Hi all,
I am doing some experiments on arsenic dissolution in alkaline environment.
We are using pure glassy As2S3 in dionized water with pH adjusted to different levels from 6 to 10 using NaOH or Na2CO3.
When I add As2S3 to the system, ORP (Eh) starts to decrease significantly and after about 2-3 hours it reaches to a number and stays on it till the end of the experiment (the ORP decreases for about 200 mV). We have measured the concentration of dissolved oxygen before and after adding the arsenic sulfide and the number was constant so we have concluded that the ORP drop is not coming from oxygen removal. (Oxygen was provided continuously).
Could you please let me know your thought and comments on the probable reasons for this weird ORP decrease?
Cheers,
generally ORP increase with oxidation and decrease with reduction and in your situation as arsenic creates a reducing environment your ORP will decrease.
you can refer to this article for more information:
Dear Mohamad,
Dissolution of As2S3 mineral even in nominal amounts might results in generation of compounds with inherently less oxidation potential, which ultimately decreases the ORP of your system.
The attached image of As-S-H2O thermodynamic system could provide a better perspective on this matter.
Dear Mehdi Golzar Ahmadi ,
Thanks for the comment :)
The thing is oxidation needs an oxidizing agent (apparently the oxidant is oxygen in our system) and the same for reduction which means we need a reducing agent.
Dissolution of As2S3 in such environment and considering the constant amount of oxygen in the system can not explain the ORP decrease. In the article you have mentioned, they have used sulfide media and the presence of HS- could explain the low ORP. But in our case we did not add any sulfide to the system and it seems sth is happening right after the addition of As2S3 to the leaching solution.
Dear Farhad Moosakazemi
Thank you for the explanation. However, I think to see a drop in ORP we should have a significant change in the quantity of a redox couple like As(V)/As(III) to explain the change in ORP. Considering the slow conversion of As(III) to As(V), I am not sure if we can say by adding the solid to the system we are producing such couple! Maybe a fast attachment of hydroxide ions on the solid surface can cause the ORP change?!
The compromised potential between reduction of O2 and oxidation of As2S3 lie within the individual equilibrium potentials for each half reactions. Do you have ORP of As2S3 in leaching solution without oxygen sparging?
Kashif Mairaj Deen
Hi Kashif,
Thanks for the comment. If I add the solid (As2S3) to DI+NaOH (pH adjusted to let say 8.0) the ORP can drop to about 150 mV (SHE). The only explanation I could find is the very fast release of sulfide or polysulfide ions from the surface as the result of attachment of hydroxide to the surface and change the charge balance. Then the oxidation of sulfide to higher oxidation numbers starts and the ORP will increase gradually by increasing the leaching time. What is your idea?
Hi Mohamad Mirazimi,
I think the surface potential of As2S3 is relatively lower (more negative) than the equilibrium potential of O2/H2O redox potential under applied conditions. Without solids the ORP value of the leach solution is controlled by the dissolved oxygen and may be higher than the ORP with solids i.e. As2S3. This clearly states the oxidation of sulfide sulfur species to high oxidation state species at the surface facilitated by the reduction of dissolved oxygen. In other words, the ORP would lie in between the O2/H2O redox potential and surface potential of As2S3. In other words, the resulting ORP after addition of solids would be lower than the O2/H2O but higher than the surface potential of As2S3 (meaning oxidation of As2S3). But in this case, the large decrease in ORP corresponds to the dissolution controlled by the supply of oxidizing species (here dissolved O2). The formation of hydroxide species is obvious but the overall charge will remain neutral. Otherwise, this may significantly influence the pH at the surface, which I assume is not the case here under applied conditions.