同理可求:E(4) = E(右)- E(左)= 0.521 – 0.153 = 0.368V
?rGm?4???zE?4?F=?2?0.368?96500?-35.51kJ?mol?1
RT?2?0.368?96500?6)?exp???1.67?10
?8.314?298.15?K?4??exp(
??rGm?4?7.25 将反应Ag(s)+
12Cl2 (g)= AgCl(s)设计成原电池,已知在25℃时,
?fHm(AgCl,s)=-127.07kJ?mol?1,?fGm(AgCl,s)=-109.79kJ?mol?1,标准电极电势
E(Ag│Ag)= 0.7994V,E +
(Cl│Cl2(g)│Pt)=1.3579V。
-
(1)写出电极反应和电池图示;
(2)求25℃时电池可逆放电2F电荷量时的热Qr; (3)求25℃时AgCl的活度积。 解:(1)电极反应和电池图示如下: 阳极:Ag(s)+ Cl - e = AgCl(s) 阴极:
--
12Cl2 (g)+ e = Cl
--
--
电池图示:Ag|AgCl(s)|Cl {a(Cl)}|Cl2(g,
p
)|Pt
(2) ?rGm???B?fGm?B?B1??fGm?AgCl,s?-?fGm?CI2,g?-?fGm?Ag,s?同理同理
2=-109.79kJ?mol?1可求:?rHm=?fHm(AgCl,s)=-127.07kJ?mol?1
?rGm??rHm-T?rSm
3?rHm-?rGm??127.07???109.79???10?rSm=???57.96J?mol-1?K?1
T298.15Qr = n T?rSm= 2×298.15×(-57.96)= -34.56kJ
?rGm?109.79?103(3)?rGm??zEF;即:E=-???1.1377V zF1?96500E = E(右)- E-
(左)= 1.3579 –E
(Cl│AgCl(s)│ Ag)
-
E(Cl│AgCl(s)│ Ag)= 1.3579 – 1.1378 = 0.2201V
+
-
解法1:设计原电池:Ag│Ag‖Cl│AgCl(s)│ Ag
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电池反应:AgCl(s) Ag + Cl
+-
??E(Cl│AgCl(s)│Ag)?E(Ag│Ag)-RTlna?Ag??a?Cl??zF
RT?=E(Ag│Ag)-lnKsp?AgCl?zF8.314?298.15lnKsp?AgCl?
1?965000.2201?0.7994-Ksp?AgCl?=1.61?10-10
解法2:根据能斯特方程:
??E(Cl│AgCl(s)│Ag)?E(Ag│Ag)?RTlna?Ag?? zFKsp(AgCl)=a?Ag?? a?Cl_? ,即:a?Ag??=Ksp(AgCl)/ a?Cl_?
?则:E(Cl│AgCl(s)│Ag)??E(Ag│Ag)?RTlnKsp(AgCl)/ a?Cl_? zF0.2201?0.7994?8.314?298.15lnKsp(AgCl)/ 1
1?96500Ksp?AgCl?=1.61?10-10
7.26 25℃时,电池Pt│H2(g,100kPa)│H2SO4(b)│Ag2 SO4(s)│Ag的标准电动势E已知E(Ag│Ag)= 0.7994V。
+
=0.627V。
(1)写出电极反应和电池反应;
(2)25℃时实验测得H2SO4浓度为b时,上述电池的电动势为0.623V。已知此H2SO4溶液的离子平均活度