Iodine from Sodium Iodide
Summary
A mild oxidizer turns colourless iodide into orange-brown elemental iodine — an oxidation1 you can confirm with the dramatic blue-black starch test. You’ll be able to explain why iodide is so easily oxidized (and chloride isn’t), what the acid is for, and how vitamin C reverses the whole thing.
History
Iodine was discovered in 1811 by Bernard Courtois while working in his family’s saltpeter factory in Paris. He accidentally added too much sulfuric acid to seaweed ash and observed a striking violet vapor that condensed into dark metallic crystals — the first isolation of iodine.
The element takes its name from the Greek ioeides (violet-colored), reflecting the color of its vapor. Within a decade, chemists had established that iodine forms a deep blue-black complex with starch — one of the most sensitive and visually dramatic color tests in chemistry, still used today in biochemistry, food science, and medical diagnostics.
This experiment demonstrates that same chemistry: a mild oxidant converts colorless iodide (I⁻) to brown elemental iodine (I₂), and the starch test then confirms what the eye can already see.
Hazards & preparation
PPE: safety glasses and gloves — iodine stains skin and clothing brown.
- Work in a well-ventilated area; don’t inhale vapours.
- Use only dilute (household) hydrogen peroxide and dilute acid — never mix concentrated oxidizers with concentrated acid.
Disposal: quench any leftover brown solution with a pinch of vitamin C to reduce the iodine back to harmless iodide, then flush down the drain with water. See the Safety page.
Materials
- Sodium iodide (NaI) — 0.5 g dissolved in 20 mL water
- Hydrochloric acid (10%) — 1.5 mL
- Hydrogen peroxide (3%, household) — 2 mL
- Cornstarch solution (optional) — 1 tsp in 50 mL hot water, cooled
- Beaker or glass — 50 mL
- Gloves, safety glasses
Procedure
- Dissolve 0.5 g sodium iodide in 20 mL water — a colorless solution.
- Add 1.5 mL dilute hydrochloric acid — still colorless.
- Add 2 mL hydrogen peroxide — the solution turns orange-brown within seconds as iodine forms.
- Optional starch test: add a few drops of the starch solution — it turns deep blue-black, confirming iodine.
- Extension: add a pinch of ascorbic acid (vitamin C) — the colour vanishes instantly as the iodine is reduced back to iodide. Add more peroxide and it returns.
What you should see
Nothing happens on adding the acid; then, seconds after the peroxide goes in, the clear liquid blooms orange-brown. A drop of starch turns it inky blue-black, and a pinch of vitamin C wipes it colourless again in an instant.
| Symptom | Likely cause | Fix |
|---|---|---|
| Stays colorless after peroxide | No acid added, or peroxide too old/weak | Add the acid first; use fresh 3% peroxide |
| Colour very faint | Too little iodide | Use a bit more sodium iodide |
| Starch gives no blue | Starch too old, or solution too warm | Make fresh starch; let the solution cool (the blue fades when hot) |
The reactions
\[\ce{2 NaI + H2O2 + 2 HCl -> I2 + 2 NaCl + 2 H2O}\]
Net ionic form:
\[\ce{2 I^- + H2O2 + 2 H+ -> I2 + 2 H2O}\]
iodide + hydrogen peroxide + acid → iodine + water
The Science
Hydrogen peroxide is a mild oxidizing agent that converts colorless iodide ions (I⁻) into elemental iodine (I₂), which dissolves in water to give the orange-brown color. The acid supplies H⁺ ions consumed in the reaction — without them the reaction is slow and incomplete.
This is a clean example of halide oxidation: iodide is oxidized (loses electrons) while hydrogen peroxide is reduced (gains electrons) to water. Iodine is the easiest halide to oxidize because iodide has the highest energy electrons of the halide series; chloride and bromide require stronger oxidants.
The starch–iodine complex forms when I₂ slips into the helical coils of amylose (the linear component of starch), shifting its absorption into the visible range and producing the characteristic blue-black color. The effect disappears on heating (the helix unwinds) and reappears on cooling — a thermochromic indicator.
Questions to Explore
Why is iodide easy to oxidize but chloride is not? The same hydrogen peroxide converts iodide to iodine but won’t touch chloride here. As you go down the halogen group (F → Cl → Br → I), what happens to the outermost electrons, and how does that make iodide a better reducing agent?
Hint / answer
Going down the group the outer electrons sit further from the nucleus and are held more loosely, so iodide gives up an electron easily — it’s readily oxidized. Chloride’s electrons are held tightly, so only a much stronger oxidizer can pull them off.
What role does the acid play? The reaction uses HCl to supply H⁺. Looking at the equation, where do these protons end up — and what happens to the rate if you leave the acid out?
Hint / answer
The H⁺ ions combine with the oxygen from peroxide to make water. They’re a reactant, so without acid the reaction is starved of protons and runs slowly and incompletely — you’d get little or no colour.
Why does the starch–iodine complex turn blue-black? Iodine threads into the helical coils of amylose. What happens to the iodine electrons when confined in the helix that causes such a dramatic colour shift?
Hint / answer
Lined up single-file inside the starch helix, the iodine forms long chains whose electrons can absorb light right across the visible range — especially reds and oranges — so the complex looks blue-black. Free iodine only absorbs a narrower band and looks brown.
Why does the blue color vanish when heated? The complex breaks up when hot and reforms on cooling. Does this suggest a strong chemical bond or a weaker association?
Hint / answer
A weak association. Heat jiggles the starch helix apart and shakes the iodine loose, and it only re-forms when things cool and settle again — a strong covalent bond wouldn’t come and go so easily with temperature.
Why did vitamin C make the color disappear instantly? Ascorbic acid bleaches the brown iodine by reducing I₂ back to I⁻. How does this relate to vitamin C’s job as an antioxidant in food?
Hint / answer
Vitamin C readily donates electrons, converting iodine back to colourless iodide. That same electron-giving is what makes it an antioxidant: it sacrifices itself to reducing agents so that oxygen and other oxidizers attack it instead of spoiling the food.
Going further
- Titrate the vitamin C. Add the brown iodine solution drop by drop to a known vitamin-C sample until the colour just persists — a simple redox titration2 for antioxidant content.
- Next in the Redox track: time a redox reaction with the sudden colour change of the Iodine Clock.