Question

Explain the reactions:

$2KCl{O}_3+I_2\to 2KI{O}_3+C{l}_2$

Answer 1

1. The species which undergoes oxidation or loss of electrons is a reducing agent.
2. And the species which undergoes reduction or gain of electrons oxidizing agent.
3. In the given reaction oxidation and reduction are taking place.
4. The reaction can be split as,

$$\begin{gathered} Cathode:2C{l}^{+5}+10e^{-}\to \underset{Chlorine}{C{l}_2\left(g\right)} \\ Anode:\underset{Iodine}{I_2\left(g\right)}\to 2I^{+5}+10e^{-} \end{gathered}$$

So, iodine is undergoing Oxidation so it is the Reducing agent and Chlorine is undergoing reduction so it is the Oxidizing agent.