103mo/LPb2+溶液 0.1 mol/LHCI supporting electrolyte The reduction of pblt at DMe: A) Pb+2e>Pb(Hg)E(Pb2+/Pb)=-0.126V Eappr Etir=Eeg+n +iR 甘 滴 汞 Open potential suppose n=0,iR=0 极 极上 appl eqdme rej,EDME= applref 已知:E参比=+0.24V 外加电压:-0,4V(0V,0.34V(-0.1V -0.44(0.2V),-0.54V(-0.3V),-0.64V(-0.4V) -0.74V(-0.5V),-084V(-0.6V),-0.94V(-0.7V 当E app 0.37VE DME 0.13V10-3 mol/L Pb2+ 溶液 0.1 mol/L HCl supporting electrolyte The reduction of Pb2+ at DME: 2 ( ) 2 Pb + e → Pb Hg + 已知：E参比= +0.24V 外加电压：-0.24V(0V), -0.34V(-0.1V), -0.44V(-0.2V), -0.54V(-0.3V), -0.64V(-0.4V) -0.74V(-0.5V), -0.84V(-0.6V), -0.94V(-0.7V), Eappl=E+iR=Eeq+ +iR Eeq Open potential suppose = 0, iR = 0 Eappl= Eeq = EDME– Eref , EDME = Eappl+Eref 当 Eappl = - 0.37V EDME = -0.13V E0 (Pb2+/Pb) = -0.126V