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