I was told by several high school students that they experienced odd results in their iodine clock reaction lab. They did several different trials, and the lower the temperature, the faster the blue color showed.

Out of curiosity, I asked about the detail, and they told me the following:

  1. Most students in most classes had the same problem, so I think it wasn't bad technique of some students that caused the problem.
  2. They did experiments of different temperature with the same solutions, so I think it wasn't that they did the lower temperature experiment with higher concentration.
  3. The temperatures were 30 °C, 20 °C and 10 °C.
  4. Their teacher only said that maybe there's something wrong with the solution. But it made me more curious: what kind of mistake could lead to lower temperature having faster results?

The only possibility I could come up with, is that iodine clock reaction isn't a simple reaction, it includes:

\begin{align} \ce{IO3^- + HSO3^- &-> I^- + H^+ + SO4^2-} \tag{R1}\\ \ce{IO3^- + I^- + H^+ &-> I2 + H2O} \tag{R2}\\ \ce{I2 + HSO3^- + H2O &-> I^- + H^+ + SO4^2-} \tag{R3} \end{align}

To my knowledge, lower temperature results in lower reaction rate. Is it possible that somehow the third equation's rate is lowered to a point that it couldn't get rid of $\ce{I2}$ fast enough, so somehow the blue color showed before bisulfite is used up?

Is this possible? If there are different explanations, I would love to hear.

  • 1
    $\begingroup$ Wang, Could you also add some realiable literature survey using Google Scholar? This reaction must be well studied. $\endgroup$
    – AChem
    Aug 14 at 19:32
  • $\begingroup$ Is there any tips for searching for certain questions on google scholar? (I tried keywords like "iodine clock low temperature", but couldn't find paper on the point.) $\endgroup$
    – Wang
    Aug 15 at 7:30

1 Answer 1


I agree that the iodine clock reaction is complex, but if you need an answer for your teacher, the explanation may center around "sulfurous acid" implied in your reaction chain, that apparently does not actually exist in water (but is found just in air). For more details, per Wikipedia, to quote:

Raman spectra of solutions of sulfur dioxide in water show only signals due to the SO2 molecule and the bisulfite ion..."

At lower temperatures, there is a notable increase in dissolved sulfur dioxide (see, for example, graph provided in this source).

Next, there is the so-called Bunsen reaction, to quote Wikipedia on the reaction:

The Bunsen reaction is a chemical reaction that describes water, sulfur dioxide, and iodine reacting to form sulfuric acid and hydrogen iodide:

2H2O + SO2 + I2 → H2SO4 + 2HI

This reaction is the first step in the sulfur-iodine cycle to produce hydrogen.

Note: The products of the above Bunsen reaction matches the cited reaction products of Equation (R3).

As a consequence, this is one possible explanation as to how an increase in sulfur dioxide gas concentration itself (from cooling) may also be beneficially impacting the iodine clock reaction as it is clearly negatively temperature dependent.

  • $\begingroup$ Thank you for your insight! But I can't understand how more SO2 ( from lower temperature) results in faster showing of blue color. Doesn't it remove I2 faster, thus delaying the blue result? $\endgroup$
    – Wang
    Aug 17 at 18:05
  • 1
    $\begingroup$ Per a source, quoting: "Molecular iodine (I2) is not easily soluble in water, which is why potassium iodide is added. Together, they form polyiodide ions... The negatively charged iodide...acts as charge donor, the neutral iodine as a charge acceptor. Electrons... excite to a higher energy level by light. The light is absorbed in the process and its complementary color is observed by the human eye" link provided below. So, conversion of the element Iodine to iodide results in a starch color change. The presence of an added Bunsen reaction facilitates this event via creating iodide. $\endgroup$
    – AJKOER
    Aug 17 at 19:38
  • 1
    $\begingroup$ Link reference, as referenced above: chemistryviews.org/details/education/10128441/…. $\endgroup$
    – AJKOER
    Aug 17 at 19:39

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