1
$\begingroup$

I'm going to do an iodine clock reaction for a project and we had to submit the materials and safety sheets for the experiment a while ago. I thought I could do one by using hydrogen peroxide, sodium thiosulfate, potassium iodide and starch, but online I'm only seeing experiments that use sulfuric acid or another strong acid of some sort which is worrying me. I talked to my teacher and it's too late to change/add materials, so I'm really hoping that it can work? I'd also like to know what the strong acid does in the reaction, and what would happen (or wouldn't happen) without sulfuric acid?

Here's a bit from wikipedia for reference about the reaction:

This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch. There are two reactions occurring in the solution.

(note: I didn't try formatting anything so the numbers after the ^ are the charges)

In the first, slow reaction, iodine is produced: H2O2 + 2I^− + 2H^+ → I2 + 2H2O

In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate: 2S2O3^2− + I2 → S4O6^2− + 2I^−

After some time the solution always changes color to a very dark blue, almost black. When the solutions are mixed, the second reaction causes the triiodide ion to be consumed much faster than it is generated, and only a small amount of triiodide is present in the dynamic equilibrium. Once the thiosulfate ion has been exhausted, this reaction stops and the blue colour caused by the triiodide – starch complex appears.

New contributor
user73318 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
2
$\begingroup$

You wrote the answer yourself: the hydrogen peroxide consumes hydrogen ion according to the reaction

$\ce{H2O2 + 2 I^- + 2\color{blue}{H^+} -> I2 + 2 H2O}$

You need the acid to provide the hydrogen ions, especially since if you allow the solution to become basic then the iodine disproportionates (see "Chemistry and Compounds" section, "Iodine Oxides and Oxoacids" subsection).

In terms of pure chemistry, maybe you could have used a moderately strong acid such as phosphoric acid or a bisulfate salt. But then the limited dissociation of such species might make the iodine formation reaction slower and thus slow down the "clock".

Looking at the Wikipedia article on this reaction, one alternative is to use a peroxydisulfate salt instead of hydrogen peroxide as your oxidant. This reacts with iodide ion according to

$\ce{S2O8^{2-} + 2I^- -> 2 SO4^{2-} +I2}$

with thiosulfate again as the reducing agent to regenerate iodide ion. Unlike hydrogen peroxide or the other oxidants, the above reaction does not consume hydrogen ions and is your best bet to run the reaction without a strong acid addition. However, peroxydisulfates have their own hazards (one of which is requiring a strong sulfuric acid solution to make it in the first place), and if you make the common choice of potassium peroxydisulfate rather than the ammonium salt you could have solubility issues.

$\endgroup$
  • $\begingroup$ Would the experiment not work at all? Would anything happen in the reactions? $\endgroup$ – user73318 Jan 13 at 15:03
  • $\begingroup$ Not sure. Wikipedia has several versions and most use a strong acid. The only one that does not, appareny, is The one with potassium peroxyisulfate. See the Wikipedia article. When I get time I will edit the answer. $\endgroup$ – Oscar Lanzi Jan 13 at 15:31

Your Answer

user73318 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.