?I3+2e,??=0.535VзӦƽⳣ
?? H3AsO3 + I3 + H2O = H3AsO4 + 3I- + 2H+
ҺPh=7ӦʲôУ
Һ[H+]=6moldm-3ӦʲôУ
10 ֪ڼԽ?H
??2PO2/P4
= -1.82V?H??2PO2/P4 = -1.18VP4-PH3ı
ƣжP4Ƿܷ᪻Ӧ
11 ԭƱжзӦܷ᪻Ӧ
(a) 2Cu
Cu + Cu2+(b)Hg
2?2Hg + Hg2+(c)2OH- + I2
IO- + I- + H2O(d)H2O + I2
HIO + I- + H+
5
12һѹǿΪ1.01310Pa缫һ90%ѹǿ1.013105Pa
У˵صĵ綯E
13 ͭˮҺŨȷֱΪ[Cu2+]=0.015moldm-3[Ni2+]=0.23moldm-3
[H+]=0.72moldm-3ȷŵʣʣ
14 ԼзӦıĦܱ仯rG?m
??Mn2+ + 2H2O + Br2bBr2 + HNO2 + H2O???2Br + (a) MnO2 + 4H+ + 2Br?????2 I + Sn4+(d)NO3??2Fe3+ NO3 + 3H+ (c)I2 + Sn2+ ? +3H+ +2Fe2+?????? Br2 +2Cl +HNO2 + H2O(e)Cl2 + 2Br? (a) MnO2 + 4H+ + 2e
??????Mn2+ + 2H2O?=1.23V;2Br?2e???? Br2
?1= nEF= 31.806kJmol ?=1.0652V;E=??=0.1648V rG?m (b) Br2 + 2e
?????2Br??=1.065V;HNO2 + H2O2e???? NO3 + 3H+
?1 ?=0.94V;E=??=0.125V rG?m= nEF= 24.125kJmol
??2 I (c)I2 + 2e????? Sn4+ ?=0.15V; ?=0.5355V ;Sn2+ 2e???1E=??=0.3855V rG?m= nEF= 74.407kJmol (d) 2Fe2++ 2e
?
????2Fe3+ ?=0.771V ; HNO2 + H2O2e???? NO3 + 3H+
?1= nEF= 32.617kJmol ?=0.94V ; E=??=0.169V rG?m?? 2Cl (e) Cl2 + 2e????? Br2 ?=1.065V ?=1.36V 2Br?2e???1E=??=0.295V rG?m= nEF= 56.935kJmol15֪ڼԽеı缫ƣ
2???Cr(OH)3(s) + 5OH(aq) CrO4(aq) + 4H2O(l) + 3e ???
??= -0.11V
??Cu(s) + 2 NH3(aq) [Cu(NH3)]+ (aq) + e ????= -0.10V
???+
ԼH2ԭCrO2rG?4[Cu(NH3)]ʱ?mK
??ȻƣrG?mKȴܴԭ CrO4fG= +31.8kJmol
?2???1K=2.710
?1??6
?3 [Cu(NH3)]+fG= +11.6kJmol
K=9.310
?
16 298KʱSn2+Pb2+ĩƽҺڵǿȵҺ[Sn2+]/
[Pb2+]=2.98֪?Pb2?/Pb= -0.126V?Sn2?/Sn Sn + Pb
2+
??Sn + Pb K
2+?=[Sn]/ [Pb]=2.98; lg K
2+2+?nE?= =
0.0591
??n(??2?0.126x)?= ; x == 0.14V;?Sn== 0.14V 2?/Sn0.05910.059117298KʱӦFe3+ + Ag+0.770V?Ag?/Ag
?? Fe2+ + Ag+ƽⳣΪ0.531֪?Fe= 3?/Fe2???n(??1?0.770x)nE??lg K= = = ; x == 0.786V; ?Ag?/Ag== 0.786V
0.05910.05910.0591?18 ͭƬгZn,Pb,Fe,Agȣͭͭе
õΪ99.99%ͭõ缫˵ͭġ
缫ƱȽϴrG?mȽСԽеijס
19 ںCdSO4ҺĵصϼѹӦĵ£ E/V 0.5 1.0 1.8 2.0 2.2 2.4 2.6 3.0 I/A 0.002 0.0004 0.007 0.008 0.028 0.069 0.110 0.192 ֽͼֽѹ
B3.02.52.0E / V1.51.00.50.000.050.100.150.20I / A óE=3.0V
20һͭУǿΪ5000AЧΪ94.5%ʾ3hСʱܵõͭkgǧˣ I=
q7 q=I t = 500094.5%33600=5.01310أ tCu + 2e == Cu 2 64 5.01310 m m = 1.6310ǧˣ
77