氧化还原反应习题及答案(课堂PPT).ppt

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,*,第五章 氧化还原反应,作 业,参 考 答 案,1,【,第,2,题,(P120)】,用氧化数法或离子电子法配平下列方程式：,(2),K,2,Cr,2,O,7,+H,2,S+H,2,SO,4,K,2,SO,4,+Cr,2,(SO,4,),3,+S+H,2,O,(7)MnO,4,-,+H,2,O,2,+H,+,Mn,2+,+O,2,+H,2,O,【,解法一：氧化数法,】,标出被氧化和被还原的元素原子的氧化数,确定氧化数增加或减少的数值（可采用,“,线桥,”,来表示）,乘以相应系数，使增、减氧化数相等,(2),K,2,Cr,2,O,7,+H,2,S+H,2,SO,4,K,2,SO,4,+Cr,2,(SO,4,),3,+S+H,2,O,2,Cr,2,(SO,4,),3,+,3,K,2,Cr,2,O,7,+,+6,+3,-2,0,H,2,S,+,H,2,SO,4,K,2,SO,4,+,S,氧化数降低,32,氧化数升高,2,3,3,K,2,Cr,2,O,7,+3H,2,S+4H,2,SO,4,K,2,SO,4,+Cr,2,(SO,4,),3,+3S+7H,2,O,+H,2,O,配平反应前后氧化数未发生变化的原子数。,3,(7)MnO,4,-,+H,2,O,2,+H,+,Mn,2+,+O,2,+H,2,O,Mn,2+,+,5,MnO,4,-,+,+7,+2,-1,0,H,2,O,2,+,H,+,O,2,氧化数降低,5,氧化数升高,12,5,5,+H,2,O,2,2,2,标出被氧化和被还原的元素原子的氧化数,确定氧化数增加或减少的数值（可采用,“,线桥,”,来表示）,乘以相应系数，使增、减氧化数相等,4,配平反应前后氧化数未发生变化的原子数。,2MnO,4,-,+5H,2,O,2,+2Mn,2+,+5O,2,6H,+,+8H,2,O,【,解法二：离子电子法,】,(2),K,2,Cr,2,O,7,+H,2,S+H,2,SO,4,K,2,SO,4,+Cr,2,(SO,4,),3,+S+H,2,O,将反应物和产物以离子形式写出,:,Cr,2,O,7,2-,+H,2,S,+,H,+,Cr,3+,+S+H,2,O,将氧化还原反应分为两个半反应，一个发生氧化反应，另一个发生还原反应：,H,2,S,S,Cr,2,O,7,2-,+H,+,Cr,3+,+H,2,O,5,分别配平两个半反应：,H,2,S-2e,-,S+2H,+,Cr,2,O,7,2-,+14H,+,+6e,-,2Cr,3+,+7H,2,O,确定两个半反应得、失电子数的最小公倍数，将两个半反应分别乘以相应系数，使其得、失电子数相等，再将两个半反应合并为一个配平的氧化还原反应的离子方程式，最后改写成化学方程式。,Cr,2,O,7,2-,+,3H,2,S,+8H,+,=2Cr,3+,+3S,+7H,2,O,3,H,2,S-,6,e,-,3,S+,6,H,+,(2,e,-,3,=,6,e,-,),Cr,2,O,7,2-,+14H,+,+6e,-,2Cr,3+,+7H,2,O,(,6e,-,1,=,6,e,-,),+),K,2,Cr,2,O,7,+3H,2,S+4H,2,SO,4,=K,2,SO,4,+Cr,2,(SO,4,),3,+3S+7H,2,O,6,(7)MnO,4,-,+H,2,O,2,+H,+,Mn,2+,+O,2,+H,2,O,将氧化还原反应分为两个半反应，一个发生氧化反应，另一个发生还原反应：,H,2,O,2,O,2,MnO,4,-,+H,+,Mn,2+,+H,2,O,分别配平两个半反应：,H,2,O,2,-2e,-,O,2,+2H,+,MnO,4,-,+8H,+,+5e,-,Mn,2+,+4H,2,O,确定两个半反应的得、失电子数相等，再将两个半反应合并即可。,5,H,2,O,2,-,10,e,-,5,O,2,+,10,H,+,(2,e,-,5,=,10,e,-,),2,MnO,4,-,+,16,H,+,+,10,e,-,2,Mn,2+,+,8,H,2,O,(,5e,-,2,=,10,e,-,),+),2MnO,4,-,+5H,2,O,2,+,6H,+,=,2Mn,2+,+5O,2,+8H,2,O,7,【,第,14,题,(P121)】,对照电极电势表：,(1),选择一种合适的氧化剂，它能使,Sn,2+,变成,Sn,4+,，,Fe,2+,变成,Fe,3+,，但不能使,Cl,-,变成,Cl,2,。,(2),选择一种合适的还原剂，它能使,Cu,2+,变成,Cu,，,Ag,+,变成,Ag,，但不能使,Fe,2+,变成,Fe,。,【,解,】,查表得,P459,附录十四得各电对的电极电势：,电极电对,Sn,4+/,Sn,2+,Fe,3+/,Fe,2+,Cl,2,/,Cl,-,电极电势,/V,0.154,0.771,1.36,电极电对,Cu,2+/,Cu,Ag,+/,Ag,Fe,2+/,Fe,电极电势,/V,0.337,0.799,-0.440,8,(1),所选择的氧化剂，其相应的电极电势必须大于,Sn,4+/,Sn,2+,=,0.154V,和,Fe,3+/,Fe,2+,=,0.771V,，但必须小于,Cl,2,/,Cl,-,=,1.36V,。根据,P459,附录十四可查得：,NO,3,-/,NO,2,=0.80V,，,NO,3,-/,HNO,2,=0.94V,，,NO,3,-/,NO,=0.96V,，,HNO,2,/,NO,=1.00V,，,Br,2,/,Br,-,=1.07V,，,ClO,4,-/,ClO,3,=1.19V,，,O,2,/,H,2,O,=1.229V,，,MnO,2,/,Mn,2+,=1.23V,，等。可见，,HNO,3,、,HNO,2,、,Br,2,、,HClO,4,、,O,2,和,MnO,2,等均能使,Sn,2+,变成,Sn,4+,，,Fe,2+,变成,Fe,3+,，但不能使,Cl,-,变成,Cl,2,。,9,(2),所选择的还原剂，其相应的电极电势必须小于,Cu,2+/,Cu,=,0.337V,和,Ag,+/,Ag,=,0.799V,，但必须大于,Fe,2+/,Fe,=,-0.440V,。根据,P459,附录十四可查得：,SO,4,2-/,H,2,SO,3,=0.17V,，,Sn,4+/,Sn,2+,=,0.154V,，,S,/,H,2,S,=0.141V,，,H,+/,H,2,=0.0000V,，等。可见，,H,2,SO,3,、,Sn,2+,、,H,2,S,和,H,2,等均能使,Cu,2+,变成,Cu,，,Ag,+,变成,Ag,，但不能使,Fe,2+,变成,Fe,。,10,【,第,16,题,(P121)】,利用电极电势表，计算下列反应在,298K,时的,r,G,：,(1)Cl,2,+2Br,-,=2Cl,-,+Br,2,，,(2)I,2,+Sn,2+,=2I,-,+Sn,4+,(3)MnO,2,+4H,+,+2Cl,-,=Mn,2+,+Cl,2,+2H,2,O,【,解,】,查,P459,附录十四得各电对的电极电势：,电极电对,Sn,4+/,Sn,2+,Br,2,/,Br,-,Cl,2,/,Cl,-,电极电势,/V,0.154,1.07,1.36,电极电对,I,2,/,I,-,MnO,2,/,Mn,2+,电极电势,/V,0.5345,1.23,11,(1),Cl,2,+2Br,-,=2Cl,-,+Br,2,正极反应：,Cl,2,+2e,-,=2Cl,-,负极反应：,2Br,-,-2e,-,=Br,2,则原电池的电动势：,E,=,Cl,2,/,Cl,-,-,Br,2,/,Br,-,=1.36-1.07=,0.29V,r,G,=-,zFE,=-2,96500Cmol,-1,0.29V,=-55970J mol,-1,=,-55.97 kJ mol,-1,0,12,(2),I,2,+Sn,2+,=2I,-,+Sn,4+,正极反应：,I,2,+2e,-,=2I,-,负极反应：,Sn,4+,-2e,-,=Sn,4+,则原电池的电动势：,E,=,I,2,/,I,-,-,Sn,4+/,Sn,2+,=0.5345-0.154=,0.3805V,r,G,=-,zFE,=-2,96500Cmol,-1,0.3805V,=-73437J mol,-1,=,-73.44 kJ mol,-1,0,14,【,第,19,题,(P122)】,过量的铁屑置于,0.050mol,L,-1,Cd,2+,溶液中，平衡后,Cd,2+,的浓度是多少,?,【,解,】,P459,附录十四得：,Fe,2+/,Fe,=-0.440V,，,Cd,2+/,Cd,=-0.403V,根据题意，该氧化还原反应为：,Cd,2+,+Fe,Cd+,Fe,2+,其正极反应为：,Cd,2+,+2e,-,Cd,负极反应为：,Fe-2e,-,Fe,2+,则原电池的电动势：,E,=,Cd,2+/,Cd,-,Fe,2+/,Fe,=-0.403V (-0.440V)=,0.037,V,15,l,g,K,=,zE,/0.0592=2,0.037/0.0592=,1.25,K,=17.783,Cd,2+,+Fe,Cd+,Fe,2+,初始浓度,/,mol,L,-1,0.050 0,平衡,浓度,/,mol,L,-1,0.050-,x,x,K,=,Fe,2+,Cd,2+,=,x,0.050-,x,=17.783,解之得：,x,=4.73,10,-2,mol,L,-1,则：平衡后,Cd,2+,的浓度为,0.050-,4.73,10,-2,=,2.7,10,-3,mol,L,-1,16,【,第,21,题,(P122)】,如果下列原电池的电动势为,0.500V(298K),：,Pt,H,2,(100kPa)|H,+,(?molL,-1,)Cu,2+,(1.0 molL,-1,)|Cu,则溶液的,H,+,浓度应是多少,?,【,解,】,P459,附录十四得：,Cu,2+/,Cu,=0.337 V,，,H,+/,H,2,=0.0000V,根据题意，该原电池的电动势：,E,=,Cu,2+/,Cu,-,H,+/,H,2,=0.337V-,H,+/,H,2,=,0.500,V,则：,H,+/,H,2,=,-0.163,V,17,对于电极反应,2H,+,+2 e,-,H,2,，根据能斯特方程有：,H,+,/H,2,=,H,+,/H,2,-,0.0592,2,lg,p,H,2,/,p,c,2,H,+,=,0.0000,-,0.0592,2,lg,1,c,2,H,+,=,-0.163,V,解之得：,c,H,+,=1.76,10,-3,mol,L,-1,即：溶液的,H,+,浓度应是,1.76,10,-3,mol,L,-1,。,18,【,第,23,题,(P122)】,已知,Ag,+/,Ag,=0.799 V,，,K,sp,(AgBr),=7.7,10,-13,。求下列电极反应的,：,AgBr+e,-,Ag+,Br,-,AgBr,Ag,+,+Br,-,，,K,sp,=,Ag,+,Br,-,=,7.7,10,-13,【,解,】,当,Br,-,=,1.0mol,L,-1,时,而,Ag,+,+e,-,Ag,Ag,+,/Ag,0.799V,AgBr/Ag,=,Ag,+,/Ag,-0.0592lg(1/,K,sp,),=0.799-0.0592lg(7.7,10,-13,),-1,=0.082 V,Ag,+,=,K,sp,/,Br,-,=,K,sp,相当于非标准状态时的,Ag,+,/Ag,，此时,c,Ag,+,与,c,Br,有关。,19,【,第,24,题,(P122)】,某根据电极电势解释下列现象：,(1),金属铁能置换,Cu,2+,，而,Fe,3+,溶液又能溶解铜。,(2)H,2,S,溶液久置会变混浊。,(3)H,2,O,2,溶液不稳定，易分解。,(4),分别用,NaNO,3,溶液和稀,H,2,SO,4,溶液均不能把,Fe,2+,氧化，但两者混合后就可将,Fe,2+,氧化。,(5)Ag,不能置换,1,molL,-1,HCl,中的氢，但可置换,1,molL,-1,HI,中的氢。,20,【,解释,】,根据,对角线关系,，将各标准电极电势,按由小到大排列后，电势表左下方的物质能和右上方的物质发生反应。,(1),金属铁能置换,Cu,2+,，而,Fe,3+,溶液又能溶解铜。,反应方向：,Cu,2+,+Fe,Cu+Fe,2+,(,金属铁置换,Cu,2+,),2Fe,3+,+Cu,2,Fe,2+,+Cu,2+,(Fe,3+,溶液溶解铜,),Fe,2+,+2e,-,Fe,Fe,2+,/,Fe,=-0.440V,Cu,2+,+2e,-,Cu,Cu,2+,/,Cu,=0.337V,Fe,3+,+e,-,Fe,2+,Fe,3+,/,Fe,2+,=0.771V,21,【,解释,】,根据,对角线关系,，将各标准电极电势,按由小到大排列后，电势表左下方的物质能和右上方的物质发生反应。,(2),H,2,S,溶液久置会变混浊。,S+2H,+,+2e,-,H,2,S,S,/,H,2,S,=0.141V,O,2,+4H,+,+4e,-,2,H,2,O,O,2,/,H,2,O,=1.229V,反应方向：,O,2,+2,H,2,S,2,H,2,O,+,2,S,（白色沉淀）,因此，,H,2,S,溶液久置会被空气中的,O,2,所氧化，生成白色的,S,沉淀而变混浊。,22,(3),H,2,O,2,溶液不稳定，易分解。,【,解释,】,氧元素的电势图（,酸性溶液,）如下：,1.229,O,2,H,2,O,2,H,2,O,0.682,1.77,由氧元素电势图可知，对于,H,2,O,2,而言，其,右,=1.77V,左,=0.682V,，所以,H,2,O,2,会发生,岐化反应,：,2H,2,O,2,O,2,+2H,2,O,。也就是说，,H,2,O,2,溶液不稳定，易分解成,O,2,和,2H,2,O,。,23,(4),分别用,NaNO,3,溶液和稀,H,2,SO,4,溶液均不能把,Fe,2+,氧化，但两者混合后就可将,Fe,2+,氧化。,【,解释,】,各标准电极电势,如下：,SO,4,2-,+4H,+,+2e,-,H,2,SO,3,+H,2,O,SO,4,2-,/H,2,SO,3,=0.17V,Fe,3+,+e,-,Fe,2+,Fe,3+,/,Fe,2+,=0.771V,NO,3,-,+4H,+,+3e,-,NO,+2H,2,O,NO,3,-,/NO,=0.96V,由此可见，,H,2,SO,4,与,Fe,2+,之间并不符合对角线关系，且二者的电极电势相差较大，因此，稀,H,2,SO,4,溶液不能把,Fe,2+,氧化。,24,此外，,NO,3,-,/NO,=0.96V,Fe,3+,/,Fe,2+,=0.771V,，,即,HNO,3,与,Fe,2+,之间符合对角线关系。虽然如此，在中性条件下,NO,3,-,/NO,=,0.407,V,左,的氧化态有：,ClO,3,、,ClO,2,-,、,ClO,-,和,Cl,2,，,它们均能发生歧化反应,。,(2),28,