无机及分析化学第六章课后习题答案(高等教育出版社,第二版)

()PtFe 2+（C1）,Fe 3+(C2)MnO4-(C3),Mn2+(C4)Pt (),E 1.51V-0.771V=0.739V3) E (+)(-) (+) - (-) =1.51V+0.0592/5 lg108 0.771V =0.83V,6-10, E E(+)E(-)1.51VE(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,6-11查表， E(Sn2+/Sn)=-0.136V, E(Pb2+/Pb)=-0.126V因为， E E(+) E(-) 0.22V=-0.136V- E(-) E(-) = E(PbSO4/Pb) =-0.356V= E(PbSO4/Pb)= E(Pb2+/Pb)+ 0.0592/2 lgC(Pb2+)= -0.126V + 0.0592/2 lgKSP (PbSO4)/C(SO42-)lgKSP (PbSO4)= - 7.77KSP (PbSO4)=1.7 10-8,6-12 反应式： Pb + Cu 2+ = Pb 2+ + Cu正极 : E( Cu2+/Cu) = E(Cu2+/Cu) + = 0.337V + 0.0592/2 lg0.5 = 0.33 V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 = nE/0.0592 = 2 0.337-(-0.126)/0.0592 = 15.64 K =4.4 1015,2) 反应式： 2H+ + Sn = Sn 2+ + H2正极 : E( H+ /H2) = E(H+/H2) + = 0.0000V负极: E( Sn2+ /Sn) = E(Sn2+/Sn) + =-0.136V + 0.0591/2 lg0.05 = -0.17 E = E(+) - E(-) = 0.0000 -(-0.17)= 0.17V lgK = nE/0.0592 = 2 0.0000-(-0.136)/0.0592 = 4.59 K =3.9 104,4) 反应式：2H+H2= 2H+H2正极 : E( H+/H2) = E(H+/H2) + = 0.0000V负极: E(H+/H2) = E(H+/H2) + = 0.0000 + 0.0591/2 (lg0.012/1) = -0.12V E = E(+) - E(-) = 0.0000 -(-0.12)= 0.12V lgK = nE/0.0592 =2 0.0000/0.0592 = 0 K =1,6-15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 = 0.4714,6-16 有关反应2 MnO4- + 5 C2O42+ 16H+ 2Mn2+ +10CO2 + 8H2O HC2O4+OH C2O42 + H2O 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,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) /99 ms =0.1000 12.00 10-3 685.6/0.1000 =0.9141,6-20Hg2Cl2(s)+ 2e = 2Hg（1）2Cl(aq） E (Hg2Cl2 /Hg) 0.28V Hg22+ + 2e = 2Hg（1） E (Hg2Cl2 /Hg) 0.80V 将上述两极反应组成原电池 ()Pt|Hg（1)| Hg2Cl2(s)|Cl Hg22+ |Hg（1) |Pt() 电池反应 Hg22+ + 2Cl = Hg2Cl