我是风儿 2014/11/16
电极反应为MnO4- + 8H+ +5e = Mn2+ + 4H2O 据能斯特方程可得: Eθ = Eo + (0.059/n)lg[氧化态]/[还原态] =Eo +(0.059/5)lg[MnO4-][H+]8/[Mn2+] =Eo + 0.012lg[MnO4-]/[Mn2+]+0.094(0.059*8/5)lg[H+] =Eo + 0.012lg[MnO4-]/[Mn2+] - 0.094pH(氢离子浓度负对数) 当[MnO4-]=[Mn2+]=1mol/L时,电对的电极电位即为条件电位,所以Eθ=Eo - 0.094pH