# How can iodine oxidise hydrogen peroxide?

The following is mentioned in NCERT (government textbook in India used in all schools) under chemical properties of $$\ce{H2O2}:$$

(iv) Reducing action in basic medium

$$\ce{I2 + H2O2 + 2 OH- -> 2 I- + 2 H2O + O2}$$

I feel the reaction is not correct. Since $$\ce{H2O2}$$ is a moderate/strong oxidizing agent, I don't think it can be oxidized by iodine and the reverse reaction should be true since $$\ce{O2}$$ is a strong oxidising agent and $$\ce{I-}$$ is a reducing agent. Can someone please verify it?

Harsh Jain, It may be a correct reaction, but it is a very complicated one.

Let us see what happens when iodine is added to a large volume of an aqueous base in the absence of $$\ce{H2O2}.$$ It forms iodide and iodate ion. If iodine is in excess, triiodide ion is also formed.

The complication arises from adding hydrogen peroxide because it can act as an oxidizing agent as well as a reducing agent. Similarly, iodate ion and iodide ion can react with each other as well as $$\ce{H2O2}.$$ For example, iodine can react with $$\ce{H2O2}$$ and form iodide ions. On the other hand, iodide ion can be oxidized by $$\ce{H2O2}$$ to free iodine.

I would suggest that you read a paper by Bray and Liebhafsky [1]. Here is the summary by the authors in their table:

### References

1. Bray, W. C.; Liebhafsky, H. A. Reactions Involving Hydrogen Peroxide, Iodine and Iodate Ion. I. Introduction. J. Am. Chem. Soc. 1931, 53 (1), 38–44. DOI: 10.1021/ja01352a006.

First, the slow action of iodine with water:

$$\ce{I2 + H2O -> H+ + I- + HOI}$$

In the presence of $$\ce{OH-},$$ the above equilibrium is moved to the right leaving $$\ce{HOI}.$$

Finally, a fast interaction with $$\ce{H2O2}$$ (which also occurs with $$\ce{HOCl},$$ (see [1, p. 1509]):

$$\ce{H2O2 + HOI -> H2O + O2(g) + H+ + I-}$$

### References

1. Connick, R. E. The Interaction of Hydrogen Peroxide and Hypochlorous Acid in Acidic Solutions Containing Chloride Ion. J. Am. Chem. Soc. 1947, 69 (6), 1509–1514. DOI: 10.1021/ja01198a074.