/2F×ln(a2(Br (4分2Ksp5.54×10- (2分5電池反應(yīng)：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10Ag(s1/2Cl2(p)E=(Cl2/Cl-)-(AgCl/Ag) 得(AgCl/Ag)=0.222V (4分) Ag++Cl-─→ (2分fG(AgCl)=-zEF=-1×1.136×F=-109.6kJ·mol-10(-)H2(p)─→2H+(0.1mol·kg-1)+(+)Hg2Cl2(s)+2e-─→2Hg(l)+2Cl-(0.1mol·kg-電池反應(yīng)：H2(p)+Hg2Cl2(s)─→2Hg(l)+2HCl(0.1mol·kg- (3分E=E-RT/2F×ln[a2(HCl)/(p(H2)/p0.3979 (2分(G/p)TVV(產(chǎn)V(反=-RT(1/p+5.30×10-G2G1RT[ln500+5.30×10-9(500-且G=- ∴E2=0.4811 (3分2(Hg2/Hg2Cl2)=(Hg2+/Hg)+RT/2F×2Ka2.6210- (2分10電池反應(yīng)：1/2H2(p)+AgCl(s)→Ag(s)+H+(a(H+))+Cl- (2分EERT/Fln[a(H+)a(Cl (2分lga(H+)=(E-E)/0.05915-lga(Cl-將第一組數(shù)據(jù)代入a(H+)=1.566×10- (2分1.507×10- (2分同 Ka2=1.4998×10- Ka3=1.4818×10-Ka(平均1.4962×10-10電池(1)的反應(yīng) Fe+HgO─→FeO+ EE$-RT/2FlnQaE$=0.9370 (3分 電池(2)的反應(yīng) H2+HgO─→Hg+ EE$-RT/2Fln[a(H2O)/a(H2E$=0.9266 (3分 由反應(yīng)(1)- Fe+H2O─→FeO+ rG$3rG$1rG$22.007kJ·mol- (2分 所以 fG$[FeO(s)]=rG$+fG$[H2O(l)]=-239.197kJ·mol- (2分 2710電池反應(yīng)：Cu(s)IH+CuI(s1/2E=-(I-,CuI/Cu)-0.05915lg{1/[a(I-平衡時(shí) 則(I-,CuI/Cu)=0.05915lg[a(I- (3分 (CuI/Cu)=(Cu+/Cu)+RT/F×(Cu+/Cu)0.668 (3分Cu、Cu2+Cu+達(dá)平衡時(shí)，有(Cu+/Cu)∴(Cu+/Cu)=(Cu2+/Cu)+RT/2F×=0.517V (3分)將(3)式代入(2)、(1)得a(I-)a(H+)=0.0028[HI]0.0028)1/2m0.05mol·kg- (1分2810Hg2Br2(s)+2Cl-(0.1mol·dm-Hg2Br2(s)+2Cl-(0.1mol·dm-(1分 (E/T)p=-1.880×10-4(1分H=zF[T(E/T)p-E]=-35.373kJ·mol- Hg2Br2(s)─→Hg2a=1)2Br(2分 (a=1)‖(2分22a(Hg2)=Ksp/[a2(Br-2(Hg2/Hg)=(Hg2/Hg)+0.05915/2×lga(Hg2 =0.799+0.05915/2×lg 2(Hg2/Hg)=()-E=0.2064V (2分2(2)式代入(1)式得：Ksp=5.51×10- (2分2910設(shè)計(jì)電池：Ag│AgCl(s)│Cl-‖Ag+│Ag (1分)電池反應(yīng)：Ag++Cl-─→AgCllgKa= Ka=5.62× Ksp=1/Ka=1.78×10-sKsp)1/21.33×10- (4分lg=(-I0.01mol·kg- (3分a(Ag+)a(Cl-)=Ksp= Ksp/()2=2.25×10-s1.45×10-5mol·kg- (2分3010電池反應(yīng) Cu2O(s)+H2(g)─→2Cu(s)+mrG$1 (2分m電池反應(yīng) H2(g)+1/2O2(g)─→mrG$2 (2分m(2)Cu2O(s)─→2Cu(s)1/2rG$=(rG$1rG$2147.3kJ·mol- (2分 mG$=-RTln(Kp/p (2分mmln[p(O2)/p]1/2=-G$/RT=-mp(O23.510×10-50 (2分3110(-)H2(g,p)+2OH-(aq)-2e-─→(+)HgO(s)+H2O+2e-─→Hg(l)+2OH-電池反應(yīng)：H2(g,p)+HgO(s)─→Hg(l)+ (2分H2(g,p)+1/2O2(g,p)─→ (1) HgO(s)Hg(l1/2 (2分mrG$(1)=-zE1F=-178.81kJ·mol-mrG$(2)=rH$(2)-TrS$(2)=-237.22kJ·mol- rG$(3)=rG$(1)-rG$(2)=58.42kJ·mol- lnK$=-rG$ K$=5.78×10- p(O23.31×10-19 (1分5 mfG$(H2O) (5分m10電池反應(yīng)：1/2H2(p)+AgBr(s)─→Ag(s)+H+(a+)+Br-(a-)E=E-RT/F×ln(a+·a-)=(AgBr/Ag)-0.1183lgm-0.1183lg 又D-H，lg=-A│z+z-│m1/2=-Am1/2 由(1)(2) E+0.1183lgm=(AgBr/Ag)-0.1183Am1/2以(E+0.1183lgm)～m1/2作圖得一直線，截距為(AgBr/Ag)=0.0728V (5分)取一組數(shù)據(jù)求 (5分10Pt│H2(p)│HCl(0.7mol·kg-E=(Cl2/Cl-)-RT/2F×ln[a2(HCl)/a(H2)a(Cl2)]=1.18rGmzEF227.7kJ·mol- (5分2HCl(p)＝Cl2(p)+ 2HCl(p(HCl)=46.4mrG$=rGm,1+rGm,2=zRT×ln(p2/p1)+(-rGm)=189.6kJ·mol-mlnK$=-rG$/RT K$=5.8210- (5分 3510HH H2(p1)─→2H+(a+)+2e-2H+(a+)+2e-─→H2(p2)HH電池反應(yīng)：H2(p1) (3分

(RT/p)dp=RT×ln(p1/p2)+(p2-E=-rGm/zF0.03796 (3分(E/T)p=R/zF×rSm=zF(E/T)p=Rln(p1/p2)=24.9J·K-1·mol-mQr=TrS =7296J> (4分m3610 (2分G=H-TS=Qp-QV+∑RTT[zF(E/T)p226.25kJ·mol- (3分EG/zF1.1723 (2分E=((Ag,Ag2O/OH-)-(H2/H+))-0.0295(Ag,Ag2O/OH0.7593 (3分3710 HgO+H2O+2e-─→Hg+2OH- H2+2OH-─→2H2O+2e-電池反應(yīng)：H2+HgO─→H2O+ (3分1上述反應(yīng)：G$=-zFE=-178.74kJ·mol- (2分1HgO─→Hg+1/2 的G$=G$-G(H2O)=58.45kJ·mol- (2分 G$=-RTlnK$=- p(O23.29×10-16 (3分 3810n(-)nHg(l)─→Hgn+(m1)+nn(+)Hgn+(m2)+ne-─→n電池反應(yīng)：Hgnm2)─→HgnnE=-RT/zF×z=RT/EF×=(8.314J·K-1·mol-1×298K)/(0.029V×96500C·mol-1)×lg10=2∴溶液中存在的是Hg 離 (各2分23910H2O(l)＝H+(a(H+))+OH-(a(OH- Kw=a(H+)·a(OH-ln[Kw(T2)/Kw(T1rH'm/R(1/T1 rH'm=56.985kJ·mol-rHm=-rH'm=-56.985kJ·mol-1rGm(298K)=2[-RT1ln(1/Kw1)-RT2ln(1/Kw2)]=-79.90rSmrHmrGm)/T76.9J·K-1·mol- (6分電池反應(yīng):已知HgO(s)+H2(g)─→Hg(l)+H2O(l) Hg(l)+1/2O2(g)─→ H2O(l)─→H+(aq)+OH-(aq) H2(g)

O2(g)=H+(aq)+OH- 2rGm,4157.5kJ·mol- fGm(OH (4分4010電池反應(yīng)：Fe3++3/2H2─→Fe+ E=-0.0363 (2分E=E0.05915/3×lg (3分 [H+]=2[S2-EE0.05915/3×{a3/2(S2-)(2a)3(S2- (3分平衡時(shí)E= 解上式得[S2-]=1×106mol·dm- (1分即H2S濃度為1×10- (1分41101由電池 E1=E$+12E2E$+0.05915/2×lga2(H+)a2(Cl (2分2平衡時(shí)， lg[a(H+)a(Cl-)]=(E$-E$ (3分 ∵a(H+)a(Cl-)=a(H+)a(OH-)a(Cl-)/a(OH-Kw[Cl(Cl-)/[OH(OH (3分將(2)式代入(1)式，并將已知數(shù)代入 Kw=1.02×10- (2分42101/2Pb(s)+1/2S2O2(a2)─→1/2PbSO4(s)+1/2SO2 (1分 E=ERT/2Flna1 (1分E=2.01V-2.01V(Pb2+/PbRT/2F×lnKsp]2.366 (2分QUWHWH+ (2分QzF(E-E E=2.425 (2分代入(1)式，a1= (2分4310a(-)2Hg(l)–2e-→Hg2+(Kapa（2分

E=0.3338V–[E$(Hg2|Hg)RTlnKap （3分

20.1271V=0.3338V–Kap=5.5×10-（5分

RTln

(0.100.772)24410（3分E=E(Hg2+|Hg22+) E(Hg2+|Hg2+)=2E(Hg2+|Hg)–E(Hg2+|Hg)=(2×0.854–0.798)V=0.910 E=(0.910–0.798)V=0.112（4分K=exp(zEF/RT)=（3分4515（2分2E=E+(RT/2F)lna(CuBr E=E–(RT/2F)ln(m)3=0.6152（4分（2分

zE$F

E=E(PbBr2|Pb)–E(Pb2+|Pb)=0.152V（4分m=0.012mol·kg-（3分46101解:(1)AgCl(s)

(2分(AgCl,Ag,Cl-)= (3分lg＝－0.509│1×(- (2分 (3分47101)Mn(OH)2─→Mn2+＋2(2)H2O─→(3)Ka=Ksp/(Kw)2=Ka=a(Mn2+)/a2(H+)=pH＝－lg(5分(5分4815 （2分 （2分-（2分⑵E=E

RTln =E+RTlna2-RTlna2 a2 Cl-CaE+RTlna

=E

RTlna

=0.3821（3分

RTlna

=E–(E

RTlna

)=(0.311–0.3821)V=-0.071

aCa2+=3.479×10-若不考慮活度系數(shù)，則cCa2+=3.479×10-3mol·dm-（6分5=gh1R' =gh2R' (2分 h20.0055 (2分=0.01078N·m- (1分15電池：Al(s)|AlCl3NaCl熔化物（2分負(fù)極：Al→Al3+a)+3e正極：Al3+a)+3e（2分 （2分E=RTlna(Al)a(Al)=exp[-（4分=（1分 G=B–(T,p)=RTlna=-2150 （2分HzEF+zFTE)p3332J（2分5電解反應(yīng)：H+(0.05mol·kg,=0.860)Br-(0.05mol·kg-1───→1/2H2p1/2 (2分E分解右-(Br2/Br(H+/H2))-RT/F×ln(a(H+)a(Br1.227 (3分5E分解陽(yáng)－陰＝[(Br2,BrRT/F×lna(Br[(H+,H2RT/Fln(1/a(H1.227 (2分E分解陽(yáng)－陰＝RT/Fln{a(Ag[0.01mol·kg-1/a(Ag+)[0.50mol·kg-1,=0.526]}=-0.0866V，E分解=0.0866 (3分5陰極上Ni2+首先還原析出Ni(s)，陽(yáng)極上OH-首先氧化成 (各2分E分解=陽(yáng)－陰=1.065 (1分5C6H5NO26H6e-C6H5NH2 (1分Q (2分W=QV=4.703×104×2=9.406×104 (2分10 =－=[(Cl/Cl-)－RTln(Cl-)]－[((H+/H)+RTln理 =1.36-0.01

ln10-71.76 (3分陰=a+blgj=0.73+0.11×lg0.1=0.62 (2分陽(yáng) (2分E分解=E理論+陰+陽(yáng)=2.38 (3分5m2.303RT/zF×lgj°(m)2.303RT/zF×lg (3分Hg(Cd)=2.303RT/zF×lgj°(Cd)/j°(Hg)=0.333 (2分5E= －不可逆= －(H+,H2)+不可 (2分E－(H+,H20.695 (3分5(Zn2+/Zn)(Zn2+/Zn)RT/2F×lna(Zn2+0.7924 (2分(Zn2+/Zn)=(H+/H2)=RT/F×lna(H+)-=RT/F×lna(H+)-(0.72+0.116 (3分10E理論＝E=E=(H+,H2O/O2)-(H+/H2)=1.229 (5分E外加＝E理論+IR+++-=2.155 (5分10據(jù)=a+bln(j/A·cm2)直線方程,分別求得a,b取平均值得 (3分 (3分=a+ lnj=lnj0+ 比較,aRT/ (2分bRT/ j04.28×10-12A·cm- (2分5F jj0exp

(2分W(H2)

j×

M(H2)

(1分

7.91042416 ×

=0.0712g·cm-2·h- (1分

×

(1分10陰極Sn2++2e-1陽(yáng)極H2O2e-

O2(g) (2分2陰=(Sn2+/Sn)RT/2F)ln(Sn2+0.170陽(yáng)=(O2/H2O)+(RT/2F)ln2(H+)+(O2)=1.612 (2分E(分解)=陽(yáng)-陰=1.78 (2分(H+/H2)-(H2)=(H=0.01+[0.1- (Sn2+)《 (2分(Sn2+2.9×10- (2分10=lnj0+(RT/ (4分由此可得<> (3分<j03.95A·cm- (3分10 j=j1+V0=p1V1/T1×T0/p0 p1p0Q=V0/22.4×4Fj1=j2j (2分5陰=平－實(shí)際，不可逆 [H+]=2×0.01mol·dm-3 (1分)H2剛析出時(shí)的電位，(H+/H2)+0.059/2lga2(H+)-0.23=(Cu2+/Cu)+0.059/2×lg解 [Cu2+]=0.108×10-23mol·dm- (4分5jQ/t2.68A·cm (2分(H2│Feablgj0.81 (1分(H+/H21.64 (2分10E分解(I2/I0.05915lga(I0.05915lg1/a(H1.127 (2分E分解=(Br2/Br-)-(H+/H2)=1.716 (2分E分解=(Cl2/Cl-)-(H+/H2)=2.010 (1分10Cu2+2e-Cu(s)OH-O2，當(dāng)有0.5mol·kg-1的Cu析出，就有2×0.5mol·kg-1的OH-氧化，溶液中有2×0.5mol·kg-1H+留下，若H++SO2→HSO [H+]=(0.5+0.01)mol·kg- (2分 (H+/H2RT/Flna(H(H20.2473 (2分H2析出時(shí)，(H+/H2=(Cu2+/Cu)-RT/2FlnaCu2+ (4分[Cu2+1.75×10-20mol·kg- (2分51molH20.0224m3，氫氣質(zhì)量：x1=(1.008×2×2×10-4)/0.0224x1x2；965001.008x2x1x2=(F×x1)/1.008儀器工作10小時(shí)，電解消耗電量Q=x2×60×10=1.03392×106C (4分)電流密度j=Q/A=1.03392×107C·m- (1分7010陽(yáng)極反應(yīng)：H2O→(1/2)O22H陰極反應(yīng)：Cu2++2e-→ (2分E外=

陽(yáng)

=[$-RT/2F×lnO2O(Cu2+/Cu)RT/zF×lna(Cu2+0.8715 (3分Cu2+濃度：0.0005mol·dm-H+濃度：0.100+0.05×0.99×2=0.199mol·dm- (2分E'外=陽(yáng)-陰=0.9481 (3分7110析出(H2)=可逆(H2)-(H2)=-0.9977 (2分I=1/2∑mizi2=0.008mol·kg-lg(Zn2+)=-Az2(Zn2+)∴析出(Zn)可逆(Zn)

12 (Zn2+)=0.8597 (4分析出(Cu)≈可逆(Cu)=0.2435 (2分∴析出(Cu)析出(Zn)析出 (2分7210I理論=[n(NaOH)mol·h-1]×F(C·mol-1)/3600(s/h)=1840.2A (2分)電流效率為:I理論/I實(shí)際×100%= (2分電能效率=理論電能/=[I理論×E可逆×t]/[I實(shí)際×E外加×t]×100%= (4分E外加=E理論+IR=E外加-E理論-IR=0.6 (2分7310(Zn2+/Zn)==-0.7628V<-0.30Zn(Cu2+/Cu)==0.3378V＞-0.30(2分Cu(2分a(Cu2+)：(Cu2+/Cu)RT/2Flna(Cu20.30a(Cu22.919×10- (3分1.00mol·kg-1Cu2+2mol·kg-12.00mol·kg-1(H+/H2析出(H+/H2,平_0.1822V0.30所以H2也在陰極析出 (3分7410陰極：Ni2+2e-可逆=+RT/2F×lna(Ni2+)=-0.25V (2分)H++e-→(1/2)H2可逆=RT/F×lna(H+)/[p(H2)/p]1/20.414不可逆=可逆+=-0.83 (2分Na++e-→ =-2.714Mg2++2e-→Mg =-2.363V 所以陰極上先析出Ni。 (1分)陽(yáng)極：H2O→1/2O2+2H++2e-不可逆=可逆+=1.7 (2分Ni→Ni2++2e- =-0.25 所以陽(yáng)極上是Ni溶 (3分 5(Fe2+/Fe)=(Fe2+/Fe)+0.05915/2×lga=-0.4429(H+/H2)=(H+/H2)+0.05915lga(H+)(1分=0.05915pH-(1分H2(Fe2+/FepH>(3分76.(Cd2+/Cd)=-0.4612(Cu2+/Cu)=0.2868(H+/H2)=-0.413所以陰極上首先是Cu2+還原成 (3分陽(yáng)極反應(yīng)：H2O2H1/2)O2[H2×0.02mol·kg-(O2/H2O)=(O2/H2O)+RT/2F×lna2(H+)=1.1463 E分解=(O2/H2O)-(Cd2+/Cd)=1.6075 (3分(Cd2+/Cd)==(Cu2+/Cu)+0.05915/2×lg[Cu2+0.103×10-26mol·kg- (2分(O2/H2O(O2/H2O)]可逆=1.415V (Cl-/Cl2)=1.4323陽(yáng)極上先發(fā)生H2O氧化反應(yīng) (2分5陰極：(H+/H2(H+/H20.05915lga(H+)-=-0.614(Na+/Na)=-2.668所以先析出 (2分(O2/OH-)=(O2/OH-)+0.05915lga(OH-)+=1.415(Cl2/Cl-)=1.314。(2分(3)E分解陽(yáng)陰1.928(1分5(Ag+/Ag)=0.3261陰極電勢(shì)低于0.3261 (2分(Cu2+/Cu)=(Cu2+/Cu)+RT/2F×lna(Cu2+)= [Cu2+]要小于0.4278mol·kg- (3分5(Na+/Na)2.813 (1分陰極：(H+/H2)=-RT/Fln1/a(H+)-(H2)=-1.58 (1分故陰極上首先是H+還原成 (1分2OH-H2O(1/2)O2+(O2/OH-)=(O2/OH-)-RT/F×lna(OH-)+=1.499 (2分2Ag++2OH-→Ag2O+H2O+(Ag2O/Ag)=(Ag2O/Ag)+RT/Fln1/a(OH-0.4622 (2分故Ag在陽(yáng)極上氧化成 (1分E分解=陽(yáng)-陰=2.0422 (2分5-0.403j=j0exp(nF/RT)=0.79=2.012.01A/cm2以上時(shí)，Cd2+5Ag+ 2CN- 0.05+ 0.1-2Ka=[a(Ag+)+a2(CN-)]/a[Ag(CN)]=x2x1.52×10-17 [Ag1.52×10-17mol·kg- (2分(Ag+/Ag)(Ag+/Ag)RT/F×lna(Ag0.196 (3分10(Cd2+/Cd)=-0.433Cd2+Zn2+濃度相同時(shí)，∵(Cd2+/Cd)故Cd先析 (2分(Zn2+/Zn)=-Zn2+Cd2+-0.793V=-0.403V+0.05915/2×lgaa(Cd2+6.5×10- (4分H2(g)Cd(H+/H2)(Cd上RT/Fln1/a(H0.48V0.89(H+/H2)(Zn上1.11(H+/H2)Cd2+/Cd)所以H2(g)不會(huì)析出 (4分10(Ag+/Ag)=(Ag+/Ag)+RT/F×lna(Ag+)=0.722(Fe2+/Fe)=-0.4994 (Cd2+/Cd)=-0.4917(Ni2+/Ni)=-0.2796(H+/H2)=(H+/H2)+RT/F×lna(H+)=-0.1775(H+/H2)(Ag)=-0.1775V-0.20V=-0.3775(H+/H2)(Ni)=-0.4175 (H+/H2)(Fe)=-0.3575(H+/H2)(Cd)=-0.4775 (各1分54組成電池Pt,Cl2│Cl-‖H+,MnO,Mn2+│Pt (1分) 2Cl-(a=1)→Cl2(101.325kPa)+2e-44正極MnO(0.1mol·kg-18H+(0.1mol·kg-15e-→Mn2+(1mol·kg-1)+4H2O44RT/zF×lna(Mn2+)/[a(MnOa8(H1.4305 (2分44電池反應(yīng)MnO+8H++5Cl-→Mn2++4H2O+ (1分4E- 0.0435V 故MnOCl-

(1分右 10[(H+/H2)]可逆=-0.035 (2分[(O2/H+)]可逆=1.195 (2分當(dāng)電流通過(guò)時(shí)，H2(H+/H2(H+/H2)]可逆0.300(O2 0.596 (2分對(duì)電池充電時(shí)氫電極的實(shí)際電勢(shì)-0.035V-0.300V=-0.335 (1分氧電極的實(shí)際電勢(shì)1.195V+0.596V=1.791 (1分故加外電壓：1.791-(-0.335)+0.20=2.33 (2分10H2OH21/2EG$/zFTS$-H$)/zF (3分 對(duì)CO2(g)+H2O(l)=HCOOH(l)+ E=-1.429 (3分PtH2O1.229VHCOOH1.4290.0591/2×lga(HCOOH a(HCOOH)=10-也需加壓1.34V。所以總是先電解H2O而非生成 (2分H2O1.229V+0.6V=1.829CO2能被還原成 (2分8710(Zn2+/Zn)(Zn2+/Zn)RT/2F)ln(Zn2+0.793 (2分(Cd2+/Cd)(Cd2+/Cd)RT/2F)ln(Cd20.433 (2分(H+/H2(H+/H2)+(RT/F)ln(H+)-(H21.014 (2分CdZn(Cd2+)=6.0×10-(4分88.(Cd2+)=6.0×10-(4分88. j陽(yáng)=1.5/3=0.5A·cm-(1分j陰=1.5/2=0.75A·cm-(1分Ga的電流效率=實(shí)際產(chǎn)量/(5分(2)W=10/(69.7/3)×26.8×3/10000.035(3分 由于(Ag+/Ag)和(H+/H2)相差很大,H2析出時(shí)AgH+(H+)10-7+0.5(2分(Ag+/Ag)+(RT/F)ln(Ag+)=(H+/H2)+(RT/F)ln(H+)-(Ag+)6.66×10-(2分(1分,9010陰極Cu2+/Cu)=(Cu2+/Cu)+(RT/F)ln(Cu2+0.308(2分(H+/H2)=(H+/H2)+(RT/F)ln(H+)-(H2)=-0.998(2分(1分陽(yáng)極Cl2/Cl-)=(Cl2/Cl-)-(RT/F)ln(Cl=1.40(2分(O2/H2O,OH-)=(O2/H2O,OH-)-(RT/F)ln(OH-)+(O2)=1.71(2分(1分 5(1)(Cr2+/