氧化还原反应习题参考答案

1、第4章氧化复原反响习题参考答案1 .解：S的氧化数分别为-2、0、2、4、5、6.2,解：(1) 3Cu + 8HNO 3(稀) 3Cu(NO 3)2+2 NO T + 4H 2O(2) 4Zn + 5H2SO4(浓)一4ZnSO4 +H 2s T+ 4H2O(3) KClO 3 + 6 FeSO4 + 3H2SO4 - KCl +3 Fe2 (SO4)3 + 3H2O(4) Cu2s + 22HNO 3 f 6Cu (NO3)2 + 3H2SO4+ 10NO T +8 H2O3.解：2MnO 7 +SO； + 2OH 2 MnO:+SO+ H2O34.解：(1)( -)Pt,I2(s) I

2、I (ci) II Cl-(C2) I Cl2(P ), Pt ( + ) 2I - 2e - - I2+)1H2O2+ 2H+ + 2e一 2H2O2I -+ H2O2+ 2H+ I2+2H2O1 Cr2O7- + 14H+ + 6e- 2Cr3+ + 7H2O+)3H2S - 2e- - S + 2H +Cr2O 7 7 3H 2s + 8H+ - 2c产 + 3S J + 72O(3) 1 ClO f + 6H+ + 6e一一 Cl- + 3H2O+)6Fe2+-e - Fe3+ClO 3+ 6Fe2+ 6H+ Cl- +6Fe(-)Zn IZnSO4 (ci) | CdSO4 (C2)

3、 I Cd ( + )2 _2_2+_2+ 5.解：由于 E (F2/HF) E (S2O8 /SO4 )E (H2O2/H2O) E (MnO 4 /Mn )E (PbO2/Pb ) E (Cl2/Cl-)E (B42/Br-)E (Ag+/Ag) E (Fe3+/Fe2+) E (I2/I-) 故氧化水平顺序为 F2 H2O2 MnO ；PbO2Cr2O72-Cl2 B2 Fe3+Iz其对应的还原产物为 HF, H2O, Mn2+ , Pb2+ , Cr3+, Cl -, Br 一, Fe2+, I 6.解：由于 E (Zn 2+/Zn) E (H+/H2) E (S/H2S) E (Sn

4、4+/Sn2+) E (SO427H 2SO3) E (Cu2+/Cu) E (I2/I-) E (Fe3+/Fe2+) E (Ag +/Ag) H2H2S SnCl2 Na2SO3 Cu KI FeC Ag KCl.7.解：(1) E (Fe3+/Fe2+) 0,该反响能自发向左进行.(3) ArGm c(Ag +)/c = +0.681VE ( Fe2+/Fe)=-0.44V, E(Ag+/Ag) - E (Fe2+/Fe) E(Cu2+/Cu) - E (Fe2+/Fe) 故Ag+先被Fe粉复原.当 Cu2+要被复原时,需 E(Ag +/Ag) = E(Cu2+/Cu), 这时 E (A

5、g+/Ag)+0.0592 V lg c(Ag +)/c = E (Cu2+/Cu)o即：0.7991 V + 0.0592 V lg c(Ag+)/c = 0.31 V , c(Ag +) = 5.0 1屋 mol L-116 .解：(1) E(Ag +/Ag) = E (Ag +/Ag) + 0.0592 V lg2(Ag+)/c = +0.74 VE(Zn 2+/Zn) = E (Zn2+/Zn) + (0.0592 V/2) lg c(Zn2+)/c = -0.78 VE = E(Ag +/Ag) - E(Zn 2+/Zn) = +1.5 V(2) lgK = z,E (Ag +/Ag

6、) - E (Zn2+/Zn)/0.0592 V , K = 5.76 便52E = E (Ag +/Ag) - E (Zn 2+/Zn) = +1.5617 VArG m=-zFE = -3.014 102 kJmol-1(3)达平衡时,c(Ag+) =x mol L 12Ag+ + Zn - 2Ag + Zn 2+平衡时浓度 c/(mol L-1)x0.30 + (0.10-x)/22 一 c(Zn )/c K 二 c(Ag )/c x = 2.5 10-27, c(Ag+) = 2.5 10?7mol L-117 .解：(1) E (MnO 42-/MnO 2) = 3 E (MnO 4/MnO 2) - E (MnO 4-/MnO 广)/2=+2.27VE (MnO 2/Mn 3+) = 2 E (MnO 2/Mn 2+) - E ( Mn 3+/Mn 2+)/1 = +1.0 V(2) MnO42-, Mn3+.(3)是 Mn2+.反响式为 Mn + 2H + Mn2+ H218 .解：(1) E (Cr2O72-/Cr2+)=0.91V; E (Cr3+/Cr2+)= -0.74