分析化学答案04氧化还原滴定

第四章 氧化还原滴定法

1. 解:E?EZn2?/Zn?=EZn2?/Zn000.0590.05902?lg?Zn2??EZn?lgZn 2?/Zn220.059?lgCZn2?/?Zn(NH3)??0.763?...

2??(?Zn2??Zn2?忽略离子浓度)

CZn2?/?Zn(NH3)1.0?10?4? 1?102.31?10?1?104.81?10?2?107.31?10?3?109.46?10?4?10?? ?3.23?10

E??0.763?2.解：

0.059lg3.23?10?10??0.763?0.280??1.04V 2E?E0Fe3?/Fe2?0?EFe3?/Fe2?0?EFe3?/Fe2??FeR???R?K?0.059lg

K??FeR??R?K?FeR?K?0.059lg?0.059lgK?FeR?3?332??3IIIII33?32?III3?Fe??0.059lg?Fe?3?2?KIII???Fe??R?3??FeR?3?332?II?Fe??R?2??FeR?3

33?2? 当FeR3?FeR3?1mol?L?1时

???? E?E3．解：（1）.

01?E0?0.059lgKII?0.771?0.38?1.15V KIII2?2?Hg2?2e?Hg,Hg2?2Cl??Hg2Cl2Hg2Cl2?2e?2Hg?2Cl0E?EHg2Cl2/Hg?0.0591?lg2Cl??? E?EHg2?/Hg?2020?EHg2?/Hg20.0592?lgHg22Ksp0.059?lg22Cl???

???

E0Hg2Cl2?0.0591lg2Cl???20?EHg2??20.0590.0591lgKsp?lg22Cl??

200?EHg2Cl2?EHg2?2/Hg?0.059lgKsp2

0.059lg1.3?10?182?0.793?0.528?0.265V?0.793?0（2）.E?EHg?2Cl2/Hg0.0591 lg?22Cl???0.265?0.0591lg2(10?2)2

?0.265?0.118?0.383V4.解： 2Fe3??H2S?2Fe2??S?2H?

E1?E0/Fe3?/Fe2?0E2?ES/H2S?Fe??0.059lg?Fe?0.059?H??lg

3?2??22?H2S?3?达平衡E1?E2E0/Fe3?/Fe2??Fe??E?0.059lg0.1??Fe?3?3?0S/H2S0.059H? ?lg?H2S?2??2?H??0.25?0.1(反应生成的)

??Fe??0.141?0.059lg0.35

有0.71?0.059lg20.10.1??Fe??Fe??2.51?10mol?L

?C?Co??E?0.059lg?0.059lg5.解：E?E?0.059lg?C?Co?23?3??11?13?1012?01Co(NH3)2?Co(NH3)3?Co3?Co2?

E01??E?0.059lg01?Co(NH?Co(NH3)3)2?

3??Co(NH?103)2??1?102.11?10?1?103.74?10?2?104.79?10?3?105.55?10?4?105.73?10?5?65.11

?103??1?12.88?54.95?61.66?35.48?5.37?0.13?171.5

?Co(NH1033.23)?1?106.7?10?1?1014?10?2?1020.1?10?3?1036.7?10?4?1030.8?10?5??6?10?1?105.7?10?101217.1?1021.7?1025.8?1027.2?1027.2?171.5E10?1.84?0.059lg27.2?1.84?1.47?0.367

10 由于 E1?E2 反应4Co2?0?0?Co2?被氧化成Co3?

?O2?2H2O?4OH??4Co3?

?3?由于Co(NH3)36浓度最大，所以最终为Co(NH3)6形式存在

??6.解： PH?10.0 H??10?10 OH-?10?4

??3???NH3??C?NH?OH??C??OH??K??b10?4?0.1??4?0.085mol?L?1 ?4.7410?10?0.80?(?0.29)?0.51V

3)?E0?E0?0.059lgAg0/AgAg0/Ag1?Ag(NH注：?Ag(NH3)?1?103.24?0.085?107.05?0.0852?1?147.7?81065.8?81214.5

?2?7.解：MnO??4H2O 4?5e?8H?Mn0.059MnO?4（1）.E?E?， lg2?5Mn01????还原一半时MnO-?L?1 Mn2??0.05mol?L?1 4?0.05mol?????E?E0??1.45V

2-（2）.CrO7?6e?14H??2Cr3??7H2O 2?0.059Cr2O7 E?E? lg3?26Cr0????? ?E?1.00? ?1.01V

0.0590.05lg 60.120?（3）.结果证明：对称电对Ey2?E 不对称电对Ey2?E 8.解：E0?E0?0.059lg10??Fe（Y）2??Fe（Y）3?

?Fe3?2??1??Y?KFe（III）Y ?Y??（Y）?Y????Y0.1?10?11.60 10.6010

?Fe（Y）?1??Y?KFe（III）Y

?Fe（Y）?1?10-11.60?1014.32?102.72

2?

?Fe（Y）?1?10?11.60?1025.1?1013. 53?102.72E?0.77?0.059lg13.5?0.77?0.059lg10?10.78?0.77?0.636?0.134V

1001009.解：由于EAg?/Ag?ECu2?/Cu

?发生反应 Cu ?2Ag??Cu2??2Ag?Cu??10 K??Ag?2??215.69

0(E1?E0(0.80?0.337)?22)?2??15.69 lgK?0.0590.059 反应进行较完全，可认为Cu2??0.025（反应完全） 代入K????Ag? 得到?Ag??2.3?10?20.025?1015.69

?9?mol?L?1

2?10．解：（1）.反应 Cr2O7?6Fe2??14H??2Cr3??6Fe3??7H2O

n 1 ? 6 n 2 ?1 n 3 ?6 0(E1?E0(1.33?0.77)?62)?n??56.95 ?lgK?0.0590.059 K?8.9?10

（2）.计量点时Cr2O756?2?1??1?Fe? ?Cr??

63?Fe?2?3?3? 反应定量进行，一般积Fe3?23?6?2???10?lg?6

21(Fe3?)Fe3?3lgK?lg?Cr??Fe??CrO??Fe??H?22?72?6?14 解得H????14110?56.95??Fe3?9?12?7Fe6??8???56.95

61412?(FeFe2?H?)6110?56.95??(0.05)89??2.9?10-26 1?(10?6)76????6?????? H???0.015m?oLl

?1?11.解：（1）.消元分数0.50 I3?2e?3I?

E?E0I3/I0.059?I3??lg

?32I??? I3??I2??0.05????V?0.51?0.05?

V?V?0.53

20.0010.0010.002221?I??1?30?0.05??2?0.05???0.05??0.05?1??0.05?.0030.0030.003333?原来 生成 与I2存在

E?0.545?0.059lg?0.545?0.038?0.507V

2132(?0.05?)332?0.05?13（2）.消元至100%时，（即消元分数=1.00时） 此时 S4O6 I??.11??02??0.025mol?L2?1

0.05?2???1??0.55mol?L22?10(E10?E2)n(0.545?0.080)?2??15.76 由lgK?0.0590.059?I??SO???I??SO?

K??I??SO??I?(2?I?)?3?32?46223?3?342?6?23

解得：I???330.55?0.025?7?1 ?3?5.65?10mol?L15.764?10?计量点时 I3?0?I3/I???1S2O32? 带入上 2??I3?0.0590.0595.65?10?7Esp?E?lg?0.0545?lg?3220.553 I?0.545?(?0.161)?0.384V2?（3）.消至1.50时，?S2O3??0.1?10.00?0.1?1

????50.005052? ?S4O6??0.05?20?0.05?2

50E?ES?O2?/SO2?46230.059S4Olg2S2O??2?62?3??20.0595?0.130?0.08?lgV

12(0.1?)250.05?

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