发布时间 : 星期二 文章物理化学课后习题答案第九章更新完毕开始阅读
负极:2I-(aI-)–2e-→ I2(s)
--电池反应:Cl2(pCl2) + 2I(aI-)==I2(s)+2Cl(aCl-)
+-(7) H2O(l)== H(aH+) + OH(aOH-)
电池:Pt(s)︱H2(pH2)︱H+(aH+)‖OH-(aOH-)︱H2(pH2)︱Pt(s)
--正极:2H2O(l) + e→2H2(pH2) + 2OH(aOH-) 负极:H2(pH2)–2e-→ 2H+(aH+)
电池反应:H2O(l)== H+(aH+) + OH-(aOH-) (8) Mg(s) + 1/2O2(g) + H2O(l)== Mg(OH)2(s)
-电池:Mg(s)︱Mg(OH)2(s)︱OH(aOH-)︱O2(pO2)︱Pt(s)
--正极:1/2O2(g) + H2O(l) + 2e→ 2OH(aOH-)
--负极:Mg(s) + 2OH(aOH-)– 2e→Mg(OH)2(s) 电池反应:Mg(s) + 1/2O2(g) + H2O(l)== Mg(OH)2(s)
(9) Pb(s) + HgO(s)==Hg(l) + PbO(s)
-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+),4+3++Sn(aSn4+)‖Tl(aTl3+),Tl(aTl+)︱Pt(s)
3+-+
正极:Tl(aTl3+) + 2e→ Tl(aTl+)
2+-4+
负极:Sn(aSn2+) –2e→Sn(aSn4+)
电池反应:Sn2+(aSn2+) + Tl3+(aTl3+) == Sn4+(aSn4+) + Tl+(aTl+) 15解:
Fe(s) + Cd2+(aq)==Cd(s)+Fe2+(aq)
?2+2+E=E– RT/2F×ln{[ Fe]/[Cd]}
??φφ(1) E=cd2+/Cd –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=φCd2+/Cd –φFe2+/Fe–
2+2+
RT/2F×ln{[ Fe]/[Cd]} = –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)
+2-正极:PbO2(s) + 4H(m)+ SO4(a SO42-) +
2e-→PbSO4(s)+2H2O(l)
?
φφ PbO2/ PbSO4 = PbO2/ PbSO4+RT/2Fln a SO42- a 4H+
2--负极:Pb(s)+SO4(a SO42-) –2e→ PbSO4(s)
??
φPbSO4/ Pb = φPb2+/Pb+RT/2Fln{Ksp/ a SO42-} 电池反应:PbO2(s) + Pb(s)+ 2H2SO4(m)== 2PbSO4(s)+2H2O(l)
?
φPbO2/ E=φPbO2/ PbSO4–φPbSO4/ Pb=
??φPbSO4–Pb2+/Pb –RT/2FlnKsp + RT/Fln a H2SO4 ?
E=E+ RT/Fln a H2SO4=2.041+ RT/Fln a H2SO4
-6-82
E/V=1.91737+56.1×10(t/℃)+1.08×10(t/℃) E=
-6-82
1.91737+56.1×10×25+1.08×10×25=1.91878V
1.919=2.041+ RT/Fln a H2SO4,ln a H2SO4= –
4.7511
32
ln a H2SO4= ln r±mH+mSO42-= – 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)
?
φφ I2/ I- =I2/ I- +RT/Fln a3
2--2-负极:2S2O3(a1) –2e→S4O6(a2)
?
φS4O62-/ =φS4O62-/ S2O32- S2O32- +RT/2Fln( a2/ a2)
2-2- 电池反应:2S2O3(a1) + I2(s)== S4O6(a2)
-+ 2I(a3)
?
△rHm=2{(2)–(1)+(3)}=2{28.786-46.735+3.43
.-1
1}= –29.04kJmol
?
△rSm=2×105.9+146.0–116.7–2×33.47=1.-1.-1
74.16 JKmol
???△rGm=△rHm–T△rSm=
3-3.-1
–29.04×10–298×174.16×10= –80.94kJmol
?E= ?3
–△rGm/nF=80.94×10/2/96484.5=0.419V