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)+2eThe 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