On reaction with stronger oxidizing agent like KIO₄, hydrogen peroxide oxidizes with the evolution of O₂. The oxidation number of I in KIO₄ changes to __________.

Solution

Chemistry ma’am · 4 steps

Here H₂O₂ is the reducing agent, so it pulls iodine down by only two steps, from +7 to +5.

1. Iodine to start. In KIO₄, K is +1 and each O is −2. So 1 + x + 4(-2) = 0, which gives x = +7. This is the highest oxidation state iodine can have, so KIO₄ can only be reduced.
2. Role of H2O2. The O in H₂O₂ is −1. Facing a stronger oxidising agent, it is oxidised to O₂ (0), so each H₂O₂ gives away 2 electrons. The periodate takes these 2 electrons and becomes iodate.

   KIO₄ + H₂O₂ → KIO₃ + H₂O + O₂

3. Iodine at the end. In KIO₃: 1 + x + 3(-2) = 0, so x = +5. The electrons balance: I goes from +7 to +5 and gains 2, and the two O atoms go from −1 to 0 and lose 2 in total.
4. So. The oxidation number of iodine changes from +7 to 5.

The common slip: 7. Reporting the starting oxidation number of iodine in KIO₄ gives 7. The question asks for the value after the reaction, in KIO₃.

The idea to keep. H₂O₂ works both ways. It is a reducing agent when the partner is a stronger oxidiser, and then its O goes from −1 to 0 as O₂. Periodate (+7) is reduced to iodate (+5).

Answer from the NTA final answer key. Solution by PrepWiser.