?2FE$??2?96485?0.2224?7K?exp???exp???3.304?10
?8.3145?298.15??RT?$?m?0.1mol?kg-1?（2）反应为2Ag(s)?2HCl???2AgCl(s)+H2(g)
?=0.809???仍有纯固体的活度为1，a(H2)?p$/p$， a(H2)p(H2)/p$所以 1/K? ?a(HCl)(??m?/m$)4$p(H2)?(??m?/m$)4p$/K$ ?[(0.809?1)4?105/3.304?107]Pa=1.30?10?3Pa
14、在273~318K范围内，下述电池的电动势与温度的关系可由所列公式表示： （a）Cu(s) | Cu2O (s) | NaOH (aq) | HgO (s) | Hg (l)
E = [461.7 – 0.144(T/K – 298) + 0.00014(T/K – 298)2]mV （b）Pt(s) | H2 (p?) | NaOH (aq) | HgO(s) | Hg(l)
E = [925.65 – 0.2948(T/K – 298) + 0.00049(T/K – 298)2]mV
??已知?fHm试分别计算HgO (s)(H2O,l)??285.83kJ?mol?1，?fGm(H2O,l)??237.13kJ?mol?1，
??和Cu2O (s)在298K时的?fGm和?fHm值。
解：电池(a)的电极反应和电池反应为
负极 2Cu(s)+2OH?(?OH?)?Cu2O(s)+H2O(l)?2e? 正极 HgO(s)+H2O(l)+2e??Hg(l)+2OH?(?OH?)
电池反应 Cu(s)?HgO(s)?Hg(l)?CuO(s) (1) 在298K时，E1(298K) = E1?(298K) = 461.7mV
?rGm(1)??zE1F??2?(0.4617V)(96486C?mol?1)??89.11kJ?mol?1
??E??rSm(1)??zF?1??2?(96485C?mol?1)(0.144?10?3V?K?1)??27.79J?K?1?mol?1
??T?p???rHm(1)??rGm(1)?T?rSm(1)?[?89.11?298?(?27.79)]J?mol?1??97.39kJ?mol?1
电池(b)的电极反应和电池反应为
负极 H2(p?)+2OH?(?OH)?2H2O(l)+2e?
?正极 HgO(s)+H2O(l)+2e??Hg(l)+2OH?(?OH)
?电池反应 H2(p?)?HgO(s)?Hg(l)?H2O(l) (2)
?
在298K时，E2(298K) = E2(298K) =925.7mV
?rGm(2)??zE2F??2?(925.7?10?3V)(96480C?mol?1)??178.66kJ?mol?1
??E??rSm(2)??zF?2??2?(96480C?mol?1)(?0.2948?10?3V?K?1)??0.0569kJ?K?1?mol?1??T?p?rHm(2)??rGm(2)?T?rSm(2)?[?178.65?298?(?0.0569)]kJ?mol?1??195.62kJ?mol?1
已知 H2(p?)?1O2(p?)?H2O(l) (3) 2?1 Jmol237.?13k?已知 ?rGm(3)??G?,?l)fm(H2O? ?rHm(3)??H?,?l)fm(H2O?1 Jmol285.?83k又 (3) – (2)得
Hg(l)?1O2(p?)?HgO(s) (4) 2??fGm(HgO)??rGm(4)??rGm(3)??rGm(2)?(?237.13?178.66)kJ?mol?1??58.47kJ?mol?1
??fHm(HgO)??rHm(4)??rHm(3)??rHm(2)?(?285.83?195.62)kJ?mol?1??90.21kJ?mol?1又 (1) + (4)得
12Cu(s)?O2(p?)?Cu2O(s) (5)
2??fGm(Cu2O)??rGm(5)??rGm(1)??rGm(4)
?(?89.11?58.47)kJ?mol?1??147.58kJ?mol?1
??fHm(Cu2O)??rHm(5)??rHm(1)??rHm(4)
?(?97.39?90.21)kJ?mol?1??187.60kJ?mol?1