在硫化钠—氢氧化钠浸出液中电沉积锑时电化学行为

来源期刊：中南大学学报(自然科学版)1982年第1期

论文作者：蒋汉瀛 舒余德 赵瑞荣 龚竹青 杨松青

文章页码：9 - 17

关键词：电化学行为； 电沉积； 极化曲线； 放电电位； 计时电位法； 氢氧化钠； 硫化钠； 浸出液； 络离子； 金属锑

摘    要：

本文用稳态法测定了极化曲线,确定了在硫化碱溶液中 Sb（Ⅲ）、Sn（Ⅳ）、As（Ⅲ）、Sb（Ⅴ）和 H⁺离子的放电电位,并分别用纯化学试剂配制的溶液和工业电解液进行了比较测定,同时和热力学计算结果进行核对。Sb（Ⅴ）的放电电位还通过计时电位法核对。

将所得极化曲线进行浓差极化修正后获得了 Tafel 直线,由此计算出有关的电极动力学参数。

应用计时电位法,我们探讨了 SbS^3-₃和 SbS^3-₄离子的阴极放电机理。结果表明;SbS^3-3₃离子放电前经历一个前置转化步骤,即:

SbS^3-₃—→SbS⁺+2S^2- （1）

生成的中间产物 SbS⁺按下列反应

SbS⁺+3e—→Sb+S^2- （2）

还原为金属锑。

SbS^3-₄离子在阴极上放电过程分两步进行,首先,SbS^3-₄离子按下式还原为 SbS^3-₃离子,

SbS^3-₄+2e—→SbS^3-₃+S^2- （3）

生成的中间产物 SbS^3-₃离子再按反应（1）和（2）式还原为金属锑。

Abstract:

When the complex Sb-Pb sulfide concentrate is leached with Na₂S-NaoH solution,the antimony enters into the solution in the form of SbS₃^3-ions, and the tin and the arsenic contained in the concentrate are dissolved too, while the lead is not dissolved and remains in the leaching residue.More- over,a part of the SbS₃^3-ions formed is oxidized by the oxygen in the air into SbS₃^3-ions.

The discharge potentials of Sb（Ⅲ）,Sn（Ⅳ）,As（Ⅲ）and H⁺ ions were obtained from polarization curves,and that of Sb（Ⅴ）by chronopotentiome- tric method.They were found to be -0.92,-1.04,-0.81.-1.12 and -0.68 volts respectively,and these values were checked up by the calculation ba- sed upon thermodynamic data.

Tafel lines were obtained after the correction for concentration polari- Zation and the kinetic parameters were determined from the Tafel lines. At 50℃,the rate constant was found to be 1.54.10^-4cm.sec^-1,the exchange current density 10^-2amp·cm^-2and the activation energy 9100 cal mole^-1.

The discharge mechanism of SbS₃^-3and SbS₃^-3ions were investigated with the help of chronopotentiometric method. It was shown that before charge transfer of the SbS₃^-3ions,a prior chemical reaction should take place as follows:

SbS₃^3-=SbS⁺+2S^2-（1）

The SbS⁺ ions formed were then reduced into metallic antimony at the cathode according to the following reaction:

SbS⁺+3e=Sb+S^2-（2）

Experiments showed that the discharge of SbS₃^3-ions at the cathode took place in two steps.It was believed that in the first step,the SbS₃^-3 ions were discharged directly as shown by the reaction:

SbS₄^3-+2e=SbS₃^3-+S^2-（3）

The SbS₃^3-ions thus formed as intermediate products were then redu- ced further into metallic antimony in accordance with equations（1）and （2）.