In the reaction 3I_2 + 6NaOH to NaIO_3 + 5NaI | vedclass

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(C) Assigning oxidation states: $3\overset{0}{I}_2 + 6NaOH 	o Na\overset{+5}{I}O_3 + 5Na\overset{-1}{I} + 3H_2O$. In this reaction,the oxidation stat

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$I_2$

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(C) Assigning oxidation states: $3\overset{0}{I}_2 + 6NaOH 	o Na\overset{+5}{I}O_3 + 5Na\overset{-1}{I} + 3H_2O$. In this reaction,the oxidation state of iodine $(I)$ changes from $0$ to $+5$ (oxidation) and from $0$ to $-1$ (reduction). Since $I_2$ undergoes both oxidation and reduction (disproportionation reaction),it acts as both the oxidizing agent and the reducing agent. Therefore,$I_2$ is the oxidizing agent.

In the reaction $3I_2 + 6NaOH \to NaIO_3 + 5NaI + 3H_2O$,which substance acts as the oxidizing agent?

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