第五章电化学

电池反应的电子转移数 z=2

∴ ΔrGm= -zFE = -2×96485×(0.9259) J·mol-1= -178.7 kJ·mol-1; ∵ΔrGm<0,∴电池反应能自发进行。

zFE$2?96485?0.9259由lnK???72.08

RT8.3145?298.15$得K$?2.01?1031

(3) Zn | Zn2+ (?Zn2+= 0.0004) || Cd2+ (?Cd2+= 0.2) | Cd

负极: Zn (s)→Zn2+ (?Zn2+= 0.0004) + 2e

正极: Cd2+ (?Cd2+= 0.2) + 2e→ Cd(s)

电池反应:Cd2+ (?Cd2+= 0.2)+ Zn (s)→ Cd(s)+ Zn2+ (?Zn2+= 0.0004) E?E$?RTRTaZn2+$B ln?a??E?lnBzFzFaCd2+B$$查表可得:?Zn??0.4028V ??0.7630V; ?Cd2+2+/Cd/Zn$$E$??Cd??Zn?[?0.4028?(?0.7630))V?0.3602V 2+2+/Cd/ZnE?E$?RTRTaZn2+$Bln?a??E?lnBzFzFaCd2+B8.3145?8.31450.0004ln]V

2?964850.2?[0.3602??0.4400V电池反应的电子转移数z=2

∴ ΔrGm=-zFE =-2×96485×(0.4400) J·mol-1= -84.91 kJ·mol-1; ∵ ΔrGm<0,

∴ 电池反应能自发进行。

zFE$2?96485?0.3602由lnK???28.04

RT8.3145?298.15$得 K$?1.51?1012

例题 5.10 用Pt作电极电解SnCl2水溶液,在阴极上因为H2有超电势,故先析出Sn(s)。在阳极上析出O2。已知?Sn2+=0.10 ,?H+=0.01,O2在阳极上析出的超电势为0.5V。 (1) 写出电极反应,计算实际分解电压。

(2) 若H2在阴极上析出时的超电势为0.5V,问刚开始析出氢气时α(Sn2+)=? (《物理化学》第五章习题5.22) 解:(1)阴极:Sn2+ + 2e– → Sn ;

?阴,析出??Sn2+/Sn??Sn2+/Sn?$RT1ln2F?Sn2+?(?0.140???0.170V8.3145?2981ln)V2?964850.1

阳极: H2O→1O(g) +2H+ (?H+) + 2e– 22

?阳,析出??O?(1.23??1.612V2/H+,H2O??O$2/H+,H2O?RT1ln2??O22F?+H8.3145?2981ln?0.5)V2?964850.012 E分解??阳,析出??阴,析出??Sn2+/Sn??O2/H+,H2O??Sn2+/Sn?(1.612?0.170)V?1.782V

(2)由于阳极上有O2析出,溶液中H+浓度会增加。当Sn2+基本析出时,溶液中H+的活度为为?H+?0.10?2?0.01?0.21。当H2开始析出时,其电极电势与Sn2+的析出电势相等。即

?Sn2+/Sn??H+/H??H2

2即 ?0.140?RT1RT1ln??ln2?0.5 2F?Sn2+2F?+H代入数据 ?0.140?8.3145?29818.3145?298ln?ln0.21?0.5

2?96485?Sn2+96485解得:?Sn2+?2.92?10?14

说明当H2开始析出时,Sn2+已基本析出完毕了。

例题 5.11 25℃时,某电解液中含Ag+(?Ag+?0.05)、Fe(?Fe2+?0.01)、Cd(?Cd2+?0.1)、

2+

2+

Ni(?Ni2+?0.1),pH=3,已知?Ag???0.4402V

>>閻忕偞娲栫槐鎴﹀礂閵婏附鐎�<<
12@gma联系客服:779662525#qq.com(#替换为@)