??rHm??fHm[H2O(l)]??285.83kJ?mol-1 (2
分) 因为
?rSm??rHm??rGm
T所以 ?[(?285.83)?(?236.97)]?103rSm?.15??163.89J?K-1?mol-1298
(2分) 而
?S??E?rm?2F???T??,则电池的温度系数为:
p???E??rSm?163.89??T??????8.493?10?4 (2
p2F2?96485V?K-1分)
Qr = TΔrSm = 298.15×( -163.89) = -48825.04 J· mol-1 (2分) ⑶ 若反应热不随温度变化,则ΔrSm也不随温度变化,故:
?E??rSm2F??T??163.892?96485?(298.15?273.15)??2.123?10?2V
E 273.15 = E 298.15 -ΔE = 1.228 - (-2.123×10-2
) = 1.249 V (3分) 52.写出下列电池的电极和电池反应式,并计算在25℃时各原电池的电动势:
(1)Zn | Zn2+(a=0.01) || Fe2+
(a=0.001),Fe3+
(a=0.1) | Pt
(2)Pt | H2(1.5p ) | HCl(b=0.1 mol·kg-1) | H2(0.5p ) | Pt
(已知E (Zn2+|Zn) = ?0.7628 V,E (Fe3+,Fe2+|Pt) = 0.771 V。)
解: (1)Zn+2 Fe3+(a=0.1)=== Zn2+
(a=0.01)+2 Fe2+(a=0.001) (1分)
17
E=0.771-(-0.7628)- V
=[1.5338-0.029 58 lg10-6] V=1.7113 V (3分)
+?
(2)(?)H2(1.5p ) 2H+2e (+)2H++2e? H2(0.5p ) H2(1.5p ) === H2(0.5p ) (4分)
E=-0.02958 lg (0.5 p /1.5p ) V= 0.01413 V 6分)
18
(
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