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在电解槽中,以Ｎａ２ＳＯ４为支持电解质,多孔石墨电极为阴极,金属铁为阳极,在１０ｍＡ·ｃｍ－２的阴极电流密度下,电解生成的过氧化氢与阳极溶解的Ｆｅ２＋进行反应,生成羟基自由基(Ｆｅｎｔｏｎ试剂),进而对有机染料进行氧化反应,使其不饱和双键断裂,从而达到有机染料降解、脱色的目的,在生成Ｆｅｎｔｏｎｎ试剂的反应中,Ｆｅ２＋被氧化为 Ｆｅ３＋,生成的 Ｆｅ３＋在阴极得电子后重新被还原为Ｆｅ２＋,因此,在电解槽中 Ｆｅ２＋起着催化剂的作用．ＣＯＤ的去除率大于８０%,脱色率达１００%．脱色速率的测定表明其为一级反应,反应速度常数ｋ等于ｌ．９８ × １０－４ｓ－１．
In the cell, with Na2SO4 as supporting electrolyte, porous graphite electrode as cathode and metallic iron as anode, the hydrogen peroxide generated by electrolysis reacted with anode dissolved Fe2 + at cathode current density of 10mA · cm-2 to produce hydroxyl (Fenton reagent), and then the organic dye oxidation reaction, the unsaturated double bond rupture, so as to achieve the degradation of organic dyes, the purpose of the formation of Fentonn reagent reaction, Fe2 + is oxidized to Fe3 +, the resulting Fe3 + In the cathode after the electron is re-reduced to Fe2 +, therefore, Fe2 + in the cell plays the role of catalyst. COD removal rate of more than 80%, decolorization rate of 100%. Determination of decolorization rate showed that it is a first-order reaction, the reaction rate constant k is equal to l. 98 × 10-4s-1.