# What causes the yellow colour in the Briggs-Rauscher oscillating reaction?

The color changes are due to different concentrations of supposedly $\ce{I2}$ (which causes the yellow/amber color), $\ce{I-}$ (colorless) and $\ce{I3-}$ (dark blue).

Since iodine ($\ce{I2}$) is barely soluble in water, I do not understand why in this case it would be.

If one was to put just iodine in water and mix it thoroughly, the color change of said water would not be as visible to the extent of the Briggs-Rauscher reaction.

• $\ce{I3-}$ is by no means dark blue. In fact, it is colored the same as $\rm I_2$. – Ivan Neretin Oct 11 '17 at 21:35
• I'm not quite sure I understand the question, but keep in mind there is an equilibrium between $\ce{I-}$, $\ce{I2}$ and $\ce{[I3]-}$ all the time, which means not all I is present at a given form ($\ce{I2}$) at a given moment, so I see no contradiction in observed color intensity and solubility of iodine in water. Also, it's not $\ce{[I3]−}$ that is dark blue, it's triiodide-starch complex. – andselisk Oct 11 '17 at 21:36
• Pardon, when interacting with starch indeed the dark blue color will be visible. – Boris Oct 11 '17 at 22:27
• My question is that how it would be possible for the mixture to have a yellow color. Many sources claim this is due to the I2(aq). My point is that I don't understand how I2 can be dissolved. – Boris Oct 11 '17 at 22:29
• Starch test is normally done with neutral iodine and neutral starch; neither ionizes in water. Pray tell us how $\ce{I3-}$ comes about in these circumstances, what is the cation that balances it, and where does the latter come from? – Ivan Neretin Oct 12 '17 at 4:39

The colour changes are due to the indicator. In the case of the Briggs-Rauscher reaction, starch is included which causes a blue colour when complexed with iodine and iodide (high concentrations of $\text{I}_3^-$. When the concentration of $\text{I}_2$ is high, the reaction mixture is yellow due to the colour of iodine.