找出以下半反应的条件电极电位。(已知jq=0.390V,pH=7.0,抗坏血酸pKa1=4.10,pKa2=11.79)
【正确答案】:半反应设为:A2-+2H++2e-=H2A
E0’=E0+(0.059/2)lg{[Ox][H+]2/[Red]}
=E0+(0.059/2)lg{cδA[H+]2/cδH2A}
=E0+(0.059/2)lg{δA[H+]2/δH2A}+lgc/c
=E0+(0.059/2)lg{ka1ka2[H+]2/[H+]2}
E0’=E0+(0.059/2)lgka1ka2
=0.390V+(0.059/2)lg[10-4.10×10-11.79]=0.390V-0.469V=-0.0788V