=-0.282V-0.5×0.0592V×lg{1.0/0.01}=-0.341V EM=E(Cl2/C)-ECo2+/Co)=1.36V-(-0.341)=1.701V,电池电动势增大 2、(1)由附表查出Ee(Cu/Cu),E(Cu/Cu),试计算Kp(Cu) (2)计算298K下反应Cul(s)=Cu(aq)+r(aq)的△，Gnm (3)若己知(2)中的反应△rHm0(298K)=84.3k.mo1,计算该反应的△r5m9 (298K). 解：(1)查得E(C/C)=0.5180r,E(Cl/C0)=-0.1858 E0(Cul/Cu)=E(Cu*/Cu)-(RT/zF)In(1/Cu*) =E(Cu*/Cu)-(RT/zF)In(1/Ksp(Cul)) E (Cul /Cu)=E"(Cu'/Cu)+0.0592VIg K(Cul) K0(C0=1.3x10-12 Or:Cu'+eCu (1)△rGma=-FEa Cul+e→Cu+r(2)△rGm°a=-fEea (2)-(1:Cul=Cu+f(3) △rGm93)=-RTlnK0 △rGm2-△Gm=△rGma) -FE2)+FE)=-RTInK Therefore,E)=E+(RT/F)InK (2)对于反应Cul(s)=Cu(aq)+r(aq)有： △rGme=-RTInKe △，G(298K)=-2.303RT1gK(Cl0=67.8kJ1mol (3)由△，G=△，H-T△，S得： A,S298K)=4H298K)AG298k0-554W/m0lK 298K= -0.282 V - 0.5×0.0592 V × lg{1.0/ 0.01} = -0.341 V EMF= E(Cl2/ Cl- ) - E(Co2+/Co) = 1.36V - (-0.341V) = 1.701 V，电池电动势增大 2、（1）由附表查出 E (Cu+ /Cu)，E (CuI/Cu)，试计算 Ksp (CuI) （2）计算 298K 下反应 CuI (s) = Cu+ （aq）+ I- (aq) 的rGm θ （3）若已知（2）中的反应 ΔrHm Θ（298K）=84.3kJ.mol-1 , 计算该反应的 ΔrSm Θ （298K）. 解：（1）查得 E Cu Cu V ( / ) 0.5180  + = ， E CuI Cu V ( / ) 0.1858  = − ， E (CuI/Cu) = E(Cu+ /Cu) – (RT/zF)ln(1/Cu+ ) = E(Cu+ /Cu) – (RT/zF)ln (1/Ksp(CuI)) → ( / ) ( / ) 0.0592 lg ( ) E CuI Cu E Cu Cu V K CuI sp    + = + 12 ( ) 1.3 10 K CuI sp  − =  Or: Cu+ + e →Cu (1) ΔrGm Θ (1) = -FEΘ (1) CuI + e → Cu + I- (2) ΔrGm Θ (2) = -FEΘ (2) (2) – (1): CuI = Cu+ + I- (3) rGm θ (3)= -RTlnKθ ΔrGm Θ (2)- ΔrGm Θ (1) = rGm θ (3) -FEΘ (2) + FEΘ (1) = -RTlnKθ Therefore, EΘ (2) = EΘ (1) + (RT/F)lnKθ （2）对于反应 CuI (s) = Cu+ （aq）+ I- (aq)有： rGm θ = -RTlnKθ (298 ) 2.303 lg ( ) 67.8 / r m sp G K RT K CuI kJ mol    = − = （3）由 r m r m r m G H T S     =  −  得： (298 ) (298 ) (298 ) 55.4 / 298 r m r m r m H K G K S K J mol K K     −   = = •