无机及分析化学第六章课后习题答案(高等教育出

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6 3 2) Zn + 4 OH

= Zn ( OH )

2 4

+ 2e

H 2 O + ClO 4)

+ 2 e = 2 OH

+ Cl

Zn + H 2 O + 2 OH

+ ClO = Zn ( OH ) 2 + Cl 4

( H 2 O 2 + 2 e = 2 OH ( 4 OH

)×32 4

+ Cr ( OH ) 4 = CrO

+ 4 H 2 O + 3e ) × 22 4

3 H 2 O 2 + 2 OH

+ 2 Cr ( OH ) = 2 CrO 4

+ 8 H 2O

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6 4 1)2 ( S 2 O82 + 2e = 2 SO 4 ) × 5

( 4 H 2 O + Mn 3)

2+

= MnO + 8 H + 5e ) × 2

4

+

2 5 S 2 O82 + 8 H 2 O + 2 Mn 2 + = 2 MnO 4 + 16 H + + 10 SO 4

Cr2 O72 + 14 H + + 6e = 2Cr 3+ + 7 H 2 O ( Fe2+

= Fe

3+

+ e )×6

Cr2 O72 + 14 H + + 6 Fe 2 + = 2Cr 3+ + 7 H 2 O + 6 Fe 3+

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6 9 1) MnO + 5 Fe + 8 H → Mn + 4 H 2O + 5 Fe E(θ+ ) > E(θ )反应正向进行 2)(-)Pt∣Fe 3)2+(C 1),Fe 3+(C 2)‖MnO4 -(C 3),Mn 2+(C )∣Pt 4

4

2+

+

2+

3+

(+)

E

= 1.51V-0.771V=0.739V

E= (+)- (-)= (+) +

00 9 . 5 V c 氧 态a ( 化 ) l g b - (-) n c还 态 ( 原 )

=1.51V+0.0592/5 lg108 –0.771V =0.83V

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6-10,

E= E(+)-E(-)=1.51V[E(Ag+/Ag) +0.0592V lg(Ag+)]-E(Zn2+/Zn) + 0.0592V lg(Zn2+)] =1.51V 由于C(Ag+)=0.10 mol/L 所以 C(Zn2+)=0.57 mol/L

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6-11

查表， E(Sn2+/Sn) =-0.136V, E(Pb2+/Pb) =-0.126V 因为， E = E(+) -E(-) =-0.356V= E(PbSO4/Pb)= 0.22V=-0.136V- E(-) E(-) = E(PbSO4/Pb) E(Pb2+/Pb) + 0.0592/2 lgC(Pb2+) = -0.126V + 0.0592/2 lgK lgKSP

(PbSO4)/C(SO42-) SP

(PbSO4)= - 7.77

K

SP

(PbSO4)=1.7 ×10-8

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6-12 1) 正极 : E(Cu2+/Cu)

反应式：

Pb + Cu 2+ = Pb 2+ + Cu

= E(Cu2+/Cu)

00 9 . 5 V c 氧 态a ( 化 ) + l g n c 还 态b ( 原 )

= 0.337V + 0.0592/2 lg0.5 = 0.33 V00 9 . 5 V c 氧 态a ( 化 ) l g n c 还 态b ( 原 )

负极: E(Pb2+/Pb) = E(Pb2+/Pb)

+

=-0.126V + 0.0591/2 lg0.1 = -0.16 E = E(Cu2+/Cu) - E(Pb2+/Pb) = 0.33-(-0.16)= 0.49V lgK K = nE /0.0592 = 2 ×[0.337-(-0.126)]/0.0592 = 15.64 =4.4 ×1015

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2) 反应式： 2H+ + Sn = Sn 2+ + H2 正极 : E( H+ /H2) = E(H+/H2)

= 0.0000V

00 9 . 5 V c 氧 态a ( 化 ) + l g n c 还 态b ( 原 )

负极: E( Sn2+ /Sn) = E(Sn2+/Sn)

00 9 . 5 V c 氧 态a ( 化 ) l g + n c 还 态b ( 原 )

=-0.136V + 0.0591/2 lg0.05 = -0.17 E = E(+) - E(-) = 0.0000 -(-0.17)= 0.17V lgK 4.59 K = nE /0.0592 = 2 ×[0.0000-(-0.136)]/0.0592 = =3.9 ×104

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4)

反应式：2H++H2= 2H++H200 9 . 5 V c 氧 态a ( 化 ) + l g n c 还 态b ( 原 )

正极 : E( H+/H2) = E(H+/H2) = 0.0000V

负极: E(H+/H2) = E(H+/H2)

+

00 9 . 5 V c 氧 态a ( 化 ) l g n c 还 态b ( 原 )

= 0.0000 + 0.0591/2 (lg0.012/1) = -0.12V E = E(+) - E(-) = 0.0000 -(-0.12)= 0.12V lgK K = nE /0.0592 =[2 ×0.0000]/0.0592 = 0 =1

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6-15有关反应 :

Ba2+ + CrO42- = BaCrO4 2CrO42+ 2H+ = Cr2O72- + H2O

Cr2O72- + 6I- + 14H+ = 2Cr3+ +7H2O +3I2 I2 + 2NaS2O3 = Na2S4O6 + 2NaI n(Ba2+) = n(CrO42- ) = 2n(Cr2O72-)=2/3n(I2) = 1/3n (S2O32-) W(BaCl2)= m(BaCl2)/ms = n(BaCl2)M(BaCl2)/ ms = 1/3n (S2O32-) M(BaCl2)/ms =[1/3 ×0.1007 ×29.61 ×10-3 ×208.3] / 0.4392 = 0.4714

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6-16

有关反应 2Mn2+ +10C

O2 + 8H2O C2O42- + H2O

2 MnO4- + 5 C2O42-+ 16H+ HC2O4-+OH -

n(MnO4- ) = 2n(C2O42-)/5 n(KHC2O4.H2O) = n(KOH) C(MnO4- ) = n(MnO4- )/V(MnO4- ) = 2n(C2O42-)/5V(MnO4- ) = 2n(KOH)/ 5V(MnO4- ) = 2 ×0.2000 ×25.20/5 ×30.00 = 0.0672 mol/L

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6-18

2CrO42- + 2H+ = Cr2O72- + H2O

2Cr2O72- + 6I- + 14H+ = 2Cr3+ +7H2O +3I2 I2 + 2NaS2O3 = Na2S4O6 + 2NaI n(Pb3O4)=n(Pb)/3=n(CrO42- )/3= 2n(I2 )/9 = n(S2O32-)/9 W(Pb3O4)= m(Pb3O4)/ms = n (Pb3O4)M (Pb3O4)/ ms= [n(S2O32-)M (Pb3O4) /9]9 ms =[0.1000 ×12.00 ×10-3 ×685.6]/0.1000 =0.9141

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6-20Hg2Cl2(s)+ 2e = 2Hg(1)+2Cl-(aq) Hg22+ + 2e = 2Hg(1) 将上述两极反应组成原电池 (-)Pt|Hg(1)| Hg2Cl2(s)|Cl- ‖Hg22+ |Hg(1) |Pt(+) )Pt| ( )| | ( ) |Pt( 电池反应 Hg22+ + 2Cl- = Hg2Cl2(s) E (Hg2Cl2 /Hg) = 0.28V E (Hg2Cl2 /Hg) = 0.80V

E = 0.80 - 0.28 = 0.52V反应平衡常数

n θ θ l K = g (E 正 E 负 ) 00 9 .5θ

lgK = 2 ×0.52 /0.0592 = 17.56 K = 1017.56 KSP

(Hg2Cl2) = 1/ K =1/1017.56 =2.8 ×10-18