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电化学过程中 RuO(x)-PdO/Ti 电极对甲基橙的降解机制。

Degradation mechanism of Methyl Orange by electrochemical process on RuO(x)-PdO/Ti electrode.

机构信息

College of Chemistry, Sichuan University, Chengdu, China.

出版信息

Water Sci Technol. 2011;63(7):1539-45. doi: 10.2166/wst.2011.414.

Abstract

The electrochemical degradation of Methyl Orange in 0.1 M NaCl solution over RuO(x)-PdO/Ti anode was investigated. Chemical oxygen demand (COD), ion chromatography (IC), Fourier Transform Infrared Spectroscopy (FTIR) and Gas chromatography-mass spectrometry (GC-MS) were employed to detect the intermediates formed during the electrochemical degradation. In the present reaction system, Methyl Orange could be effectively degraded. After 1 h treatment, the discoloration could reach 97.9% with COD removal of 57.6%. The results indicated that in the presence of chloride, the electrolysis was able to oxidise the dye with partial mineralisation of carbon, nitrogen and sulfur into CO(2), NO(-)(3) and SO(2-)(4), respectively. After 8 h electrolysis, 62% of sulfur contained in Methyl Orange was transformed to SO(4)(2-), and 17.6% of nitrogen changed to NO(3)(-). The intermediates during electroprocess were detected to be low molecular weight compounds, chlorinated compounds, derivatives of benzene and long chain alkanes. Based on these data, a possible degradation mechanism of Methyl Orange was proposed.

摘要

在 0.1 M NaCl 溶液中,RuO(x)-PdO/Ti 阳极上电化学降解甲基橙。采用化学需氧量(COD)、离子色谱(IC)、傅里叶变换红外光谱(FTIR)和气相色谱-质谱联用(GC-MS)检测电化学降解过程中形成的中间产物。在本反应体系中,甲基橙能有效降解。经过 1 h 处理,其脱色率可达 97.9%,COD 去除率为 57.6%。结果表明,在氯离子存在的情况下,电解能够氧化染料,部分将碳、氮和硫矿化为 CO(2)、NO(-)(3)和 SO(2-)(4)。经过 8 h 电解,甲基橙中 62%的硫转化为 SO(4)(2-),17.6%的氮转化为 NO(3)(-). 检测到电过程中的中间产物为低分子量化合物、氯化化合物、苯衍生物和长链烷烃。基于这些数据,提出了甲基橙的可能降解机制。

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