?rGm??ZEF?71.796kJ?mol?1?rHm??ZEF?T?rSm?60.287kJ?mol3）
?1
E?E??0.059161lg2?(Ac?)2?(Cu2?)
E??E?0.02958lg[?(Ac?)2?(Cu2?)??0.3011V4）E?(Ac/AgAc/Ag)?E?(Cu2?/Cu)?E??0.6381V 5）
E?(Ac/AgAc/Ag)?E?(Ag?/Ag)?0.05916lgKspKsp?1.88?10?3
正极 ：2AgAc(s)+2e->2Ag(s)+2Ac-.
负极： 2Ag(s)-2e->2Ag+.
电池反应：AgAc->Ac-+Ag+.
E??E正?E负?RTlnksp?0.6381?0.7994??0.1613F
?ksp?1.88?10?33． 解：1）+）AgCl(s)+e→Ag(s)+Cl- -）0.5H2(100kPa)-e→H+
电池 AgCI(s)+0.5H2(100kPa)=Ag(s)+HCI(b=0.50mol.kg1) 2) +）AgCl(s)+e→Ag(s)+Cl-…
-）Ag-e?Ag?
电池 AgCl(s)?Ag??Cl?
?E??E?(Cl?/AgCl/Ag)?E?(Ag?/Ag)?0.05916lgKsp E?(Cl?/AgCl/Ag)?E?(Ag?/Ag)?0.05916V1gKsp
=0.7994V+0.05916V1g1.75?1010
9
=07994V+0.05916V?(?9.757) =0.2222V
3)EMF??E?(Cl?/AgCI/Ag)?0.2222V
E=E??0.05916Iga(HCl) 得: 0.276V=0.2222V-0.05916Ig? a(HCl)=0.1232
a?(HCl)=a(HCl)0.5=0.351
??=a?/(b?/b?)?0.351/0.462?0.760
4. 解 单个分子在表面吸附层占据的面积为
As?1 L?2(HCl)
式中?2为溶质的表面吸附量。 稀溶液中吉布斯吸附公式为 ?2??联立上面两式得 c??cd?
RTdcRTd?
LAsdc由题给条件知 d???0.55dc即代入上式得： c?dc??1.818 d?8.314?2926.022?1023?0.5?10?9??2?1.818?0.0147mol?m?3
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