内容发布更新时间 : 2025/1/1 7:25:04星期一 下面是文章的全部内容请认真阅读。
? +?左=?Zn2?,Zn0.0592Zn2??lg???
20.0592 lg0.1 =-0.764+
2=-0.793V
右：
Ag??e?Ag???0.0592lg?Ag??
?右???Ag?,Ag?0.799?0.0592lg0.01
=0.681V
E??右??左
?0.681???0.793?
=1.474V
8.解：
E?0，所以是原电池。
左边：
2HA?2e?H2?2A?
??
?左??H，H?0.0592lg??H???2????左??H?0.0592lgH???，H?2
=0.0592lg??H???
E??右??左
?0.413?0.244?0.0592lg?H???
?0.0592lg??H????0.169 ??3??H?1.4?10mol/l ??
HA?H??A?
?? ????HA????K??HA?1.4?10?3?0.116 =
0.215 =7.6?10
?4
9.解：E??右??左
0.921?0.2443??左
?左??0.6767V
左边：CdX2?4?2e?Cd?4X?
??左??Cd2?,Cd?0.05922lg??Cd2??? Cd2??4X?CdX2?4 K??CdX2?4??稳? ??Cd2?????X??4? ?Cd?2??2?????CdX4??
K??4稳??X?
??左?2?,Cd?0.0592??CdX2??4?Cd2lg?
KX??4稳????0.6767??0.403?0.05920.22lgK4稳?0.150?K稳?7.0?1011
10-.解： E??右??左
?左??0.648V7
??左??Cd2?,Cd?0.05922lg??Cd2??? Cd?2X??CdX2
K2?2sp???Cd????X???
?K?Cd2????sp??X??2
?0.893?0.2443??左
??左??Cd2?,Cd?Ksp0.0592lg?22?X???
?0.6487??0.403?Ksp0.0592lg22?0.02?
Ksp?2.0?10?12
第十二章 电位分析法
10.（1）E=E右-E左
=E
θ
+RT/ZFln[Fe3+]/[Fe2+]-ESCE
=(0.771+0.059lg0.025/0.015)-0.2443 =0.784-0.2443 =0.540 （2）E=E右-E左
=E
θ
+RT/ZFln[Zn2+]-ESCE
=(-0.764+0.059/2lg0.00228)-0.2443 =-0.842-0.2443 =-1.085 （3）I3-+2e=3I- E=E右-E左
=E
θ
+RT/ZFln[I3+]/[I+]-EAg/AgCl
=(0.545+0.059/2lg0.00667/0.004333)-0.194 =0.690-0.194 =0.496 12.解：E=E右-E左
=EθAg2CrO4/Ag-0.059/2lgCrO42--ESCE
θ
=EAg2CrO4/Ag+0.059/2P(CrO4)-ESCE
代入数字得：0.386=0.446+0.059/2P(CrO4)-0.2443 解得：P(CrO4)=6.25 13.解：E
?EpH?EAg/AgCl
?K-0.0592pH
?Es?K-0.0592pHsEx?K-0.0592pHxEs?Ex
0.0592
pHx?pHs?
?4.006??0.2094???0.2806?
0.0592 =5.21 14．解：E?ECu2??ESCE
?K??K,?0.0592lgaCu2??ESCE 2
0.0592pMg 2
2?Es?Ex?pMx?pMs?
0.0592??lg?3.25?10?3???3.772??0.124?0.086? 0.0592
17．解：依题意有：
(1)当aNa+=0.1000mol/L时，E1=67mv=0.067V 当aK+=0.1000mol/L时，E2=113mv=0.113V
E=K+0.059lg（ak++Kk+，Na+·an/aNa+） 又因为：n=1，a=1
所以：E1=K+0.059lg（Kk+，Na+·aNa+）=0.067V E2=K+0.059lg（ak+）=0.113V
得：ΔE=E1-E2=0.059lg(Kk+，Na+·aNa+)/ak=0.067-0.113=-0.046V 解得：Kk+，Na+=0.166
（2）误差=KK?,Na??aNa?aK???11?100%
=16.6%
第十三章电解与库伦分析法
7解：?Ag???????0.0592lgAg?????Ag Ag,Ag?0.799?0.0592lg0.01?0
=0.681V
??Cu??Cu
0.05922??lg?Cu??Ag
??20.0592?0.337?lg2
22??,Cu =0.346V
??Ag??Cu?Ag先析出,?Ag???Ag
(2)Ag析出完全时电位为：
?,Ag??5??0.0592lg?Ag?10??Ag ????
?0.799?0.0592lg?0.01?10?5??0
=0.385V ??Ag ??Ag ?Ag8.解： ?Cu,,??Cu
??Cu
?和Cu2?可完全分离。阴极控制电位在0.385V或0.385-0.346V之间
???Cu?2?,Cu
0.05922??lg?Cu??Cu
??20.0592?0.337?lg0.1?0
20.0592lg0.1??Sn 20.0592??0.136?lg0.1?0
2 =0.307V
? ?Sn??Sn?2?,Sn
=-0.166V ??Cu??Sn
?阴极上Cu先析出
Cu析出完全时，阴极电位为：
0.0592,?lg0.1?10?5??Cu ?Cu??Cu2?,Cu?20.0592 =0.337-6?+0
2
?? =0.159V
,?Cu??Sn ?Cu和Sn可完全分离
?阴极电位控制在0.159V或-0.166?0.159V之间，即可完全分离Cu和Sn。
9.解：it?it?10?kt