中文摘要：

采用钛基氧化物涂层材料（Ti/SnO2-Sb2O5）为阳极,石墨为阴极,在含Fe2＋和不含Fe2＋2种电解质条件下研究了苯胺的电化学氧化降解效率和机制.结果表明,Ti/SnO2-Sb2O5阳极氧化降解有机物的电位约2.0 V±0.1 V,而石墨阴极还原O2生成H2 O2的优化电位为-0.65 V.H2 O2单独作用不能氧化苯胺,当电解质中不存在Fe2＋时,苯胺的降解主要依赖于阳极氧化作用;而当电解质中存在Fe2＋时,阴极电Fenton氧化和阳极氧化（双极电化学氧化）均能有效降解水中苯胺,但前者作用更大.在阴极电位为-0.65 V、初始pH值3.0、初始Fe2＋浓度为0.5 mmol.L-1条件下,处理180 mg.L-1苯胺水溶液10 h,COD的去除率为77.5%,去除COD的电流效率达到97.8%.该研究表明,以Ti/SnO2-Sb2O5为阳极,控制合理的阳极和阴极电位,可以实现双极电化学氧化有效降解水中有机物,并且获得较高的电流效率.

英文摘要：

The efficiency and the mechanism of aniline degradation by an electrochemical oxidation process using a Ti/SnO2-Sb2O5 electrode as the anode and a graphite electrode as the cathode,were studied in two aqueous electrolytes with/without Fe2＋.The results showed that the reasonable anodic potential was about 2.0 V±0.1 V for Ti/SnO2-Sb2O5 electrode to oxidize organic compounds,while the optimum cathodic potential was-0.65 V for graphite electrode to reduce O2 generating H2O2.The oxidation degradation of aniline could not take place only by the single action of H2O2.Anodic oxidation was accounted for the degradation of aniline in the absence of Fe2＋,while in the presence of Fe2＋ both electro-Fenton oxidation and anodic oxidation（dual-electrode electrochemical oxidation） could degradate aniline effectively,and in this case the former was the main mechanism.Under the conditions of-0.65 V cathodic potential,pH 3.0 and 0.5 mmol·L-1 Fe2＋,the removal rate of COD was 77.5% after 10 h treatment and a current efficiency of 97.8% for COD removal could be obtained.This work indicates that the dual-electrode electrochemical oxidation is feasible for the degradation of organic compounds with a high current efficiency by using Ti/SnO2-Sb2O5 as anode as well as the reasonable anodic and cathodic potentials.