# Which reaction between Lugol and Hydrogen peroxide?

The Lugol solution is a mixture of $\ce{(K+, I- )}$ and $\ce{I2}$ such that $\ce{I3-}$ are formed by:

$$\ce{I- + I2 -> I3-}$$

I know that the brown color comes from $\ce{I2}$, but in this solution one have $\ce{I3-}$ not $\ce{I2}$, why is it brown?

When I mix Lugol with $\ce{H2O2}$ without the presence of $\ce{H3O+}$ there is a variation in the conductivity of the mixture but the brown color doesn't disappear. If I add $\ce{H3O+}$ the solution becomes transparent after few minutes.

My guess is that either $\ce{I2}$ is reduced by $\ce{H2O2}$ (hence the disappearing of the brown color) or a redox reaction between $\ce{H2O2}$ and $\ce{I-}$, but in the later case I don't understant why the brown color disapear instead of appearing.

Can you enlighten me please, thanks.

I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:

• $\ce{I+ , IO3-}$ don't absorb in the visible region
• $\ce{I3-}$ I think should be red
• $\ce{I2}$ is blue if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see Hildebrand (1909). In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.

Because there is an equilibrium:

$$\ce{I_2 + I^- <=> I_3^-}$$

So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water). Adding $H_2O_2$ should oxidize the iodide in this way: $$\ce{2I^- + 2H^+ + H_2O_2 -> I_2 + 2H_2O}$$ It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter, you can make the solution transparent reducing the colored compounds. If you have a look to the iodine clock reactions iodine can be reduced by bisulfite: $$\ce{I_2 + HSO_3^- + H2O -> 2I^- + HSO4^- +2H^+}$$

But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:

FIRST: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$), see Piranha solution (I think this could be very dangerous please take all the precautions!). But now I don't think that the equation that if I wrote before (see J. Am. Chem. Soc., 1931, 53 (1), pp 38–44 and J. Am. Chem. Soc., 1978, 100 (1), pp 87–91):

$$\ce{I_2 + 5H2O2 -> 2H^+ + 2IO3^- + 4H2O}$$

could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!

SECOND: simply you are modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the light absorption to take place.

I'm sorry this is now quite not an answer, but I hope it helps!

Iodine color reaction are wonderful (but very complex)!

• Hi I can't comment your question so I ask you here, what acid did you add? I will improve my question when I will have more infos! – G M Sep 11 '13 at 21:31
• (see comments above: he used H2SO4) – ManishEarth Sep 12 '13 at 12:17
• I was told in a french forum that the brown color comes from $I_{3}^{-}$ and that when we put $I_2$ in water there is an equilibrium $2I_2 \leftrightarrow I^+ + I_{3}^{-}$ which gives a slight brown color, slight because the equation is much more shifted to the left then to the right. Things change when we put $I_2$ in $(K^+, I^-)$, the brown color increases because of $I_2 + I^- \leftrightarrow I_{3}^{-}$ equilibrium. NOW I really still don't understand which reaction occures and why brown color disappear, is that because of $I_2$ disappearing or $I^-$ disappearing? – Pedro Sep 12 '13 at 12:59
• @pedro I never seen $2I_2↔I^+ +I_3^−$ equilibrium. I've seen $I^+$ in $I_2↔I^+ +I^-$ (J. Am. Chem. Soc., 1978, 100 (1), pp 87–91)) but this doesn't matter. I don't think however is more to the left because $I_2$ is not so soluble in water. I think $I^-$ could act only indirectly because is color less, I think something happen to $I_2$ or $I_3^-$. – G M Sep 13 '13 at 9:36
• I appreciate all these details, since I'm not chemist (but theoretical physicist so far from chemestry) I think that this experiment is far from being well suited for my classroom :-) I'll try to find another (less complex) slow reaction to study with my students. – Pedro Sep 13 '13 at 14:05