电池：Pb(s)︱PbO(s)︱OH-(aOH-)HgO(s)︱Hg(l) 正极：HgO(s) + H2O(l) + 2e-→ Hg(l) + 2OH-(aOH-) 负极：Pb(s) + 2OH-(aOH-) –2e-→PbO(s) + H2O(l) 电池反应：Pb(s) + HgO(s)==Hg(l) + PbO(s) (10) Sn2+(aSn2+) + Tl3+(aTl3+) == Sn4+(aSn4+) + Tl+(aTl+)
电池：Pt(s)︱Sn2+(aSn2+)，Sn4+(aSn4+)‖Tl3+(aTl3+)，Tl+(aTl+)︱Pt(s) 正极：Tl3+(aTl3+) + 2e-→ Tl+(aTl+) 负极：Sn2+(aSn2+) –2e-→Sn4+(aSn4+)
电池反应：Sn2+(aSn2+) + Tl3+(aTl3+) == Sn4+(aSn4+) + Tl+(aTl+) 15解：
Fe(s) + Cd2+(aq)==Cd(s)+Fe2+(aq)
E=E?
– RT/2F×ln{[ Fe2+]/[Cd2+]}
(1) E=φ?cd2+/Cd –φ?Fe2+/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=φ?Cd2+/Cd –φ?Fe2+/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) φ PbO2/ PbSO4 = φ?PbO2/ PbSO4+RT/2Fln a SO42- a4
H+ 负极：Pb(s)+SO42-(a SO42-) –2e-→ PbSO4(s) φPbSO4/ Pb = φ?Pb2+/Pb+RT/2Fln{K?sp/ a SO42-}
电池反应：PbO2(s) + Pb(s)+ 2H2SO4(m)== 2PbSO4(s)+2H2O(l)
E=φPbO/ PbSO= φ?PbO–RT/2FlnK?
24–φPbSO4/ Pb2/ PbSO4–φ?Pb2+/Pb sp + RT/Fln a H2SO4E=E?
+ RT/Fln a H2SO4=2.041+ RT/Fln a H2SO4
11
E/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=1.91878V 1.919=2.041+ RT/Fln a H2SO4，ln a H2SO4= – 4.7511 ln a H2SO4= ln r±3mH+2mSO42-= – 4.7511 3lnr±+2lnmH++lnmSO42-= –4.7511
3lnr±+2ln2.0+ln1.0= –4.7511
lnr±= –2.0458，r±=0.129 24解：
正极：I2(s) +2e-→2I-(a3)
φI2/ I- =φ?I2/ I- +RT/Fln a3
负极：2S2O32-(a1) –2e-→S4O62-(a2) φS4O62-/ S2O32- =φ?S4O62-/ S2O32- +RT/2Fln( a2/ a2)
电池反应：2S2O32-(a1) + I2(s)== S4O62-(a2) + 2I-(a3)
△rH?
m=2{(2)–(1)+(3)}=2{28.786-46.735+3.431}= –29.04kJ.mol-1
△rS?
m=2×105.9+146.0–116.7–2×33.47=174.16 J.K-1.mol-1
△rG?
m=△rH?
m–T△rS?
m= –29.04×103–298×174.16×10-3= –80.94kJ.mol-1 E?
= –△rG?
m/nF=80.94×103/2/96484.5=0.419V
E?
= φ?I2/ I- — φ? S4O62-/ S2O32-=0.419V φ?S4O62-/ S2O32 =φ?I2/ I- — 0.419=0.535–0.419=0.116V 25解：
Pt(s)︱H2(p?
)︱H2SO4(0.010mol.kg-1)︱O2 (p?
)︱Pt(s) 正极：O2 (p?
) + 4H+(aH+) + 4e-→2H2O(l) φO?2/H2O= φO2/H2O +RT/4Fln{(pO2/p?
)aH+4} 负极：2H?
2(p) – 4e-→4H+(aH+)
φ H+/H2 = φ?H+/H2 –RT/4Fln{(pH2/p?
)2/aH+4}
12
电池反应：O??
2 (p) + 2H2(p) == 2H2O(l)
E= φ?
?
O2/H2O –φ H+/H2= φ?O2/H2O –φ?H+/H2 + RT/4Fln{(pO2/p) (pH2/p)2} E=φ?
O2/H2O –φH+/H2= φ?O2/H2O –φ?H+/H2 = E =1.227V ∵ 2H?
2O(g) ==O2 (g) + 2H2(g) Kp=9.7×10-81 K?
?
?
?
p= ( pH2/p)2(pO2/p)/( pH2O/p)2 =9.7×10-81 ( p?
?
H2/p)2(pO2/p)/( 3.200/100)2 =9.7×10-81
( p?
?
H2/p)2(pO2/p)=( 3.200/100)2×9.7×10-81=9.933×10-84 ∴O?
?
2 (p) + 2H2(p) == 2H2O(l)
K?
=1/{( p?
H22/p)(pO2/p?
) }=1/9.933×10-84=1.0068×1083 E?
=RT/nFln K?
=8.314×298/4/96484.5×ln1.0068×1083 =6.4196×10-3×(6.7770×10-3+191.11456)=1.227V
26解：
Pb(s)︱PbCl2(s)︱HCl(0.010mol.kg-1)‖H2(10kPa)︱Pt(s) 正极：2H+(aH+) + 2e-→H2(10kPa) φ ?
H+/H2 = φ? H+/H2 –RT/2Fln{(pH2/p)/aH+2} 负极：Pb(s) + 2Cl-( aCl-) –2 e-→PbCl2(s) φ?
PbCl2/Pb =φ?Pb2+/Pb+RT/2Fln(Ksp/ aCl-2)
电池反应：2HCl(aHCl) + Pb(s) == H2(10kPa) + PbCl2(s) E=φ?
?
H+/H?2–φPb2+/Pb=φ?H+/H2 –φPb2+/Pb –RT/2FlnKsp –RT/2Fln{(pH2/p)+RT/FlnaHCl
mPbCl.2=0.039 molkg-1
I =1/2×{0.039×22 +2×0.039×(–1)2}= 3×0.039mol.kg-1
lg r ? ? ? A z ? z ? I ? ? 0 . 509 2 ? ? 1 3 ? 0 .039 ? ? 0 .3482
r±=0.706
K?
sp=aPb2+aCl-2=a?
±3=( r±m±/m)3=( r±3mPb2+ mCl-2)
= 0.7063×0.039×(2×0.039)2=8.35×10-5
13
mHCl=0.010 mol.kg-1
I =1/2×{0.01×12 +0.01×(–1)2}= 0.010mol.kg-1
lg r ? ? ? A z ? z ? I ? ? 0 . 509 1 ? ? 1 0 . 01 ? ? 0 .0509
r±=0.950
E=φH+/H2–φPb2+/Pb=φ?H+/H2 –φ?Pb2+/Pb –RT/2FlnK
?
sp –RT/2Fln{(pH2/p
?
)+RT/FlnaHCl
= 0.126–0.012839×ln(8.35×10-5) –0.012839×ln0.1+0.025678×ln(0.9502×0.012)
=0.126+0.12057+0.029562 –0.23914=0.037V 27解：
Cu(s)︱CuAc2(0.10mol.kg-1)∣AgAc(s)︱Ag(s)
(1) 正极：AgAc(s) + e-→Ag(s) + Ac-( aAc-) φ?
Ag+/Ag = φ?Ag+/Ag +RT/FlnKsp/ aAc- 负极：1/2Cu(s) – e-→1/2Cu2+( aCu2+) φCu2+/Cu = φ?Cu2+/Cu +RT/2Fln aCu2+
电池反应：2AgAc(s) + Cu(s)== 2Ag(s) + CuAc2(aq) E= φAg+/Ag –φCu2+/cu= φ?Ag+/Ag – φ? Cu2+/Cu +RT/FlnK
?
sp – RT/Fln( aAc- aCu2+1/2
)
(2) △rGm= – nFE= – 2×96484.5×0.372= –71.784kJ.mol-1
( ?E ) ? E (T 2 ) ? E (T 1 ) ? 0 .374 ? 0 .372 ? 2 ? 10 ?4 ?TpT2?T1308?298△rSm=nF ( ?? ET ) p=2×96484.5×2×10-4=19.297J.mol-1
△rHm=△rGm+T△rSm= –71.784 + 298×19.297×10-3= –66.033 kJ.mol-1 (3) E=φ?
Ag+/Ag –φCu2+/cu= φ?Ag+/Ag – φ? Cu2+/Cu +RT/FlnK
sp – RT/Fln( aAc-aCu2+1/2
)
E=φ?Ag+/Ag –φCu2+/cu= φ?
Ag+/Ag – φ?
Cu2+/Cu +RT/FlnK
sp – RT/Fln( mAc-mCu2+1/2)
0.372= φ?Ag+/Ag – φ? ?
Cu2+/Cu+0.025678lnK
sp – 0.025678ln( 0.2×0.11/2
)
0.372=0.799–0.337+0.025678lnK?
sp– 0.025678ln0.2–0.01284ln0.1 lnK?
?
sp = – 6.2697，Ksp = 1.89×10-3 28解：
Hg22+(aHg22+) + 2e-→Hg(l)
14
φHg22+/Hg = φ? Hg22+/Hg + RT/2FlnaHg22+
Hg2SO4(s) + 2e-→Hg(l) + SO42-( aSO42-) φ?Hg2SO4/Hg =φ?Hg22+/Hg+ RT/2Fln( Ksp? / aSO42-) φ?Hg?
2SO4/Hg =φ?Hg22+/Hg+ RT/2Fln Ksp
=0.789 + 0.01284ln(8.2×10-7)=0.609V
29解：
(1) Ag(s) + Fe3+( aFe3+)== Ag+( aAg+) + Fe2+( aFe3+) 电池：Ag(s)︱Ag+(aAg+)‖Fe3+( aFe3+)，Fe2+( aFe2+)︱Pt(s) 正极：Fe3+( aFe3+) + e-→Fe2+( aFe2+) 负极：Ag(s)– e-→Ag+(aAg+)
电池反应：Ag(s) + Fe3+( aFe3+)== Ag+( aAg+) + Fe2+( aFe3+) E?
=RT/nFlnK?
，K?
=exp(nFE?
/RT )
E?= φ?Fe3+/Fe2+ – φ?Ag+/Ag =0.771–0.799= –0.008
lnK?
=nFE?
/RT，lnK?
=1×96484.5×( –0.008)/(8.314×298)= –0.3115 K?=0.73
(2) Hg2Cl2(s)== Hg22+(aHg2+) + 2Cl-(aCl-)
电池：Hg(l)︱Hg22+(aAg+)‖Cl-(aCl-)︱Hg2Cl2(s)︱Hg(l) 正极：Hg2Cl2(s) + 2e-→2Hg(l) + 2Cl-(aCl-) 负极：2Hg(l)–2e-→Hg2+(aHg22+)
电池反应：Hg2Cl2(s)== Hg22+(aHg2+) + 2Cl-(aCl-) E?
=RT/nFlnK?
sp，K?
?
sp=exp(nFE/RT ) E?= φ?Hg2Cl2/Hg– φ?Hg2+/Hg =0.268–0.793= –0.525
lnK?
=nFE?
/RT，lnK?
sp=2×96484.5×( –0.525)/(8.314×298)= –40.89 K?sp=1.74×
10-18 (3) HBr(aHBr)==H+(aH+)+Br-(aBr-)
电池：Pt(s)︱H2(p?
)‖HBr(aHBr)︱AgBr(s)︱Ag(s)
15