1、8 氧化还原反应习题解答 (p222-226)1. 用氧化值法配平下列各氧化还原方程式。(1)3Cu 2S +22HNO3 = 6Cu(NO3)2 +3H2SO4 +10NO+8H2O(2)NH 4NO2 = N2 + 2H2O(3)(NH 4)Cr2O7 = N2 + Cr2O3 +4H2O(4)3As 2S3 + 28HNO3+4 H2O = 6 H3AsO4 + 9H2SO4 +28 NO(5)K 2Cr2O7 + 3H2S +4 H2SO4 = Cr2(SO4)3 + 3S + 7H2O + K2SO4(6)2Na 2S2O3 + I2 = Na2S4O6 + 2NaI(7)2 MnO
2、 4 +3 Mn2+ +2 H2O =5 MnO2 +4 H+(8)4Co(NH 3)62+ + O2 +2H2O = 4 Co(NH3)63+ + 4OH-2. 用离子-电子法配平下列方程式。 (1-11 )为酸性介质, (12-14)为碱性介质(1) Cr2O72 + 3H2O2 + 8H+ = 2 Cr3+ + 3O2 +7 H2O(2) 2 MnO4 +5 SO32 +6 H+ = 2Mn2+ + 5 SO42 +3 H2O(3) ClO3 + 6I- + 6H+ = Cl- +3 I2 + 3H2O(4) 5NaBiO3(s) + 2Mn2+ +14 H+ = 5Na+ +5 Bi3
3、+ + 2 MnO4 +7 H2O(5) H2S +2 Fe3+ = S +2Fe2+ + 2H+(6) 3P4(s)+20HNO3(浓)+8 H 2O =12H3PO4+20NO(7) 2FeS2+30HNO3 Fe2(SO4)3+30NO2+H2SO4+14H2O(8) 7PbO2+2MnBr2+14HNO3 7Pb(NO3)2+2Br2+2HMnO4+6H2O(9) 28HNO3+3As2S3+4H2O 9H2SO4+6H3AsO4+28NO(10) As2S5+10NO3-+10H+ 2H3AsO4+10NO2+5S+2H2O(11) 2Bi3+3S2O32-+3H2O Bi2S3+3
4、SO42-+6H+(12) Cl2 + 2OH-= Cl- + ClO-+ H2O(13) 2Cr(OH)4- + 3H2O2 + 2OH-= 2 CrO42 +8H2O(14) SO32 + Cl2 +2OH-=2 Cl- + SO42 + H2O3. (1)逆向进行(2) )(0=/Sn+24VE15./Mg- 37)(2.-+ V5正向进行(3) )1.09(=/Br+-2VE7/Fe-+3 )(32.-正向进行(4) )(=/Pb+2VE13.0n/Z- 76)(-.=-+ V63.0正向进行4. (1)nE5.10.)(lg592.01)M/O( 4824 (2) V677Ce3(3
5、)VcKHA 17.05.10lg92.l.)/H()Ac/(22 (4)gEClsp 34.850g159gl 2 (5) VccE 40.185.0)1.(0lg4592.80 )/(42O/S)O/S(62 6OHOS22332 23(6)Vcp564.0)1.(/lg592.)/(lg9/H)/( 34OH22 2 5. 解答:(1)2Ag +(0.10 molL-1) + Cu(s) =2Ag(s) + Cu2+(0.010 molL-1)电池符号: (-) Cu|Cu2+(0.010 molL-1)|Ag+(0.10 molL-1)|Ag (+)电极反应:(+) Ag+e = Ag
6、(s)(-) Cu-2e = Cu2+电动势 VcgEAAg 740.1.lg0592.7l10592./)()/( CuCu 283222 V46.8.)/()/((2)MnO 2(s)+ 2Cl-(12 molL-1) + 4H+(12 molL-1) = Mn2+(1.0 molL-1) + Cl2(100kPa) + 2H2O(l)电池符号:(-) Pt, Cl2(100kPa)|Cl-(12.0 molL-1)|Mn2+(1.0molL-1),H+(12.0 molL-1)|MnO2(s),Pt(+)电极反应:(+) MnO2+ 4H+2e= Mn2+ 2H2O (-) 2Cl2e=
7、 Cl2电动势 VcEMn 36.102lg59.31/)(lg2059.)/nO()/Mn( 44H22 2 cpCl .l0.6)/(.)/l()/Cl( 22l22 V06.31.(3)H 3AsO3(0.10 molL-1)+ I2(s) + H2O(l) = H3AsO4(1.0 molL-1) + 2I-(0.010 molL-1) +2H+(0.10 molL-1)电池符号:(-) Pt|H3AsO4 (1.0 molL-1), H+(0.10 molL-1), H3AsO3(0.10 molL-1)|I-(0.010molL-1)|I2(s), Pt (+)电极反应:(+) I
8、2+2e =2I-(-) H3AsO3+ H2O-2e = H3AsO4+2H+电动势 3432HAsO43432(/)(/0.59(AsO/)(AsO/)lg0.5921.( .lg.ccEEV 222 2I0.591(I/)(I/)lg60654(/).)c E = 0.654-0.529=0.125V(4)Cr 2O72 (1.0 molL-1) + 6Fe2+(0.10 molL-1) +14H+(1.0 molL-1) = 2Cr3+(0.10 molL-1) + 6Fe3+(1.0 molL-1) + 7H2O(l)电池符号:(-)Pt|Fe3+(1.0 molL-1), Fe2+
9、(0.10 molL-1)|Cr2O72 (1.0molL-1), Cr3+(0.10molL-1), H+(1.0 molL-1)|Pt (+)电极反应:(+) Cr2O72 +14H+6e = 2Cr3+ 7H2O(-) Fe2+-e = Fe3+电动势 27314CrOH232377 2(/)(/0.59(Cr/)(r/)lg6ccE 1420.59.1lg.60V32323Fe/1.0(Fe/)(e/F)l0.759lg83cVE=1.35-0.830=0.52V6. 解答:(1) E=E(+)-E(-) cc/lg2059.4/lg2059.)/Fe()/eF() 22 FeFe22
10、 VcEE 82.01.lg2059.763/lg2059.)/Zn()/Zn() Zn22 2 0.293= -0.440+(0.0592/2)lgc(Fe2+)+0.822 c(Fe2+)=1.010-3(molL-1)(2)( -)AgAg +(y molL-1)Ag +(1.010 -1 molL-1)Ag (+) E=0.0592V,求负极Ag+的浓度。 VgEE 740.1/.lg0592.71./lg10592.)/A()/) A A( ccc(Ag+)=0.10(molL-1)y.0.592l.(3) (-)Pt,Cl2(P )Cl -(x molL-1) (1.0 molL-
11、1), H+(xmolL-1), Cr3+(1.0 molL-1)Pt 27CrO(+) 27314CrO2323 1477 2/)/0.590.592()CrO/)(r/)lg3lg()6(6ccEE x 2Cl22. .ll 1l(/)pcx欲使反应按原电池的方向进行,则应由:, x1.42,即 HCl 浓度至少为 1.42molL-11420.590.5913lg()36lg6()x7. 解答: Ag+e =Ag=-nF E (Ag+/Ag)=-77.124 kJ.mol-1 mfrGE (Ag+/Ag)=-77.124103/96487=0.7992V8.解答: 222 sp,PbI0
12、.59(PbI/)(/Pb)lgK-0.3672=-0.126 I,=7.110-92sp,PbI(.367.1)lg8.K2sp,PbI9. 解答:E (Ag3PO4/Ag)= E (Ag+/Ag)+0.0592lg(K sp,Ag3PO4)1/3 =0.7991+ 0.0592lg 31604.=0.486V10. 解答:pH=3.00 的 H2S 饱和溶液中,C H+=1.010-3H2S = 2H+ + S2- K= Ka1 Ka2 = H+2S2-/ H2S S2- = Ka1 Ka2 H2S/ H+2 115318 mol.L03.)0.(.15.9 (或用分布系数 )212SH2
13、2SaacV201.235.10lg9.7.0 Slg10592.)/A()mol.L.)(/A)S/(1491522 spKgESE11. 解答:E (HCN/H2)= -0.545V=0.0592lg K (HCN) ,K (HCN)=6.210-1012. 解答:Ka1Ka2, 所以二级解离可以忽略H+=7.110-33431 10.7POHaKE (H3PO4/H2)= E(H+/H2)= E (H+/H2)+0.0592lgCH+=0.0592lg(7.110-3)=-0.127V13. 解答:(1)电极反应式原电池符号: (-)Zn|Zn2+(0.30molL-1)|Ag+(0.1
14、0molL-1)|Ag (+)(+) Ag+e =Ag(-) Zn-2e =Zn2+(2) V74.01lg592.80lg0592.)/A()/g( A cE8376ZnZn2Zn2 该原电池的电动势 E=0.74-(-0.78)=1.52V(3)反应 2Ag+ + Zn2Ag + Zn 2+的平衡常数 52)()( 10.5092.)80592.lg KK(4)达平衡时,溶液中所剩 Ag+的浓度 12722 Lmol6.Ag3AgZn1. 14. 解答:半反应Cu2+ Cl- +e = CuClCu + Cl- -e = CuCl V50.12.lg059.1lg0592.)/Cu()/u
15、lC( 6Culsp,22 KE 7.l,EK =3.81058.)17(059.)(lg() ZK15. 解答: V4.01.lg092.lg0592.)/A()/( 6A cgE 3534Cuu2Cu22lgK = =19.6, K =3.64 1019 ZE0.0592 设放电终止时, ,则2+/2+=E(+) =E(-)E (Pb2+/Pb) = E (Cu2+/Cu)+0.059222+ +0.059222+代入数据可得, =2.29101619. 解答:(1) E1 =?+:2+4+4+=22 E2 =0.401V-:2+4+22=44H2O E3 =0.0592lg 4+4=1(
16、 ) 4 可得, 4 E1 -4 E2 = E3 所以, E1 = E2 - 0.25E3 =1.23V(2)2 +2e=2 H2+2-/ / + =E(2H2)=E(+H2)0.05922( ) 20.829(3)酸性溶液中,电池反应:(- ) H22=2+(+) 2+4+4+=22= / =1.23VEE1( +H2)碱性溶液中,电池反应:(-)2 +2e=2 H2+2-(+) 2+4+22=4/ =0.401-(-0.829)=1.23VE=E2E(2H2)(4)如果改变 PH,其电动势不变。先假设在酸性条件下形成电池/ / )E=E(22)+0.05924 lg+4( +H2 +0.0
17、5922 lg+2= / / )E(22) ( +H2由此可见,电池的电动势为定值,与 PH 无关。碱性条件下也如此。20. 解答:(1)因为 ,所以反应向左进行。(3+/2+)(2/)(2) 2+2()2+2=2+2()3E( )=1.84+0.0592lg 0.151V()3/()231.610441.61015=所以,可以制取。(s)2+2()2+2=2+2()321. 解答: 24+32+22=52+4+Ecell=+0.05926(4)2(2+)3(+)4 =1.701.23+0.059261.0106(2+)311020当 Ecell=0 时, 不能再被还原。2+=2.852+ 1021因为 很小,所以 已经除尽。2+浓 度 2+22. 解答:(1)In + E(右) E(左) 能发生歧化反应Tl+ E(右)E左35.解答:Ecell=(+/)(4/)=0.220-0.546= -0.326V (4/)= + (4/) (2+/)0.05922(24)0.326= 0.13+ ,得 ( )=1.19 0.059220.054 108