Question

Which of the following species are oxidised and reduced in the following (unbalanced) reaction?
${CN}^{-}+O{Cl}^{-}⟶N_2+H{CO}_3^{-}+{Cl}^{-}$

• A. Oxidized- $C$, $N$ in $C{N}^{-}$, reduced - ${OCl}^{-}$
• B. Oxidized- $N$, ${CN}^{-}$, $H{CO}_3^{-}$; $N_2$ reduced - $C$ ${OCl}^{-}$
• C. Oxidized- $C$, $N$, $H{CO}_3^{-}$; $N_2$ reduced - ${OCl}^{-}$ , ${CN}^{-}$
• D. None of these

Answer 1

The correct option is A Oxidized- $C$, $N$ in $C{N}^{-}$, reduced - ${OCl}^{-}$

Oxidation number of C in $C{N}^{-}$ is +2, N in $C{N}^{-}$ is -3, O in $OC{l}^{-}$ is -2, Cl in $OC{l}^{-}$ is +1, N in $N^2$ is 0, C in $HC{O}_3^{-}$ is +4 and of Cl in $C{l}^{-}$ is -1.

Oxidation of $C$ ($O.N=+2$) and $N$ ($O.N=-3$) of ${CN}^{-}$ into $H{CO}_3^{-}$ ($O.N$ of $C=+4$) and $N_2$ ($O.N$ of $N=0$) respectively, takes place.

Reduction of ${OCl}^{-}$ ($O.N$ of $Cl=+1$) into ${Cl}^{-}$ ($O.N$ of $Cl=-1$) occurs in the given reaction.
Hence, option A is correct.