北京化工大学：《物理化学》课程教学资源（双语练习题）第七章 电化学英文习题及参考答案 electrochemistry problems

第七章电化学美文司题及家考答素 1.Use the information of standard potential table to calculate the standard potential of the cell AglAgNO(aq)Fe(NO3)(aq)Fe and the standard Gibbs energy and enthalpy of the cell reaction a 25C.Estimate the value ofa 35C.(5m △，Hn=-89.1U,mo') Solution:R:Fe2(aq)+2e- →Fc(s) L:2Ag(s) -2Ag(aq)+2e The cell reaction:2Ag(s)+Fe(aq) →2Ag(ag+Fe(s) E=E-E=(-0.44-(0.80=-1.24W △，G=-zFE0=-2×96500×(-1.24)J·mol-=239kJmo △，Hg=-2A,H(4g°，ag)-△，H(Fe2(ag》 =2×105.58-(-89.1)kJ-mol-1 =300.3kJ-m0 9,=-d.4C-AL.29-03m 个 29815K =-0.206kJ.mo.K herefore:△，G(308K)≈[239+10K×(-0.206K-]k·mol=237kJmol 2.Determine the standard potential of a cell in which the reaction is Co(aq)+3Cl(aq)+3Ag(s) -·3AgCl(s+Co(s) from the standard potentials of the couples: Eg14ae=0.22,E8c=1,8IV,E8*a=-028w Solution:We need toobtain Efor the couple (3)Co"(aq)+3e- -→Co(s) from the values ofE for the couples (1)Co(aq)+e- --Co(aq) E°=1.81Ψ (2)Co"(aq)+2e- Co(s) Eg=-0.28 We see that(3)=(1)+(2):therefore E,=E+,空.1181W+2x-0282.042y Then: R:Co"(aq)+3e- —→Cos)ER=0.421Ψ

1 第七章 电化学英文习题及参考答案 1. Use the information of standard potential table to calculate the standard potential of the cell Ag|AgNO3(aq)||Fe(NO3)2(aq)|Fe and the standard Gibbs energy and enthalpy of the cell reaction at 25℃. Estimate the value of q DrGm at 35℃. ( 1 ( ( )) 105.58 - D = × H + kJ mol m Ag aq f q , 1 ( ( )) 2 89.1 - D = - × H + kJ mol m Fe aq f q ) Solution：R：Fe2+(aq)+2e-————→Fe(s) L：2Ag(s)————→2Ag+ (aq)+2e￾The cell reaction：2Ag(s)+Fe2+(aq)————→2Ag+ (aq)+Fe(s) E = ER - EL = (-0.44V) - (0.80V) = -1.24V q q 1 2 96500 ( 1.24) - D G = -zFE = - ´ ´ - J × mol r m q q 1 239 - = kJ × mol 2 ( , ) ( ( )) 2 rHm f H m Ag aq f Hm Fe aq + + D = D - D q q q 1 2 105.58 ( 89.1) - = ´ - - kJ × mol 1 300.3 - = kJ × mol 1 1 1 0.206 298.15 (239 300.3) ( ) - - - = - × × - × = D - D = -D = ¶ ¶D kJ mol K K kJ mol T G H S T G r m r m p r m r m q q q q Therefore： 1 1 (308 ) [239 10 ( 0.206 )] - - D G K » + K ´ - K kJ × mol r m q 1 237 - = kJ × mol 2. Determine the standard potential of a cell in which the reaction is Co3+(aq)+3Cl- (aq)+3Ag(s)————→3AgCl(s)+Co(s) from the standard potentials of the couples： E V Cl AgCl Ag 0.22 / , - = q ， E V Co Co 3 2 1.81 / + + = q ， E V Co Co 0.28 / 2+ = - q . Solution：We need to obtain Eθ for the couple (3) Co3+(aq)+3e-————→Co(s) from the values of Eθ for the couples (1) Co3+(aq)+e-————→Co2+(aq) E1 = 1.81V q (2) Co2+(aq)+2e-————→Co(s) E2 = -0.28V q We see that (3)=(1)+(2)；therefore 3 1 1.81 2 ( 0.28 ) 3 1 1 2 2 3 E E V V E ´ + ´ - = + = n n n q q = 0.42V Then： R：Co3+(aq)+3e-————→Co(s) ER = 0.421V q

L:3Ag(s)+3CI(aq)- -→3AgCI(s+3eE=0.22y the cell reaction:Co(aq)+3CI(aq)+3Ag(s) 一3 AgCI(s)+Cos) E°=Eg-E=0.42V-0.22V=0.20

2 L：3Ag(s)+3Cl- (aq)————→3AgCl(s)+3e- EL = 0.22V q the cell reaction：Co3+(aq)+3Cl- (aq)+3Ag(s)————→3AgCl(s)+Co(s) E = ER - EL = 0.42V - 0.22V = 0.20V q q q