物化打印版测验题及答案课件

Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,11/7/2009,#,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,*,电化学测验题答案,1,电化学测验题答案1,自测题,填空和选择填空,1.某一电池反应,r,S,m,(298K)0，,则25原电池可逆工作时是吸热还是放热？_因为_,2.用同一电导池分别测定浓度为0.01和0.1,molkg,-1,的两个电解质溶液，其电阻分别为1000,500,。,则,m,之比为（）,(,A)1:5 (B)5:1 (C)10:5 (D)5:10,吸热,Q,r,=,T,r,S,m,B,2.解,2,自测题填空和选择填空吸热Qr=TrSmB2.解2,3.下列电解质溶液中(0.01,molkg,-1,)，,最大的是(),(,A)NaCl (B)CaCl,2,(C)LaCl,3,(D)CuSO,4,4.,对于同一电解质水溶液，当其浓度逐渐增加时，何种性质将随之增加,(),(,A),稀溶液范围内的,(,B),m,(C),(D),K,cell,5.,某一电解质,M,+,A,-,，,则其,a,与,a,之间关系是(),(,A),a,=a (B),a,=,a,2,(C),a,=,a,(D),a,=,a,1/,A,A,D,3.解：根据,ln,=-Az,+,z,-,I,1/2,右式中(,NaCl),的数最小。,3,3.下列电解质溶液中(0.01molkg-1)，最大,6.1,mol kg,-1,K,4,Fe(CN),6,溶液的离子强度,(),(,A)10 (B)7 (C)4 (D)15 mol kg,-1,7.,将两铂丝插入,m,(Sn,2+,)=0.2,m,(Sn,4+,)=0.02 mol kg,-1,的溶液中构成电池，则,E,=(),(A),E,y,+0.059/2 (B),E,y,-0.059/2 (C)0 (D),E,y,+0.059,8.,将,Ag(s),Ag-Au(,合金,a,Ag,=0.120),设计成电池：,_,则该电池在25,时的电动势,E,=_,A,C,Ag(s)|Ag,+,|Ag-Au(s),-0.0591,lg,a,=0.054V,6.解：,I,=(4,1,2,+1,4,2,)/2=10 mol kg,-1,4,6.1mol kg-1 K4Fe(CN)6溶液的离子强度,9.按书写习惯，下列电池,E,=()V,Pt|H,2,(,p,y,)|HI(,m,1,=0.01molkg,-1,)|AgI|Ag-Ag|AgI|HI(,m,2,=0.001molkg,-1,)|H,2,(,p,y,)|Pt,(A)0.118 (B)-0.059 (C)0.059 (D)-0.118,10.,已知,Cu,2+,+2e,Cu,1,y,=0.337V,Cu,+,+e,Cu,2,y,=0.521V,求,Cu,2+,+e,Cu,+,的,3,y,=()V,-0.118,0.153,9.解：,电池反应：,HI(,m,2,),HI(,m,1,),E,=2,0.059lg(0.001/0.01)=0.118V,10.解：,F,3,y,=2,F,1,y,F,2,y,=0.153V,5,9.按书写习惯，下列电池E=()V,11.,电池,Hg(,l,)-Hg,2,Cl,2,(s)|HCl(,a,)|Cl,2,(,p,y,)-Pt,在25,a=,0.1,时的电动势,E,=1.135V,a,=0.01,时的电动势,E,=_V,1.135,C,11.解：,电池反应：2,Hg(,l,)+Cl,2,(,p,y,),Hg,2,Cl,2,(s),E=E,y,12.解：,电池反应：,H,2,(g,p,y,)+1/2O,2,(g,p,y,),H,2,O(,l,),E=E,y,12.,有两个电池的电动势,正确的是（）,(1),H,2,(,p,y,)|KOH(0.1molkg,-1,)|O,2,(,p,y,),E,1,(2)H,2,(,p,y,)|H,2,SO,4,(0.01molkg,-1,)|O,2,(,p,y,),E,2,(A),E,1,E,2,(C),E,1,=E,2,(D),不能确定,6,11.电池Hg(l)-Hg2Cl2(s)|HCl(a)|C,13.,已知电池,(1),Cu|Cu,2+,(,a,2,)|Cu,2+,(,a,1,)|Cu,E,1,(2)Pt|Cu,2+,(,a,2,),Cu,+,(,a,)|Cu,2+,(,a,1,),Cu,+,(,a,)|Pt,E,2,=(),(A),E,1,=E,2,/2(B),E,1,=2,E,2,(C),E,1,=,E,2,(D),E,1,E,2,2,E,1,13.解,:,电池反应：,(1),Cu,2+,(,a,1,),Cu,2+,(,a,2,),(2),Cu,2+,(,a,1,)+Cu,+,(,a,),Cu,2+,(,a,2,),+Cu,+,(,a,),7,13.已知电池2E113.解:电池反应：7,15.,已知25,，,p,y,下,A(s)+2BD(,aq,)=AD,2,(,aq,)+B,2,(g),在电池中可逆进行，系统做电功150,kJ，,放热80,kJ。,则该反应,r,H,m,=_,(A),80(B),230(C),232.5(D),277.5kJ,14,.有,电池反应,(1)1/2,Cu(s)+1/2Cl,2,(,p,y,),1/2,Cu,2+,(,a,=1)+Cl,-,(,a,=1),E,1,(2)Cu(s)+Cl,2,(,p,y,),Cu,2+,(,a,=1)+2Cl,-,(,a,=1),E,2,E,1,和,E,2,的关系为（）,(,A),E,1,=,E,2,/2 (B),E,1,=,E,2,(C),E,1,=,2,E,2,(D),E,1,=4,E,2,B,230,kJ,15.解:,r,H,m,=,r,G,m,+,T,r,S,m,=,15080=,230kJ,8,15.已知25，py下 A(s)+2BD(aq)=AD,16.,将铁粉，镉粉丢入含,Fe,2+,(0.1mol,kg,-1,),和,Cd,2+,(0.001mol,kg,-1,),的溶液中,已知,y,(Fe,2+,/Fe),=,0.44V,y,(Cd,2+,/Cd),=,0.40V，,正确答案为（）,(,A),铁粉，镉粉皆会溶解；,(,B),铁粉，镉粉皆不会溶解；,(,C),铁粉溶解，镉粉不溶；,(,D),镉粉溶解，铁粉不溶。,D,电极电势越低,越易被氧化!,9,16.将铁粉，镉粉丢入含Fe2+(0.1molkg-1),18.当发生极化时，阳极上发生 _反应，电极电势将_；阴极的电极电势将_。,17.电解水溶液时，在铜阳极上会发生(),(,A),析出,O,2,(B),析出,Cl,2,(C),析出铜(,D)Cu,极,溶解.,D,氧化,升高,降低,10,18.当发生极化时，阳极上发生 _反应，电极电势将,第七章测验题答案,一、,(25分),25时有,电池 (,Pt)H,2,(,p,y,)|HI(,m,)|AuI(s)-Au(s),(1),写出电极反应式和电池反应式；,(2)若,m,=10,-4,mol,kg,-1,时，,E,=0.97V (,=1),m,=3.0 mol,kg,-1,时，,E,=0.46V,求,m,=3.0mol,kg,-1,时的,(3)已知,y,(Au,+,/Au)=1.68V，,求,AuI,的活度积,解,:(1),电极反应,(,)1/2,H,2,e,H,+,(+)AuI+e,Au(s)+I,电池反应,1/2H,2,+AuI(s),Au(s)+HI(,m,),11,第七章测验题答案一、(25分)解:(1)电极反应,(2),-,-,:0.51=0.1182,lg(,3.0,10,4,),=0.69,(3),从,式可得：,E,y,=,y,(AuI/Au)=0.97+0.1182 lg 10,-4,=0.497V,根据,y,(AuI/Au)=,y,(Au,+,/Au)+(,RT/F,)ln,K,ap,0.0591 lg,K,ap,=,y,(AuI/Au),y,(Au,+,/Au),K,ap,=,9.6,10,-21,12,(2)-:0.,二、,(15分)要从某溶液中析出锌直到锌离子浓度不超过,10,-4,mol,kg,-1,同时在,析出锌的过程中不会有,H,2,析出。求算,pH,值至少应为若干？已知25时在锌电极上,H,2,析出的过电势为0.72,V，,y,(Zn,2+,/Zn)=0.763V,解：,25时,Zn,析出电势为,(Zn,2+,/Zn)=,y,(Zn,2+,/Zn)+(0.0591/2)lg Zn,2+,=,0.763 0.1182,=0.8812V,要使,H,2,不析出,(H,+,/H,2,),(Zn,2+,/Zn),(H,+,/H,2,)=0 0.0591 pH 0.72 0.8812V,pH2.73,13,二、(15分)要从某溶液中析出锌直到锌离子浓度不超过10-,三、,(35分)25时有下列数据：,y,/V,d,y,/,dT,Cu,+,+e =Cu,0.52,0.0020,Cu(NH,3,),2,+,+e =Cu+2 NH,3,-0.11,0.0030,Cu,2+,+2e =Cu,0.34,0.0035,1.求,f,G,m,y,(Cu,+,),f,G,m,y,(Cu,2+,),2.,求,Cu,2+,+e =Cu,+,的,y,3.求,Cu+Cu,2+,=2Cu,+,的,K,y,4.,计算反应,Cu,+,+2 NH,3,=Cu(NH,3,),2,+,在,25,时的,r,G,m,y,，,r,S,m,y,，,r,H,m,y,，,Q,r,14,三、(35分)25时有下列数据：y/Vdy/,解：1.,Cu,+,+e =Cu,y,=0.52V,r,G,m,y,=,f,G,m,y,(Cu,+,)=,F,y,=50.18 kJ,mol,-1,f,G,m,y,(Cu,+,)=,F,y,=50.18 kJ,mol,-1,同理：,Cu,2+,+2e =Cu,y,=0.34V,f,G,m,y,(Cu,2+,)=,2F,y,=65.6 kJ,mol,-1,求,Cu,2+,+e =Cu,+,的,y,r,G,m,y,=,f,G,m,y,(Cu,+,),f,G,m,y,(Cu,2+,),=15.42 kJ,mol,-1,=,F,y,(Cu,2+,/Cu,+,),y,(Cu,2+,/Cu,+,),=0.16V,或,r,G,m,y,=,F,y,2F,y,=,F,y,(Cu,2+,/Cu,+,),y,(Cu,2+,/Cu,+,)=,2,y,y,=0.16,V,15,解：1.Cu+e =Cu y,3.反应,Cu+Cu,2+,=2Cu,+,的,r,G,m,y,r,G,m,y,=,2,f,G,m,y,(Cu,+,),f,G,m,y,(Cu,2+,),=34.8,kJ,mol,-1,=,RT,ln,K,y,K,y,=7.9,10,-7,16,3.反应 Cu+Cu2+=2Cu+的rGmy,4.,反应,Cu,+,+2 NH,3,=Cu(NH,3,),2,+,=,r,G,m,y,=,r,G,m,y,r,G,m,y,=,F,y,+,F,y,=,F,(0.52+0.11),=,60,.8,kJ,mol,-1,r,S,m,y,=,r,S,m,y,r,S,m,y,=96.5,J,K,-1,mol,-1,r,H,m,y,=,r,G,m,y,+,T,r,S,m,y,=,89,.6,kJ,mol,-1,Q,r,=,T,r,S,m,y,=28,.8,kJ,17,4.反应 Cu+2 NH3=Cu(NH3)2,四、,(25分),已知,NaCl,KNO,3,NaNO,3,在稀溶液中的,解：(1)对于强电解质：,m,(NaCl)=,m,(Na