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本文用经典极化曲线的方法和交流电测定电极表面状态的方法,对锑在氢氧化钾、硫酸和盐酸中的阳极过程进行研究,获得如下的结果: 1.锑的阳极电位φ与极化电流密度i及pH之间的关系可由公式表示:φ=a+ blogi—b′pHo在1—13.4N氢氧化钾溶液中,b=116毫伏,b′=115毫伏;在1—14N硫酸中,b=24毫伏,b′=48毫伏;在1—12N盐酸中,b=24毫伏,b′=—48至—54毫伏。a为常数。 2.溶液pH对锑的阳极溶解速度的关系,λ=((?)logi/(?)pH)_φ,在氢氧化钾溶液中,λ=1;在硫酸中,λ=2;在盐酸中λ′=((?)logi/(?)pH)_φ=+2:即溶解速度在氢氧化钾或盐酸中随浓度增长而加速;在硫酸中则相反。 3.假设锑在这些介质中的阳极反应具共同的基本步骤:Sb+O~-→SbO_吸+e~-,在氢氧化钾溶液中,过程的进行依靠类似的步骤,使SbO继续氧化而逐步形成Sb_2O_3,Sb_2O_4;后者在溶液中因不稳定而歧化为Sb_O_3及Sb_2O_5。O~-假定通过如下步骤形成:OH~-→OH_吸+e~-,OH_吸+OH~-→O~-+H_2O。前者如为控制步骤则与实验数据相符合。在硫酸中则以共同步骤形成的中间物SbO失去电子而成SbO~+为控制步骤。在盐酸中因氧化膜SbO被氯离子破坏而生成络离子SbCl_4~-为控制步骤。在酸中,O~-假定是由水分子的单电子放电产生:H_2O→O~-+2H~++e~-。 4.由于氧化产物(Sb_2O_3,Sb_2O_5及吸附氧原子)在电极表面的积累,不论在酸或碱中,形成3—5个分子层,才导致表面钝化。氧的逸出发生在稳定钝化区的电位,这可能是O~-在钝化表面继续放电的结果。
In this paper, we studied the anodic process of antimony in potassium hydroxide, sulfuric acid and hydrochloric acid by the method of classical polarization curve and the method of measuring the surface state of the electrode by alternating current. The results are as follows: 1. The anodic potential of antimony and the polarization current The relationship between density i and pH can be formulated as: φ = a + blogi-b’pHo in 1-13.4N potassium hydroxide solution, b = 116 mV, b ’= 115 mV; in 1-14N sulfuric acid , b = 24 mV, b ’= 48 mV; in 1-12N hydrochloric acid, b = 24 mV, b’ = -48 to -54 mV. a is a constant. 2. Relationship between solution pH and antimony anodic dissolution rate, λ = ((?) Logi / (?) PH) _φ, λ = 1 in potassium hydroxide solution; λ = 2 in sulfuric acid; λ ’= ((?) logi / (?) pH) _φ = + 2: The rate of dissolution is accelerated with increasing concentration in potassium hydroxide or hydrochloric acid; in sulfuric acid the opposite is true. 3. Assuming that the anodic reaction of antimony in these media has a common basic step: Sb + O ~ - → SbO - + e - -, the process proceeds in a potassium hydroxide solution by a similar procedure to continue the oxidation of SbO While gradually forming Sb_2O_3, Sb_2O_4; the latter in the solution due to instability and disproportionate Sb_O_3 and Sb_2O_5. O ~ - is presumed to be formed by the following steps: OH ~ - → OH - sucking + e ~ -, OH - sucking + OH ~ - → O ~ - + H_2O. The former is the same as the experimental data if it is the control step. In sulfuric acid is the common step to form the intermediate SbO lost electrons SbO ~ + as the control step. In hydrochloric acid due to oxide film SbO chloride ions are destroyed to generate complex ions SbCl_4 ~ - as the control step. In acid, O ~ - is assumed to be generated by the single electron discharge of water molecules: H 2 O → O ~ - + 2H ~ + + e ~ -. Due to the accumulation of the oxidation products (Sb 2 O 3, Sb 2 O 5 and adsorbed oxygen atoms) on the surface of the electrode, 3-5 molecular layers are formed in the acid or alkali to cause surface passivation. Oxygen evolution occurs at the potential of the stabilized passivation region, which may be the result of O - - continuing to discharge on the passivated surface.