The Hydrogen Peroxide Rainbow

Demonstration Kit

Introduction

Produce a rainbow of colors by demonstrating how hydrogen peroxide can both oxidize and reduce. Hydrogen peroxide, a colorless solution, will cause another colorless solution to turn yellow, then red-orange. It also has the ability to make a purple solution go through a color change with the end solution being colorless.

Concepts

  • Oxidation–reduction
  • Oxidizing agent
  • Reducing agent

Materials

Hydrogen peroxide solution, H2O2, 6%*
Potassium iodide solution, KI, 0.12 M, 50 mL*
Potassium permanganate solution, KMnO4, 0.004 M, 50 mL*
Sulfuric acid solution, H2SO4, 1 M, 18 drops*
Beakers, 250-mL, 2
Graduated cylinders, 50- or 100-mL, 2
Overhead projector (optional)
Stirring rod or magnetic stirrer
*Materials included in kit. 

Safety Precautions

Treat the sulfuric acid solution as you would concentrated sulfuric acid; severely corrosive to skin, eyes and other tissue. Hydrogen peroxide solution is an oxidizer and corrosive to skin, eyes and other tissue. Potassium permanganate solution is an oxidizing agent and strong skin irritant; common cause of eye accidents. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron. Please review current Safety Data Sheets for additional safety, handling and disposal information.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. The solution from Part A may be reduced according to Flinn Suggested Disposal Method #12a. The solution from Part B may be neutralized according to Flinn Suggested Disposal Method #24b.

Procedure

Part A

  1. Using a graduated cylinder, measure out 50 mL of potassium iodide solution and transfer it to a 250-mL beaker.
  2. Add 18 drops of sulfuric acid solution to the beaker. Stir the contents with a stirring rod.
  3. Add the hydrogen peroxide solution dropwise to the beaker, stirring after each drop, until the solution turns red-orange. Note the color changes from colorless to yellow to red-orange. Approximately 40 drops of hydrogen peroxide solution will be needed to produce the color changes.

Part B

  1. Using a graduated cylinder, measure out 50 mL of potassium permanganate solution and transfer it to a clean, 250-mL beaker.
  2. Add 18 drops of sulfuric acid solution to the beaker and stir to mix the contents.
  3. Add the hydrogen peroxide solution dropwise to the beaker, stirring after each drop, until the solution becomes colorless. Note the color changes from purple to brown to colorless. Approximately 10 drops of hydrogen peroxide solution will be needed to produce the color changes.

Student Worksheet PDF

14126_Student1.pdf

Teacher Tips

  • Adding the hydrogen peroxide drop by drop, and stirring after the addition of each drop, helps students better observe the color changes that occur in both parts.
  • Make sure that the students can see all of the color changes, especially in Part B where the brown is a precipitate and the purple becomes lighter. Doing the demonstration in the vicinity of the students or on the overhead projector will make the demonstration more dramatic. Stir after the addition of each drop.
  • The KI solution may have an initial pale yellow tinge. This merely indicates the presence of trace I2.

Correlation to Next Generation Science Standards (NGSS)†

Science & Engineering Practices

Analyzing and interpreting data
Constructing explanations and designing solutions
Engaging in argument from evidence
Obtaining, evaluation, and communicating information

Disciplinary Core Ideas

MS-PS1.B: Chemical Reactions
HS-PS1.B: Chemical Reactions

Crosscutting Concepts

Cause and effect
Stability and change

Performance Expectations

MS-PS1-2. Analyze and interpret data on the properties of substances before and after the substances interact to determine if a chemical reaction has occurred.
HS-PS1-2. Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties.

Answers to Questions

  1. Describe what happened in this demonstration.

    First, hydrogen peroxide was added dropwise to a beaker containing potassium iodide solution and sulfuric acid solution. The color in the beaker changed from colorless to yellow to red-orange. Then, hydrogen peroxide was added dropwise to a beaker containing potassium permanganate solution and sulfuric acid solution. The color in this beaker changed from purple to brown to colorless.

  2. Write a balanced chemical equation for each of the following reactions.
    1. Hydrogen peroxide reacting with iodide ion in the presence of hydrogen ion in Part A

      H2O2(aq) + 2H+(aq) + 2I–(aq) → 2H2O(l) + I2(aq)

    2. Iodine combining with Iodide ions in Part A

      I–(aq) + I2(aq) → I3–(aq)

    3. Hydrogen peroxide reacting with permanganate in the presence of hydrogen ion in Part B

      5H2O2(aq) + 2MnO4–(aq) + 6H+(aq) → 8H2O(l) + 5O2(g) + 2Mn2+(aq)

  3. What substance is responsible for the intermediate yellow color of the solution in Part A?

    The free iodine produced by the initial reaction between hydrogen peroxide and iodide ion is responsible for the intermediate yellow color of the solution.

  4. Was hydrogen peroxide an oxidizing agent or a reducing agent in Part A? What about Part B? Explain.

    Hydrogen peroxide is an oxidizing agent in Part A and a reducing agent in Part B. In Part A, the hydrogen peroxide was reduced, meaning it gained electrons, and became water. In Part B, the hydrogen peroxide is oxidized, meaning it lost electorons.

Discussion

The reactions that take place in Parts A and B are oxidation–reduction (redox) reactions. Hydrogen peroxide can act as both a strong oxidizing agent and a weak reducing agent. A reducing agent is oxidized in a redox reaction. This means it loses electrons, causing its oxidation number to increase. An oxidizing agent is reduced in a redox reaction, thereby gaining electrons and causing its oxidation number to decrease.

Hydrogen peroxide acts as an oxidizing agent in Part A. Hydrogen peroxide is reduced to water and the iodide ion is oxidized to iodine. The solution changes from colorless to yellow, indicating the presence of free iodine (I2) in solution. The iodine combines with iodide ions (I–) to form red-orange I3 – ions.

Part A Reactions

{14126_Discussion_Equation_1}
In Part B, hydrogen peroxide, in the presence of strong oxidizer permanganate, MnO4–, now acts as a reducing agent, reducing Mn(7+) to Mn2+. Mn2+ is not initially produced. Mn(7+) is first reduced to Mn(4+) producing a brown precipitate in solution. Mn(4+) is then reduced to Mn2+, giving a clear solution.

Part B Reactions
{14126_Discussion_Equation_3}

References

Special thanks to Jim and Julie Ealy, The Peddie School, Hightstown, NJ, who provided Flinn Scientific with the instructions for this activity.

Bray, W. C. J. Am. Chem. Soc. 1921, 43, 1262.

Porterfield, W. W. Inorganic Chemistry, 1984, 338.

Stone, C. H. J. Chem. Educ. 1944, 21, 301.

†Next Generation Science Standards and NGSS are registered trademarks of Achieve. Neither Achieve nor the lead states and partners that developed the Next Generation Science Standards were involved in the production of this product, and do not endorse it.