In the reaction: I_2 + 2S_2O_3^{2-} to 2I^{-} | vedclass

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(B) In the given reaction: $I_2 + 2S_2O_3^{2-} 	o 2I^{-} + S_4O_6^{2-}$$1$. The oxidation state of $I$ in $I_2$ changes from $0$ to $-1$ in $I^{-}$.

Vedclass · Accepted Answer

$I_2$ is an oxidising agent and $S_2O_3^{2-}$ is a reducing agent.

Vedclass · Answer

(B) In the given reaction: $I_2 + 2S_2O_3^{2-} 	o 2I^{-} + S_4O_6^{2-}$$1$. The oxidation state of $I$ in $I_2$ changes from $0$ to $-1$ in $I^{-}$. Since the oxidation state decreases,$I_2$ undergoes reduction and acts as an oxidising agent.$2$. The oxidation state of $S$ in $S_2O_3^{2-}$ changes from $+2$ to $+2.5$ in $S_4O_6^{2-}$. Since the oxidation state increases,$S_2O_3^{2-}$ undergoes oxidation and acts as a reducing agent.

In the reaction: $I_2 + 2S_2O_3^{2-} \to 2I^{-} + S_4O_6^{2-}$

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