Final answer:
Chlorine is reduced (oxidation number decreases from +5 to -1), while bromine is oxidized (oxidation number increases from -1 to 0). HBr serves as the reducing agent and KClO₃ is the oxidizing agent in the given reaction.
Step-by-step explanation:
In order to identify the elements that are oxidized and reduced in the reaction KClO₃(aq) + 6HBr(aq) → KCl(aq) + 3Br₂(l) + 3H₂O(l), we need to look at the changes in the oxidation numbers of the elements involved in the reaction.
First, the oxidation number of chlorine (Cl) in KClO₃ is +5 (since K has +1 and O has -2 for a total of -6 in O₃, the chlorine must be +5 to balance it out). In KCl, the oxidation number of Cl is -1. Therefore, chlorine has been reduced (gain of electrons), as its oxidation number has decreased from +5 to -1.
On the other hand, the oxidation number of bromine (Br) in HBr is -1. In Br₂, each bromine has an oxidation number of 0 because it is in its elemental state. Therefore, bromine has been oxidized (loss of electrons), since its oxidation number has increased from -1 to 0.
The substance that gets oxidized acts as the reducing agent, while the substance that gets reduced acts as the oxidizing agent. Thus, HBr is the reducing agent and KClO₃ is the oxidizing agent in this reaction.