氧化還原和電化學(xué)一

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一、選擇題

1.(0701)(C)7.(0713)(B)

8.(0714)(C)2.(0702)(C)整個分

9.(0715)(D)子是中性的。其中Cl為-1，

10.(0716)(A)O為-2，H為+1。

3.(0703)(A)11.(0717)(D)4.(0704)(C)12.(0718)(A)5.(0711)(B)13.(0719)(B)6.(0712)(B)14.(0720)(C)二、填空題(共16題46分)21.(0705)

配平的方程式為：

15.(0767)(C)16.(0768)(D)17.(0769)(C)18.(0781)(C)19.(0782)(A)20.(0791)(D)

(1)7PbO2+2MnBr2+14HNO37Pb(NO3)2+2Br2+2HMnO4+6H2O(2)2CrI3+27Cl2+64KOH2K2CrO4+6KIO4+54KCl+32H2O22.(0706)

N2H4(-2)NH2OH(-1)NCl3(+3)N2O4(+4)23.(0710)

(1)>(4)>(2)>(3)24.(0721)小，小。25.(0724)

(3)>(1)>(2)26.(0725)大于小于

10.0591由于??(A)=??(Ag+/Ag)+lg71.1?101??(B)=??(Ag+/Ag)+

=

=

0.0591lg(1.6?1010)1??(C)=??(Ag+/Ag)+0.059/2·lg9.0?10?12所以??(A)>??(B)??(B)1.08求得lg[H+]=-4.55pH1.59所以應(yīng)控制：1.590所以自發(fā)進行.

lgK?=2?0.47/0.0591=15.91K?=8.1?1015

49.(0738)

[Co(NH3)6]3++e?

=[Co(NH3)6]

2+

??([Co(NH3)6]3+

/[Co(NH3)6]2+

)=??(Co3+

/Co2+

)-0.0591lgK穩(wěn)([Co(NH3)6]3?)K穩(wěn)([Co(NH3)6]2?)3?所以0.0591lgK穩(wěn)([Co(NH3)6])K2?=1.84-0.062=1.78(V)

穩(wěn)([Co(NH3)6])所以

K穩(wěn)([Co(NH3?3)6])K2?穩(wěn)([Co(NH3)6])=1.31?1030

K穩(wěn)([Co(NH3)6]3+)=1.31?1030?1.3?105=1.7?103550.(0739)

反應(yīng)方程式：Co(OH)2+12Br2+OH?=

Co(OH)3+Br?

(1)??(Co(OH)3/Co(OH)2)=??(Co3+/Co2+

)-0.0591lg

Ksp(Co(OH)2)K

sp(Co(OH)3)=1.84-0.0591lg1.6?10?141.6?10?44

=1.84-1.71=0.13V

(2)E?=??(Br2/Br?)–??(Co(OH)3/Co(OH)2)=1.07-0.13=0.94(V)

?lgK?=

zE?1?0.940.05910.0591=15.91

K?=8.1?1015

51.(0741)

??(Cu2+

/CuI)=??(Cu2+/Cu+

)+0.0591lg[Cu2?][Cu?]由于[C