lg????Az?z??? =0.967
I8.1?1?40??0.509?1?b?1??0.0 1449 HA ======= H+ + A-
平衡浓度beq/mol·kg
-1
0.01-0.0810×0.01 0.0810×0.01 0.0810×0.01
= 0.00919 = 0.00081 = 0.00081
b?20.000812)(0.967?)2?aHA(已电离)aH?aA?a??b1K??????6.68?10?5baHA0.00919?10.009190.00919(未电离)?HAHAb?(?? 说明: 近似认为未电离的HA的活度系数 ?HA = 1 习题5.11
解:由5-13式得 ?(盐) = 0.01482 – 0.00015 = 0.01467 S/m
?m? (盐) =( 2×59.46 +2×80.0) ×10-4 = 0.02789 Sm2/mol
C = ?(盐)/?m? (盐) = 0.01467/0.02789 = 0.5260 mol/m3 = 5.26×10-4 mol/L
习题5.12
解:由5-21式得 I =0.5×(0.3×12+0.1×22+0.1×12) =0.4 mol · kg-1
第六章 电化学
习题6.1
(1)解: (-) Cu (s) → Cu2+(a Cu2+) + 2e (+) 2Ag+ (a Ag+) + 2e→ 2Ag (s)
电池反应: Cu (s) + 2Ag+ (a Ag+) → Cu2+ ( a Cu2+) + 2Ag (s)
(2)解: (-) H2 ( pH2) → 2H+ (a H+) +2 e (+) 2Ag +(a Ag+) + 2e → 2Ag (s)
电池反应: H2 ( pH2) +2 Ag+ (a Ag+) → 2H+ (a H+) + 2Ag (s) (3)解: (-) Ag (s) + Br(a Br-) → AgBr (s) + e
-
(+) AgCl (s) + e → Ag (s) + Cl- (a Cl-)
电池反应: Br (a Br-) + AgCl (s) → AgBr (s) + Cl- (a Cl-)
-
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(4)解: (-) Sn2+ (a Sn2+) → Sn4+(a Sn4+) + 2e (+) 2Fe3+ (a Fe3+) +2e → 2Fe2+ (a Fe2+)
电池反应: Sn2+ (a Sn2+) + 2Fe3+(a Fe3+) → Sn4+(a Sn4+) + 2Fe2+ (a Fe2+) (5)解: (-) Pb (s) + SO42-(a SO42-) → PbSO4 (s) + 2e (+) Cu2+(a Cu2+) +2e→ Cu (s)
电池反应:Pb (s) + Cu2+ (a Cu2+) + SO42-(a SO42-) → PbSO4 (s) + Cu (s)
习题6.2
(1)解:Zn (s)│ZnSO4 (aq)║CuSO4 (aq)│Cu (s)
(2)解:Ag (s)︱AgI (s)│I-(a I)║Cl- (a Cl-)│AgCl (s)︱Ag (s) (3)解:Pt (s)│Fe2+(a Fe2+),Fe 3+(a Fe3+)║Ag+(a Ag+)│Ag (s) (4)解:Pt (s)︱H2 ( pH2)│H+(a H+)│O2 ( pO2)︱Pt (s) (5)解:Ag (s)︱AgCl (s)│Cl-(a Cl-)│Cl2 ( pCl2)︱Pt (s)
习题6.3
解: 设293K时电池电动势为E:
E = ? 4.16×10-5(293 ? 273 ) + 1.0186V = 1.0178 V
设反应中 n = 2
ΔrGm = ? nFE = ?2×1.0178 ×96485 = ?196.40 kJ/mol ΔrSm = 2×96485 ×(?4.16×10-5) = ? 8.03J/K· mol
ΔrHm =ΔrGm +TΔrSm = ?196.40 + 293×(?8.03)×10-3 = ?198.75 kJ/mol QR = TΔrSm = 293K×(? 8.03)×10-3= ? 2.35 kJ/mol W ′max = ΔrGm = ?196.40 kJ/mol
ΔrUm = QR + W ′max = ?2.35 ? 196.40 = ? 198.75 kJ/mol
习题6.4
解:负极(氧化反应):Zn (s) → Zn2+(a Zn2+) + 2e
正极(还原反应):Cu2+(a Cu2+) + 2e → Cu (s)
电池反应: Zn (s) + Cu2+(a Cu2+) → Zn2+(a Zn2+) + Cu (s) (1)反应中 n = 2
E? = φ+?? φ-? = 0.337 ? (?0.763) = 1.100V
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RTaZn2?RT0.1E?E?ln?1.100?ln?1.100V
nFaCu2?nF0.1?(2)ΔrGm = ? nFE = ?2×1.100×96485 = ?212.27 kJ/mol
ΔrG? = ? RT lnK? = ? nFE?
lnK ? = 2×1.100×96485 /(8.314×298 ) = 85.675
K ? = 1.62×1037
习题6.5
解:负极(氧化反应):Zn (s) → Zn2+ (a Zn2+) + 2e
正极(还原反应):2Tl+ (a Tl+) + 2e → 2Tl (s)
电池反应: Zn (s) + 2Tl+ (a Tl+) → Zn2+(a Zn2+) + 2Tl (s) (1)E? = φ+?? φ-? = ? 0.3363 V ? (?0.763V) = 0.4267V (2)反应中 n = 2
aZn2?RT8.314?2980.95E?E?ln?0.4267?ln?0.4255V
2F(aTi?)22?964850.932? 习题6.6
解:负极(氧化反应): 0.5H2 ( p? ) → H+(a H+) + e
正极(还原反应): AgBr (s) + e→ Ag (s) + Br-(a Br-) 电池反应: 0.5H2 ( p? ) + AgBr (s) → Ag (s) + HBr (a) 反应中 n = 1
E?E??RT2RT2RTblnaHBr?E??ln???ln? FFFbF(E??E)b96485?(0.07103?0.200)0.100?ln???ln??0.208 ln???2RTb2?8.314?2981r?= 0.812
习题6.7
解:(1)?玻??玻??2.303RT?logaH???玻?0.05916PH FpHx = pHS + (Ex ?ES)/0.05916
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= 4.00 +(0.2065 ? 0.1120)/0.05916 = 5.60 (2) Ex = ES ? 0.05916 (pHS ? pHx)
= 0.1120 ? 0.05916 (4.00 ? 2.50) = 0.0233V
习题6.8
解:(1)设计电池:Pt (s)︱H2 ( p?)│HI (a ±=1)│I2 (s)︱Pt (s)
(2)反应中 n = 2
E? =φ+?? φ-? = 0.5362 V
lnK? = 2×96485×0.5362/(8.314×298)= 41.76
K? = 1.37×1018
(3)反应中 n = 1, E?相同,K? (2)= [K? (1)]1/2 =1.17×109
电池的E和E?与电池反应的书写方式无关,而K?与电池反应的书写方式有关。
习题6.9
解:(1)设计的电池为: Pt (s)︱H2 ( p?)│H2SO4 (0.1mol · kg-1)│Ag2SO4 (s)︱Ag (s)
负极(氧化反应): H2 ( p?) → 2H+(a H+) + 2e
正极(还原反应):Ag2SO4 (s) + 2e→ 2Ag (s) + SO42-( a SO42- )
电池反应: H2 ( pH2 ) + Ag2SO4 (s) → 2Ag (s) + 2H+(a H2 ) + SO2-( a SO2-) (2) E?0.627?0?8.314?298 ln0.1?0.22?0.6982?96485(3)设计的电池为: Ag (s)│Ag+(a Ag+)║SO42-( a SO42- )│Ag2SO4 (s)︱Ag (s)
负极(氧化反应): 2Ag (s) → 2 Ag+(a Ag+) + 2e
正极(还原反应): Ag2SO4 (s) + 2e→ 2Ag (s) + SO42-( a SO42- ) 电池反应: Ag2SO4 (s) → 2 Ag+(a Ag+)+ SO42-( a SO42- )
E? = 0.627V ? 0.799V = ? 0.172V lnKsp???2?96485?(?0.172)??13.396 58.314?298?6 Ksp?1.52?10
习题6.10
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解:首先明确两个电极反应,然后根据电极反应写出电池表示式。电池的最高电压就是电池的电动势,可以由参与电池反应的物质的标准化学势数据求出。介质酸碱性对电池电动势的影响,根据介质改变后电池反应的变化而定。如果电池反应改变了,反应的Gibbs自由能就会发生改变,电池的最高电压也将发生改变。
电解质溶液为酸性时:
负极(氧化反应): CH4 ( p) + 2H2O (l) → CO2 ( p) + 8H+(aH+) + 8e 正极(还原反应): 2O2 ( p) + 8H+(aH+) + 8e→ 4H2O (l) 电池反应: CH4 ( p) + 2O2 ( p) → CO2 ( p) + 2H2O (l) 电池表示式: Pt (s)︱CO2 ( p) , CH4 ( p)│H+(a H+ )│O2 ( p)︱Pt (s) 电池电动势: ?fGm??nFE
??fGmE??nF??[?394.38?2?(?237.19)?(?50.79)]?1.06V
8?96485电解质溶液为碱性时:
负极(氧化反应):CH4 ( p) + 10OH-(a OH-) → CO32- (a CO32-)+ 7H2O (l) + 8e 正极(还原反应): 2O2 ( p) + 4H2O (l) + 8e→ 8OH-a OH-)
电池反应: CH4 ( p) + 2O2 ( p) + 2OH-(a OH) → 3H2O (l) + CO32-(a CO32-) 电池表示式:Pt (s)︱CH4 ( p)│CO32-( a CO32-=1)║OH-(a OH-=1)│O2 ( p)︱Pt (s) 电池电动势: E = ?Δf G? / nF
= ? [(?528.10) +3×(?237.19) ? (?50.79) ?2×(?157.27)]×103/(8×96485)= 1.13V 由计算可知,电解质溶液不同,电池反应也不同,因而反应的ΔrG发生改变,电池的标准电动势也发生改变。
虽然电解质溶液为碱性时电动势较高。但是,电解质溶液为碱性时,碱液作为原料不断被消耗掉,变成副产物碳酸盐,从收益及效益两方面综合分析,电解质溶液为酸性时更具有实用价值。
习题6.11
解:负极(氧化反应): Pb (s) → Pb2+ (a = 0.01) + 2e 正极(还原反应): Cl2(p?) +2e → 2Cl- (a =0.5)
电池反应: Pb (s) + Cl2(p?) → Pb2+ (a = 0.01) + 2Cl- (a = 0.5)
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