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 iodine atom
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $I{O}_3^{-}$ is oxidising agent
B $Cr(OH{)}_3$ is oxidised
C $6e^{-}$ are being taken per iodine atom

Oxidation number of chromium in reactant side of reaction is $+3$, whereas in product side it has $+6$ oxidation state. Hence, we can conclude that $Cr(OH{)}_3$ is oxidized.
$I$ has oxidation $+5$ in reactant side of reaction and $-1$ on the product side, so it has reduced in the reaction. By this we can say that $I{O}_3^{-}$ is the oxidising agent
Electron gain $=$ (initial oxidation state of $I$ $-$ final oxidation state of $I$) $=$ $6$.