Disappearing iodine

Safety

• Put on protective gloves and eyewear.
• Conduct the experiment on the tray.

FAQ and troubleshooting

The iodine stain did not disappear. How can I fix it?

Try to add about 5-10 drops of Na₂S₂O₅ solution onto the stain. Wait a little longer than you did the first time (about 1 minute). Dry the stain with filter paper.

Step-by-step instructions

1. Pour a few drops of iodine 2,5% water solution onto a piece of paper.

2. Take the bottle with 0,5M sodium disulfite (Na₂S₂O₅) solution as a reductant and apply 5-10 drops onto the iodine stain.

3. Wait for 10 seconds.

4. Blot the stain with filter paper. The iodine stain disappears!

5. As a follow up, try to restore the stain: apply 1 drop 2М of hydrochloric acid HCl and 2 drops of 3% Hydrogen peroxide H₂O₂ solution to an experimental area.

Expected result

The iodine stain fades, as disodium disulfite (Na₂S₂O₅) reduces brown iodine (I₂) to colorless iodide ions (I^-).

Disposal

Dispose of solid waste together with household garbage. Pour solutions down the sink. Wash with an excess of water.

Scientific description

Why do iodine stains disappear?

We use water solution of disodium disulfite Na₂S₂O₅ to decolorize iodine stain. The following reaction takes place when disulfite interacts with iodine in water:

Na₂S₂O₅ + I₂ + H₂O → NaHSO₄ + HI

Molecular iodine acts as an oxidizer, and disodium disulfite as a reductant.

When iodine I₂ disappears from the system, brown-orange color of stains disappears as well.

More details:

This reaction can be divided into two parts: reduction of iodine and oxidation of disulfite:

S₂O₅^2– + 3H₂O – 4e^– → 2SO₄^2– + 6H⁺

I₂ + 2e^– → 2I^–

If we take into account the number of electrons in the first part, we should multiply the second one by two (to achieve so-called electronic balance):

2I₂ + 4e^– → 4I^–

When we express a reaction like we did it right above, separated into two parts (taking into account electrons, not just molecules or ions), we call these parts half-reactions.

Compare how the reaction equation looks before and after we have applied the half-reaction method:

Na₂S₂O₅ + I₂ + H₂O → NaHSO₄ + HI

Na₂S₂O₅ + 2I₂ + 3H₂O → 2NaHSO₄ + 4HI

What is oxidation?

Redox, or reduction-oxidation, reactions involve electron transfer between two particles (molecules, atoms or ions). An oxidizer (iodine) wants to get some electrons from a reductant (disulfite). Oxidation is an electron transfer process, in which an oxidizer (an element with relative electron deficiency) takes electrons away from an outer electron shell of a reductant (an element with relative electron excess). Therefore, an oxidizer is being reduced (gains electrons) during this process and a reductant oxidized (loses electrons).

More details:

We should introduce another concept to talk about redox processes: the oxidation state of an atom. Oxidation state is a positive or negative number attributed to an element in a compound. This number is a formal charge of an atom in the given compound.

Oxidation state is equal to zero in elementary substances such as iodine I₂. When we talk about a one-atom ion (for example, I^–, Mn⁺², H⁺, Na⁺, and K⁺), oxidation state is equal to the charge of this ion.

The sum of oxidation states of all the atoms in a neutral molecule is always equal to zero. The quantity of atoms should also be taken into account! For example, KMnO₄ has a total of 6 atoms: K⁺, Mn⁺⁷, and 4 atoms of O^-2. Hence, the sum is: +1 + (+7) + 4*(–2) = 0.

In a compound ion, this sum is equal to its total charge. For instance, HSO₄^– has H⁺, S⁺⁶, and 4 atoms of O^-2, so the sum is: +1 + (+6) + 4*(–2) = –1.

That’s interesting!

Is it possible to remove other stains with disulfite?

First of all, we want to remind you about safety rules. We highly recommend not conducting any experiments with substances that you are unfamiliar with.

Let us analyze which type of stains we can remove with disodium disulfite. In our case, iodine colors the stain. It acts as an oxidizer and is reduced by disulfite to iodine ion I^–. The stain becomes colorless. Hence, if we want to remove a stain with disodium disulfite, the stain color should be determined by an oxidizer (such as I₂). The oxidizer must be reduced by disulfite to a colorless or highly soluble in water (better both the same time!) form. For example, we can decolorize a purple potassium permanganate (KMnO₄) stain or a brown manganese dioxide (MnO₂) stain (see «Chemical eraser» experiment). Both of these compounds are reduced to soluble (and usually colorless) manganese Mn²⁺ salts:

MnO₂ + Na₂S₂O₅ + NaHSO₄ → MnSO₄ + Na₂SO₄ + H₂O

KMnO₄ + Na₂S₂O₅ + NaHSO₄ → MnSO₄ + Na₂SO₄ + H₂O

Does iodine dissolve in water?

Molecular iodine in its pure form is a highly volatile crystalline substance with deep violet color. I₂ is poorly soluble in water (approximately 0.3 gram per 1 liter). Therefore, iodine concentration in water is about 0.08%. However, we use 2.5% iodine aqueous solution in our experiment. How is it possible?

It turns out that when we add iodide ions to water (e.g. in form of KI salt), the solubility of iodine increases dramatically. This can be explained by the formation of I₃^– complex:

I₂ + I^– → I₃^–

This complex is highly soluble in water. At the same time, it is not very stable, so iodine is free to participate in reactions as I₂. For example, it can be reduced to iodide ion just as it happened in our experiment.