第九章 氧化還原反應(yīng)P218

1、第九章 氧化還原反應(yīng) P 218一.(1NH 3+O 2NO 2+H 2O(g N -3+4 7 4 O 0-2 22 4 74NH 3+7O 24 NO 2+6H 2O(g (2Ca 3(PO 42+C +SiO CaSiO 3+P 4+CO P +50 54 20 1 C 02 2 102Ca 3(PO 42+10C +6SiO 26CaSiO 3+P 4+10CO (3Ca(OH2+Cl 2Ca(ClO 32+CaCl 2+H 2O Cl 0+5 25 10 1 Cl 0-1 21 2 5 6Ca(OH2+6Cl 2 Ca(ClO 32+5CaCl 2+6H 2O(4P +CuSO 4+

2、H 2O Cu 3P +H 3PO 4+H 2SO 4P 0-3 3 6 5 Cu +2+1 31 3P 0+5 5 5 6 11P +15CuSO 4+24H 2O 5Cu 3P +6H 3PO 4+15H 2SO 4(5AgN 3+H 2S Ag 2S +S +(NH 42S +N 2(1N -13 0 2 13 23 83 2S -20 2 N -13 -3 283 163 12AgN 3+4H 2S Ag 2S +2S +(NH 42S +2N 2 (2 N -13 0 (0+13 23=2 4321765-13 -3 (-3+1323=-16S -20 (0+2=2 4567123

3、10AgN 3+12H 2S 5Ag 2S +4S +3(NH 42S +12N 2 8AgN 3+12H 2S 4Ag 2S +5S +3(NH 42S +9N 2 6AgN 3+12H 2S 3Ag 2S +6S +3(NH 42S +6N 2 4AgN 3+12H 2S 2Ag 2S +7S +3(NH 42S +3N 2 16AgN 3+12H 2S 8Ag 2S +S +3(NH 42S +21N 214AgN 3+12H 2S 7Ag 2S +2S +3(NH 42S +18N 2 12AgN 3+12H 2S 6Ag 2S +3S +3(NH 42S +15N 2 4AgN 3+

4、4H 2S 2Ag 2S +S +(NH 42S +5N 222AgN 3+36H 2S 11Ag 2S +16S +9(NH 42S +24N 2 6AgN 3+32H 2S 3Ag 2S +21S +8(NH 42S +N 2 10AgN 3+28H 2S 5Ag 2S +16S +7(NH 42S +8N 2(6FeS 2+O 2Fe 3O 4+SO 2Fe +2+83 (83-23=2 32 1S -1+4 (4+123=30O 0-2 (-2-02=-4 83FeS 2+8O 2 Fe 3O 4+6SO 2二.(1MnO 4-+H 2O 2Mn 2+O 2(酸性MnO 4-+8H +

5、5e Mn 2+4H 2O 2H 2O 22H +O 2+2e 52MnO 4-+16H +5H 2O 22Mn 2+10H +8H 2O +5O 22MnO 4-+6H +5H 2O 22Mn 2+8H 2O +5O 2(2Zn +NO 3-Zn(OH42-+NH 3(堿性Zn +4OH - Zn(OH42-+2e 4NO 3-+6H 2O +8e NH 3+9OH -14Zn +7OH -+ NO 3-+6H 2O 4Zn(OH42-+NH 3 (3MnO 2+O 2MnO 42-+H 2O (堿性 MnO 2+4OH -MnO 42-+2H 2O +2e 22H 2O +O 2+4e4O

6、H - 12MnO 2+4OH -+O 22MnO 42-+2H 2O(4 MnO 42-MnO 4-+MnO 2 (中性MnO 42- MnO 4-+e 2MnO 42-+2H 2O +2e MnO 2+4OH -13MnO 42-+2H 2O 2MnO 4-+MnO 2+4OH - (5Cu 2S +NO 3-Cu 2+NO +S 8 (酸性8Cu 2S 16Cu 2+S 8+32e 3 4H +NO 3-+3e NO +2H 2O 3224Cu 2S +32NO 3-+128H +48Cu 2+32NO +3S 8+64H 2O3.(1F 2Ag +Cu +Cu 2+Na +K +(2C

7、u 2+e -Cu +(3F 2, Ag +4. (1Zn +Cl 22+2Cl -(22AgCl +Cd 2+2Ag +2Cl -(3Pb +SO 42-+2H PbSO 4+H 2(42Fe 2+Cl 23+2Cl -5.(1 (-Pt,Cl 2(101.3kPa Cl -(1mol L -1MnO 4-(1mol L -1Mn 2+(1mol L -1,H +(6mol L -1Pt(+242/ 1.51 1.360.15MnO Mn Cl ClE V -+-+-=-=-=-=24/ 1.584 1.360.224MnO Mn E V -+-=-=-=10965000.224215.8/

8、G nFE kJ mol =-=-=- 0E 0G 反應(yīng)正向進(jìn)行10n =(2 (-Cu Cu 2+(0.1mol L -1H +(0.01mol L -1H 2(91.2kPa,Pt (+CuV += 2/0.00H HV += 0.34E V =-Cun Cu +=-=-=(222H H HP P H +=-=-=-22/0.4270H H Cu Cu E +=-=- 反應(yīng)逆向進(jìn)行(2965000.42782.4/G nFE kJ mol =-=-= 2n =6.解:(1G nFE =- 40/G k J m o l 3E V -= 不受濃度左右即0.2070.207V E V - (2P

9、b(s+2H +SO 4PbSO 4(s+H 242/0.00(0.3560.356PbSO PbH H E V lg 05E E =- 2lg 0.35612.04740.05910.0591n E =121.115310MnMn VMnO H -+-+=-=(2122211.11531042H P P xx xH SO +-=56.0710/x mol L -當(dāng)24H SO 濃度大于56.0710/mol L -時(shí),反應(yīng)正向7.解:(-Pb(s+SO 42- PbSO 4+2e(+PbO 2+SO 42-+4H +2e PbSO 4+2H 2O24/40.05911lg 0.3562(Pb

10、PbPbSOPbsp V k PbSO +-=-=-24PbSO V +=E +-=-=1.658(0.356 2.014V -=6921.1710n K E K = (-Pb,PbSO 4SO 42-SO 42-,H +PbSO 4, PbO 2 (+8.解:2Na+2H 2O 2Na +2OH -+H 2(-極:Na + Na / 2.713Na Na V +=- (+極: 2H 2 H 2+2OH - 22/0.8277H O H V =-0.8277( 2.713 1.8853E V+-=-=-=6326.3110n K E K = 63ln 8.314298ln(6.3110363.

11、97/G RT K kJ mol =-=-=-9.解:(+Cr 2O 72-+14H +6e -2Cr 3+7H 2O (-AgCl(s+e -Ag(s+Cl -23142270.05910.05910.12lg 1.33lg 1.32660.001Cr V H Cr O +-=-=-=10Agsp V k AgCl +-=-=1.320.22 1.10E V +-=-=-=1.330.22 1.11E V +-=-=-=11264.910n K E K = 10.解:MnO 2+4H +2Cl -Mn 2+2H 2O+Cl 2(g222/ 1.23 1.360.13MnO Mn ClClE

12、V +-=-=-=-523.9910n K E K -=-=-= 設(shè)平衡時(shí): H +=xmo l L -1MnO 2+4H +2Cl -Mn 2+2H 2O+Cl 2平衡 x x+2 1(225424224/1K x x H c Cl c x x +-+-=+=2(816(2158.3x x x -+-=11.解:以22(/A O H O 2.052E V =-=12.解:+ AgCl(s+e Ag(s+Cl - -2HA+2e H 2+2A -*0.568E V +-=-10AgAgsp V k AgCl +-=-=-= 0.348-=-j- = j H q + / H2 0.0591 PH

13、 2 / P - lg 2 2 H + q -0.348 = 0.00 - 2 0.0591 1 lg 2 2 H + H + = 1 . 2 160 9 - H + = 1.672 10 -12 2 H + 1.672 10-12 Ka = = = 3.345 10-12 HA 0.5 13.解：(-Pt,H2H2O,OH H2O ,H H2,Pt(+ +: 2H +2e -: H2O+2e + + H2 H2+2OH - q j+ = -0.00V q j- = -0.8227V + - Eq = 0 . 8 2 2 7 V 2H +2OH 2H2O lg K q = 2 0.8227 =

14、 27.8409 0.0591 K q = 6.93 1027 Kw = 1 1 = = 1.2 10-14 Kq 6.93 1027 Os+ 2OsCl63- 14.（1）Os2+歧化 3 Os2+12Cl- Eq = 0.85 - 0.4 = 0.45V lg K q = 2 0.45 = 15.2284 0.0591 K q = 1.69 1015 （2） jOsO4 / Os = q 1.0 4 + 0.85 + 0.4 + 0.85 2 = 0.869V 4 +1+1+ 2 （3）在 HCl 中，OsO4Os 反應(yīng)可以生成 OsCl62- ，OsCl63Os2+歧化，不會(huì)生成 Os2

15、+ 但由于 Os 過(guò)量，Os 與生成的 OsCl62-繼續(xù)反應(yīng)直至均生成 OsCl6315.解：對(duì)于反應(yīng) H3AsO4+2I-+2H+=H3AsO3+I2+H2O q E = j+ - j- = (j H3 AsO4 / H3 AsO3 - H3 AsO3 - jq 0 0.0591 lg 2 I2 / I - 2 H3 AsO4 H + 0.56 - 0.0591 1 lg - 0.535 0 2 2 1 H + 反應(yīng)正向進(jìn)行 反應(yīng)逆向進(jìn)行 PH 0.56 - 0.535 = 0.423 0.0591 PH 0.423 六解：查表知 ksp (Cu (OH 2 = 2.2 10-20 ksp

16、 (CuOH = 1.0 10-14 Cu+在酸性介質(zhì)中不能存在 q jCu + / Cu q jCu 2+ / Cu 歧化反應(yīng) 2Cu+ Cu2+ Cu q q Eq = jCu + / Cu - jCu 2+ / Cu = 0.521 - 0.152 = 0.369V lg K q = n 1 Eq = 0.369 = 6.2437 0.0591 0.0591 K q = 1.75 106 4Cu3+2H2O q jCu 3+ 4Cu2+4H+O2 q jO / H O = 1.23V 2 2 / Cu 2+ q - jCu 2+ / Cu = 1.80V Eq = 1.8 - 1.23 = 0.57 lg K q = 4 0.57 = 38.579 0.0591 K q = 3.79 1038 (2 Cu (OH 2 +e- CuOH + OH - q j1q = jCu (OH 2 / Cu ( OH Cu (OH 2 Cu 2+2 OH - k spq (Cu (OH 2 = Cu 2 + OH - 2 CuOH Cu 2+eq jCu (OH Cu + OH Cu + q = jCu 2+ / Cu + - k spq (CuOH = Cu + OH -