Aqueous solutions of Iodine and Sodium hydroxide were mixed in a round bottom flask at 70 C. Following chemical reaction was carried out. 3I2 + 6 NaOH → NaIO3 + NaI + H2OThis reaction is termed as:

Explanation

To determine whether such a reaction is a redox reaction, the oxidation states of the reactants and products is compared. The oxidation state of Na remains +1 throughout, as does the oxidation states of Hydrogen (+1) and Oxygen (-2). Iodine undergoes a change in oxidation state. Its oxidation state is 0, in its molecular form, on the reactant side. In NaI, iodine's oxidation state is -1 so, Iodine has undergone reduction. In NaIO3, the oxidation states of Na remain +1 and oxygen's also remains -2 so, we can determine that the oxidation state of iodine is +5 in this reaction. (NOTE: For oxidation numbers, the total oxidation state of a compound is zero so, Iodine's oxidation number must be +5 to ensure NaIO3's oxidation state is zero). Since Iodine underwent oxidation AND reduction, we can conclude that this is a redox reaction (disproportionation to be precise, as the same species has undergone oxidation and reduction) so, option A is correct. Option B is incorrect, as the two salts produced are soluble in water. Option C does not describe this reaction. Option D is incorrect as no free radicals are produced during this reaction.