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,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,*,氧化还原反应与电化学,3.用离子-电子法配平下列半反应,(1)NO,3,-,NO,2,-,(,酸性介质),(2)Br,2,BrO,3,-,(,碱性介质),(3)H,2,O,2,H,2,O,(,酸性介质),(4)H,2,O,2,O,2,(,碱性介质),(5)CrO,4,2-,CrO,2,-,(,碱性介质),解：(1),NO,3,-,+2H,+,+2e,-,=,NO,2,-,+H,2,O,(2)Br,2,+12OH,-,=2BrO,3,-,+6H,2,O+10e,-,(3)H,2,O,2,+2H,+,+2e,-,=,2H,2,O,(4)H,2,O,2,+2OH,-,=,O,2,+2H,2,O+2e,-,(5)CrO,4,2-,+2H,2,O+3e,-,=,CrO,2,-,+4OH,-,1,4,.,用离子-电子法配平下列反应式：,解：,(1),5,),Fe,2+,=Fe,3+,+,e,+,1,),MnO,4,-,+8H,+,+,5,e,=Mn,2+,+4H,2,O,得,：,5,Fe,2+,+,MnO,4,+8H,+,=,Mn,2+,+,5Fe,3+,+4H,2,O,(2),2,),Cr,3+,+8OH,=CrO,4,2-,+4H,2,O,+,3e,+3,),H,2,O,2,+,2e,=2OH,得：,2,Cr,3+,+,3,H,2,O,2,+10OH,-,=2CrO,4,2-,+8H,2,O,(3),5,),H,2,O,2,=O,2,+2H,+,+,2e,+2,),MnO,4,-,+8H,+,+,5,e,=Mn,2+,+4H,2,O,得,：,2,MnO,4,+,5,H,2,O,2,+6 H,+,=2,Mn,2+,+,5O,2,+8H,2,O,(4),4,),Mg=Mg,2+,+,2e,+,1,),2NO,3,-,+10H,+,+,8,e,=N,2,O+5H,2,O,得：,2NO,3,+4,Mg,+10H,+,=N,2,O+4,Mg,2+,+5H,2,O,2,5,.将下列氧化还原反应组成原电池，写出原电池符号和两极反应。,(1),Pb,+2Fe,2+,=2Fe,3+,+Pb,2+,(2),Ni,+,Sn,4+,=,Ni,2+,+Sn,2+,解：(1)负极：,Pb,=Pb,2+,+2e,-,正极：,Fe,3+,+e,-,=Fe,2+,原电池符号：,(-),Pb,|Pb,2+,(c,1,)|Fe,3+,(c,2,),Fe,2+,(c,3,)|,Pt,(+),(2)负极：,Ni,=,Ni,2+,+2e,-,正极：,Sn,4+,+2e,-,=Sn,2+,原电池符号：,(-),Ni,|,Ni,2+,(c,1,)|Sn,2+,(c,2,),Sn,4+,(c,3,)|,Pt,(+),3,6,.写出下列各电池的化学反应式，并计算298.15,K,时电池的电动势。,(1)(-),Fe,|Fe,2+,(0.001mol,dm,-3,)|,Ni,2+,(0.1mol,dm,-3,)|,Ni,(+),(2)(-),Pb,|Pb,2+,(1mol,dm,-3,)|,Ag,+,(1mol,dm,-3,)|,Ag,(+),4,6.解：(1),Fe+Ni,2+,=Fe,2+,+Ni,+,=,(,Ni,2+,/,Ni,)+(0.059/2),lg0.1,=-0.23-0.0295=-0.26V,-,=,(,Fe,2+,/,Fe,)+(0.059/2),lg0.001,=-0.4402-0.0885,=-0.53V,E=,+,-,-,=,-0.26-(-0.53)=0.27V,(2),Pb,+2Ag,+,=Pb,2+,+2Ag,E=,+,-,-,=,(,Ag,+,/,Ag,)-,(,Pb,2+,/,Pb,),=0.7996-(-0.1263)=0.9255V,5,7.将下列反应组成原电池,Sn,2+,+2Fe,2+,=Sn,4+,+2Fe,2+,(1),写出原电池符号。,(2),计算在298.15,K,标准态时原电池的电动势和反应的吉布斯函数变。,(3),若改变,Sn,2+,浓度，使其,c(,Sn,2+,)=1.0,10,-3,mol,dm,-3,，,则原电池的电动势为多少？,(4),该原电池在使用若干时间后，电池的电动势和两极的电极电势如何变化？,6,7.解：,(1),原电池符号：,(-),Pt,|,Sn,2+,(c,1,),Sn,4+,(c,2,),|,Fe,2+,(c,3,),Fe,3+,(c,4,),|,Pt,(+),(2),E,=,(,Fe,3+,/Fe,2+,)-,(,Sn,4+,/Sn,2+,),=,0.770-0.15=0.62V,G,=,-,nFE,=,-2,964850.62,=,-119641 Jmol,-1,=-119.641 kJmol,-1,(3),(-)=,(,Fe,3+,/Fe,2+,)+(0.059/2)lg(1/1.0,10,-3,),=0.15+0.0885=0.238,V,E=,(+)-,(-),=0.770-0.238=0.532V,(4),使用后：,c(Fe,3+,),，,c(Fe,2+,),，,(+)；,c(Sn,2+,),，,c(Sn,4+,),，,(-)；,E,减小。,使用若干时间后,E,减小直至,E,=0。,7,8.,下列电池反应中，当,Cu,2+,浓度为何值时，该原电池的电动势为零？,Ni+Cu,2+,(x,moldm,-3,)=Ni,2+,(1.0,moldm,-3,)+Cu,9,.,由镍电极和标准氢电极组成原电池。,若,c(,Ni,2+,)=0.01,mol,dm,-3,时，原电池的电动势为0.288,V，,其中,Ni,为负极。试计算镍电极的标准电极电势。,8,8,.,解:,(+)=,(,Cu,2+,/Cu,)+(0.059/2)lgc(,Cu,2+,),=,0.3402+(0.059/2)lgc(,Cu,2+,),(-)=,(,Ni,2+,/Ni,)=-0.23V,由,E=,(+)-,(-)=0,得：,(+)=,(-)，,即,0.3402+(0.059/2)lgc(,Cu,2+,)=-0.23,c(,Cu,2+,)=4.69,10,-20,moldm,-3,9.,解:,E=,(+)-,(-)=,(,H,+,/H,2,),-,(,Ni,2+,/Ni,),=,(,Ni,2+,/Ni,),=,(,Ni,2+,/Ni,)-(0.059/2)lg0.01,=0.288,(,Ni,2+,/Ni,)=,0.229V,9,10.,由两个氢电极,H,2,(1.01325,10,5,Pa,),|H,+,(0.1 mol,dm,-3,),和,H,2,(1.01325,10,5,Pa,),|H,+,(x mol,dm,-3,),组成原电池，测得该原电池的电动势为0.016,V。,若电极,H,2,(1.01325,10,5,Pa,),|H,+,(x mol,dm,-3,),作为该原电池的正极，问组成该电极的,H,+,浓度为多少？,11.,今有一种含有,Cl,-,、Br,-,、I,-,三种离子的混合溶液，欲使,I,-,氧化为,I,2,而又不使,Br,-,、,Cl,-,氧化，在常用的氧化剂,Fe,2,(SO,4,),3,和,KMnO,4,中，选用哪一种能符合上述要求？,10,10.解:,E=,(+)-,(-),=(0.059/2)lgx,2,/1-(0.059/2)lg0.1,2,/1,=0.059(,lgx,+1),=0.016,lgx,=-0.729,即:,x=0.19,mol,dm,-3,11.解:,已知：,(I,2,/I,-,)=+0.535V,(Br,2,/Br,-,)=+1.065V,(Cl,2,/Cl,-,)=+1.3583V,(,Fe,3+,/Fe,)=+0.770V,(,MnO,4,-,/Mn,2+,)=+1.491V,即:,(,I,2,/I,-,),(,Fe,3+,/Fe,),(Br,2,/Br,-,),(,Cl,2,/Cl,-,),(,MnO,4,-,/Mn,2+,),选择,Fe,2,(SO,4,),3,合适。,11,12.,通过实验得到如下现象,(1)在水溶液中，,KI,能与,FeCl,3,反应生成,I,2,和,FeCl,2,，,而,KBr,则不能与,FeCl,3,反应；,(2)溴水能与,FeSO,4,溶液作用生成,Fe,3+,和,Br,-,离子，而碘则不能与,FeSO,4,反应。,根据以上现象定性地推断,Br,2,/Br,-,、,I,2,/I,-,和,Fe,3+,/Fe,三电对电极电势的大小。,解:,(,I,2,/I,-,)(,Fe,3+,/Fe,)0,即,Fe,3+,为氧化剂，,Cu,为还原剂,所以反应正向进行。,(,Cu+2Fe,3+,Cu,2+,+2Fe,2+,),17,(3),(,Sn,4+,/Sn,2+,)=+0.15V,(,Fe,3+,/Fe,2+,)=+0.770V,E,=,(,Sn,4+,/Sn,2+,),-,(,Fe,3+,/Fe,2+,),=0.15-0.770=-0.62V0,即,Fe,3+,为氧化剂，,Sn,2+,为还原剂,所以反应逆向进行。(,Sn,4+,+2Fe,2+,Sn,2+,+2Fe,3+,),(2),(,Fe,3+,/Fe,2+,)=+0.770V,(,Cr,2,O,7,2-,/Cr,3+,)=+1.33V,E,=,(,Fe,3+,/Fe,2+,)-,(,Cr,2,O,7,2-,/Cr,3+,),=0.770-1.33=-0.56V0,即,Cr,2,O,7,2-,为氧化剂，,Fe,2+,为还原剂,所以反应逆向进行。,即：,6,Fe,3+,+2Cr,3+,+7H,2,O,Cr,2,O,7,2-,+6Fe,2+,+14H,+,18,17.,试求下列电池反应在298.15,K,时的,G,值。说明反应是否能从左至右自发进行。,(1)1/2,Cu+1/2Cl,2,(101325 Pa)=1/2Cu,2+,(,1mol,dm,-3,)+,Cl,-,(,1mol,dm,-3,),(2),Cu+2H,+,(0.0,1mol,dm,-3,)=,Cu,2+,(0.,1mol,dm,-3,)+,H,2,(0.9,101325 Pa),19,17.解:(1),+,=,(,Cl,2,/,Cl,-,),=,+1.3583V,=,(,Cu,2+,/Cu,),=,+0.3402V,(,附录数据，错误；正确应为,0.337,V,),G,=-,nFE,=-1,96485(,1.3583-,0.3402,),10,-3,=-98.23 kJ,mol,-1,0,反应不能自发进行。,20,17.解:(1),+,=,(,Cl,2,/,Cl,-,),=,+1.3583V,=,(,Cu,2+,/Cu,),=,+,0.337,V,G,=-,nFE,=-1,96485(,1.3583-,0.337,),10,-3,=-98.54 kJ,mol,-1,0,反应不能自发进行。,21,18.,在,298.15,K,时，由,Co,|Co,2+,(1mol,dm,-3,),和,Cl,2,(101325Pa)|,Cl,-,(,1mol,dm,-3,),组成原电池，测得其电动势为1.63,V，,此时钴电极为负极。现已知氯电极的标准电极电势为1.36,V，,问：,(1),此电池反应的方向如何？,(2),钴电极的标准电极电势为多少？,(3)当,氯气的压力增大或减小时，原电池的电动势将如何变化？,(4),当,Co,2+,浓度降低时，原电池的电动势将如何变化？,22,18.解:(1),Co+Cl,2,(g)=Co,2+,+2Cl,(2),E,=,+,-,=,(,Cl,2,/,Cl,)-,(,Co,2+,/,Co,),(,Co,2+,/,Co,)=,(,Cl,2,/,Cl,)-E,=1.36-1.63=-0.27V,(3),Cl,2,+2e,
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