Disappearing iodine

Disodium disulfite, a reducing agent, removes iodine stains

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


Disappearing iodine - disodium disulfite, a reducing agent, removes iodine stains

by MEL Science


Put on gloves before beginning the experiment. Perform the experiment on a tray.

Always follow general safety recommendations. Please note that conducting chemistry experiments you must comply with the relevant legal procedures in your country.

Perform this experiment

Reaction formula

Na2S2O5 + H2O + I2 → NaHSO4 + HI

Step-by-step instruction

  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!

Scientific background

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:

S2O52-+ 3H2O -4e- → 2SO42- + 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 O2-. 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+, S6+, and 4 atoms of O2-, so the sum is: +1 + (+6) + 4*(–2) = –1.


Published on 16 January 2016

  • Fire
  • Heating with fire
  • Explosion
  • Poisoned gas
  • Organic
  • Electricity
  • Solution
  • Oxidation reduction
  • Color change
  • Precipitate
  • Gassing
  • Catalyst