Physical chemistr lectrochemistry (4) Xuan Cheng Xiamen University
1 Electrochemistry(4) Xuan Cheng Xiamen University Physical Chemistry
pp pical chemistry ectrochemid四 QUIZ (3 points each) (15分) Given that AGo=-2127 kJ. mol-l for the cell reaction in the Daniel cell at 25C and that m( cuso4 =10×103 molkg and m(ZnSO4)=3.0×103 mol kg, calculate (1)the ionic strength of the solutions (2 )the mean ionic activity coefficients in the respective compartments ()the reaction quotient (4)the standard cell potential (5)the cell potential (take y+=Y=y+ in the respective compartments
2 Given that rGo = -212.7 kJmol-1 for the cell reaction in the Daniel cell at 25℃ and that m(CuSO4 ) = 1.010-3 molkg-1 and m(ZnSO4 ) = 3.010-3 molkg-1 , calculate (1) the ionic strength of the solutions (2) the mean ionic activity coefficients in the respective compartments (3) the reaction quotient (4) the standard cell potential (5) the cell potential Physical Chemistry Electrochemistry QUIZ (3 points each) (15分) (take + = - = ± in the respective compartments)
pp pical chemistry ectrochemid四 QUIZ Answers 1=-∑mz (1分) for cuso solution 2 0×103×2+(-2)=40×103m01-kg(1分) for zn so∠ solution 0×10~3×2+(-2 1.2×10ml.h-1 (1分)
3 Answers. (a) for CuSO4 solution for ZnSO4 solution Physical Chemistry Electrochemistry QUIZ = 2 2 1 I mz ( ) 3 2 2 3 1 1.0 10 2 ( 2) 4.0 10 2 1 − − − I = + − = mol k g ( ) 3 2 2 2 1 3.0 10 2 ( 2) 1.2 10 2 1 − − − I = + − = mol k g (1分) (1分) (1分)
pp pical chemistry ectrochemid四 QUIZ Answers b) in very dilute aqueous solution at25°C(1分) /2 hny±=0.5102+|=-Vm/m /2 hny±(CO4)=-0.510×2×|-2140×103/1 y(CuSO4=0.74(1分) /2 hny(ZhSO4)=-0.510×2×|-211.2×10-2/ y4(znSo4=0.60(1分)
4 Answers. Physical Chemistry Electrochemistry QUIZ (b) in very dilute aqueous solution at 25℃ ( ) 1/ 2 ln 0.510 | | / o = − z + z − I m m ( ) 1/ 2 3 ln ( 4 ) 0.510 2 | 2| 4.0 10 /1 − CuSO = − − ± (CuSO4 ) = 0.74 ( ) 1/ 2 2 ln ( 4 ) 0.510 2 | 2 | 1.2 10 /1 − ZnSO = − − ± (ZnSO4 ) = 0.60 (1分) (1分) (1分)
pp pical chemistry ectrochemid四 QUIZ Answers (c) The cell reaction:(1分) Cuz+(m1)+ so42(m1)+Zn>Cu+ Zn +(m2 )+so42(m2) SO2 (zn2+) Q=∏a"= ZnSO (SO4 )ZnSO4(SO4 )DmSO4 (ac12 I Sir Q)(cn2+mcn2-)((02) CuSo Ince )m(SO2 )CuSO4 Y+( Cu2+)CuSO 4 =Y (SO4-)CuSO4=Y+( CuSO4)=0.74 Y+(Zn *)ZnSO 4=Y(S042)ZnSO4 =Y+(ZnSO4)=0.60 Also My=m( Cu2+)CuSO4=m(S042)CuSO4=1x10 mo·ka1 m2=m (Zn+)ZnSO4 =m(So4)ZnSO4=3x10-3mol-kgs
5 Answers. Physical Chemistry Electrochemistry QUIZ (c) The cell reaction: Cu2+ (m1 ) + SO4 2- (m1 ) + Zn → Cu + Zn2+ (m2 ) + SO4 2- (m2 ) = = = + + − − + + − − + − + − ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 4 2 4 4 2 4 2 2 4 2 4 4 2 4 2 2 4 2 4 2 4 2 4 2 ( ) ( ) ( ) ( ) ( ) ( ) CuSO CuSO ZnSO ZnSO CuSO ZnSO Cu Cu SO SO Zn Zn SO SO Cu SO v Zn SO m m m m a a a a Q a Since + (Cu2+)CuSO4 = - (SO4 2- )CuSO4 = ± (CuSO4 ) = 0.74 + (Zn2+)ZnSO4 = - (SO4 2- )ZnSO4 = ± (ZnSO4 ) = 0.60 Also m1 = m (Cu2+)CuSO4 = m (SO4 2- )CuSO4 = 110-3 molkg-1 m2 = m (Zn2+)ZnSO4 = m (SO4 2- )ZnSO4 = 310-3 molkg- 1 (1分) (1分)
pp pical chemistry ectrochemid四 QUIZ Answers Therefore, Q (0.60×3.0×103) (074×10×10-3)2≈59 (1分) △,GO=-nFEO (1分) G 2127×1000 E +1.102(2分) 2×96500 (e)E=EO、RT In Q (1分) 8.314×298 E=+1.102 2×96500 n59=+1079(2分)
6 Answers. Physical Chemistry Electrochemistry QUIZ 5.9 (0.74 1.0 10 ) (0.60 3.0 10 ) 3 2 3 2 = = − − Therefore, Q (1分) o o (d) r G = −nFE (1分) V nF G E o o r 1.102 2 96500 212.7 1000 = + − = − = − (2分) (e) Q nF RT E E o = − ln (1分) E ln 5.9 1.079V 2 96500 8.314 298 1.102 = + = + − (2分)
pp pical chemistry ectrochemid四 Electrode processes The electric charges are transferred at a certain rate within the metals(two electrodes) and solution interfaces when there is a current passing through the electrochemical system 在电化学系统中有电流通过时(电化学电源或 电解池)在两个电极的金属和溶液界面间以一定 速率进行着电荷传递过程,即电极反应过程
7 Electrode Processes 在电化学系统中有电流通过时(电化学电源或 电解池)在两个电极的金属和溶液界面间以一定 速率进行着电荷传递过程，即电极反应过程。 The electric charges are transferred at a certain rate within the metals (two electrodes) and solution interfaces when there is a current passing through the electrochemical system. Physical Chemistry Electrochemistry
Phsical Chemistr Elcctrochemiary The electrode processes M+e 正反应叫阴极过程或阴极 E 反应,设其反应速率为U; Ec M++e 逆反应叫阳极过程或阳极 反应,设其反应速率为Ua 电极上进行的阴极与阳极过程 U>U时,电极作为阴极; Cathode U>U时,电极作为阳极: Anode U=U时,电极反应处于平衡。 Equilibrium
8 正反应叫阴极过程或阴极 反应，设其反应速率为υ c； 逆反应叫阳极过程或阳极 反应，设其反应速率为υ a 。 = 时，电极反应处于平衡。 时，电极作为阳极； 时，电极作为阴极； c a a c c a The electrode processes M+ + e- M c a 电极上进行的阴极与阳极过程 Ec Ea M a c M++e M 电极金属 Physical Chemistry Electrochemistry Cathode Anode Equilibrium
pp pical chemistry Eleckrochemidta Electrochemical reaction rate and current densit 电化学反应速率与电流密度 电极反应的反应速率定义为 1 ds 式中,A电极的截面积,单位为m2; 5一反应进度,单位为mol。 U电化学反应速率,定义为单位时间内,单位面积 的电极上,反应进度的改变量。单位为molm2sl
9 电化学反应速率与电流密度 电极反应的反应速率定义为 式中，A⎯电极的截面积，单位为m2； ξ⎯反应进度，单位为mol。 υ⎯电化学反应速率，定义为单位时间内，单位面积 的电极上，反应进度的改变量。单位为molm-2 s -1 。 Physical Chemistry Electrochemistry dt d A v 1 Electrochemical reaction rate and current density
pp pical chemistry Eleckrochemidta 在电化学中,易于由实验测定的量是电流,所以常用 电流密度j(单位电极截面上通过的电流,单位为Am2) 来表示电化学反应速率v的大小,汽的关系为 ZFu 阴极过程:=ZFU 阳极过程:i=ZFUn 阴极上 Ja =c-a 阳极上 平衡电极上 Ja =je j6叫交换电流密度
10 在电化学中，易于由实验测定的量是电流，所以常用 电流密度 j (单位电极截面上通过的电流，单位为Am-2) 来表示电化学反应速率υ的大小，j与υ的关系为 j = Z Fυ (12 -14) : : a a c c = = j ZF j ZF 阳极过程 阴极过程 = = = − = − c a 0 a c a c c a c a j j j j j j j j j j j j j 平衡电极上 阳极上 阴极上 j0叫交换电流密度。 Physical Chemistry Electrochemistry