Question

In the following reaction:
$Cr(OH{)}_3+(OH{)}^{-}+{I{O}_3}^{-}\to {Cr{O}_4}^{2-}+H_{2}O+I^{-}$

• A. ${I{O}_3}^{-}$ is oxidising agent
• B. $Cr(OH{)}_3$ is oxidised
• C. $6e^{-}$ are being taken per $I$ atom
• D. All of the above are correct

Answer 1

The correct option is D All of the above are correct

The redox reaction is shown below:
$Cr(OH{)}_3+(OH{)}^{-}+{I{O}_3}^{-}\to {Cr{O}_4}^{2-}+H_{2}O+I^{-}$
The oxidation number of $I$ changes from $+5$ to $-1$.
Thus, each iodine takes $6$ electrons and is reduced.
Hence, ${I{O}_3}^{-}$ is the oxidizing agent.
The oxidation number of chromium changes from $+3$ to $+6$.
Thus, $Cr(OH{)}_3$ is oxidized.