电化学原理_(李狄_著)北航出版社_课后1-7章习题参考答案
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电化学原理第一章习题答案
1、解:2266KCl KCl H O H O 0.001141.3
1.01014
2.31010001000
c K K K K cm 11λ−−−−×=+=+=+×=×Ω溶液 2、解:E V F
i i =λ,F
E V i i λ=,,, 10288.0−⋅=+s cm V H 10050.0−⋅=+s cm V K 1
0051.0−⋅=−s cm V Cl 3、解:
,62.5501
2
1
,,,,2−−⋅Ω=−+=eq cm KCl o HCl o KOH o O H o λλλλ
2O c c c ,c 1.004H H +−====设故,2
,811c
5.510cm 1000
o H O λκ−−−=
=×Ω
4、(1)
1
2
1
,,Cl ,t t 1,t 76.33mol (KCl o KCl o Cl cm λλλλλ−−−−+−+−=++=∴==Ω⋅∵中)121121121,K ,Na ,Cl 73.49mol 50.14mol 76.31mol (NaCl o o o cm cm cm λλλ++−−−−−−−=Ω⋅=Ω⋅=Ω⋅同理:,,中)
(2)由上述结果可知: 1
2
1
Cl ,Na ,1
2
1
Cl ,K ,mol 45.126mol 82.142−−−−⋅Ω=+⋅Ω=+−+−+cm cm o o o o λλλλ,在KCl 与NaCl 溶液中
−
Cl ,o λ相等,所以证明离子独立移动定律的正确性;
(3) vs cm vs cm u vs cm u F u a o o l o l o i o /1020.5,/1062.7,/1091.7,/2
4N ,24K ,24C ,C ,,−−−×=×=×==++−−λλ5、解:Cu(OH)2== Cu 2++2OH -,设=y ;2Cu c +OH c −=2y 则K S =4y 3
因为u=Σu i =KH 2O+10-3[y λCu 2++2y λOH -]以o λ代替λ(稀溶液)代入上式,求得y=1.36×10-4mol/dm 3所以Ks=4y 3=1.006×10-11 (mol/dm 3)3
6、解: ==+,令=y ,3AgIO +Ag −3IO Ag c +3
IO c −=y ,则=y S K 2,K=i K ∑=+(y O H K 23
10−+Ag λ+y −3
IO λ)
作为无限稀溶液处理,用0λ代替,
=+y O H K 23
10−3AgIO λ
则:y=436
51074.1104
.68101.11030.1−−−×=××−×L mol /;
∴= y S K 2=3.03
8
10
−×2)/(L mol 7、解:HAc o ,λ=HCl o ,λ+NaAc o ,λ-NaCl o ,λ=390.7,1
2
1
−−⋅Ωeq cm HAc o ,λ=9.02
1
2
1
−−⋅Ωeq cm ∴α0/λλ==0.023,==1.69
αK _
2
)1/(V αα−5
10−×8、解:由欧姆定律IR=iS KS l ⋅
=K il
,∵K=1000c λ,∴IR=1000il c
λ⋅=V 79.05.0126101010533≈××××− 9、解:公式log ±γ=-0.5115||||+Z −Z I (设25)
C °
(1)±γ=0.9740,
I=212i i z m ∑,I=2
12
i i c z ∑,=()±m ++νm −−νm ν1
(2)±γ=0.9101,(3)±γ=0.6487,(4)±γ=0.8114
10、解:=+H a ±γ+H m ,pH=-log =-log (0.209+H a 4.0×)=1.08
电化学原理第二章习题答案
1、 解:
()+2326623Sb O H e Sb H O ++++ ,()−236H H +6e + ,电池:2322323Sb O H Sb H O ++
解法一:0
0G E nF ∆=−83646F =0.0143V ≈,E=+0E 2.36RT F 223
2323log H Sb O Sb H O
P a a a ==0.0143V
0E 解法二:0
60
2.3 2.3log log 6Sb Sb H H RT RT a a F F
ϕϕϕ+++=+
=+; 2.3log H RT
a F
ϕ+−=
∴0
00.0143Sb E E ϕϕϕ+−=−===V
2解:⑴,(()+22442H O e H O +
++ )−224H H +
4e + ;电池:22222H O H O +
2220
02
2.3log 4H O H O P P RT E E E F
a =+= 查表:0ϕ+=1.229V ,0ϕ−=0.000V ,00
1.229E V ϕϕ+−∴=−= ⑵视为无限稀释溶液,以浓度代替活度计算
()242Sn Sn e ++−+ ,(),电池:32222Fe e Fe ++++ 23422Sn Fe Sn Fe 2+++++ +
23422022.3log 2Sn Fe Sn Fe C C RT E E F C C ++
++
=+=(0.771-0.15)+22
0.05910.001(0.01)log 20.01(0.001)××=0.6505V ⑶(),,
(0.1)Ag Ag m e +
−+ ()(1)Ag m e Ag +
++ (1)(0.1)Ag m Ag m +
+
→电池:(1)0
(0.1)
2.3log Ag m Ag m a RT E E F a ++=+
,(其中,=0) 0
E 查表:1m 中3AgNO 0.4V γ±=,0.1m 中3AgNO 0.72V γ±=, 2.310.4
log
0.0440.10.72RT E V F
×∴==× 3、 解:
2222|(),()|(),Cl Hg Hg Cl s KCl m Cl P Pt ()2222Hg Cl Hg Cl e −−++ ,()222Cl e Cl −++ ,222Hg Cl Hg Cl 2+ 电池: