Question

Consider the following unbalanced redox reaction:

$H_{2}O+AX+BY⟶HA+OY+X_{2}B$

The oxidation number of $X$ is $-2$ and neither $X$ nor water is involved in the redox process. The possible oxidation states of $B$ and $Y$ in $BY$ are, respectively :

• A. +1, -1
• B. +2, -2
• C. +3, -3
• D. All of these

Answer 1

The correct option is D All of these

Oxidation state of $X=-2\Rightarrow \stackrel{+2}{A}\stackrel{-2}{X}$
Oxidation state of $O=-2\Rightarrow \stackrel{-2}{O}\stackrel{+2}{Y}$
Oxidation state of $X=-2\Rightarrow \stackrel{-2\times 2}{X_2}\stackrel{+4}{B}$
$e^{-}+A^{2+}⟶A^{1-}$ (reduction)
$\stackrel{+a}{B}\stackrel{-a}{Y}⟶B^{4+}+Y^{2+}$ (oxidation)
Therefore, $B$ or $Y$ or both might have been oxidised.
In $BY$, the oxidation state of $B\le +4$
Oxidation state of $Y\le +2$.
If the oxidation state of $B$ is $+1$ and that of $Y$ is $-1$, then both will be oxidised.
If the oxidation state of $B$ is $+2$ and that of $Y$ is $-2$, then both will be oxidised.
If the oxidation state of $B$ is $+3$ and that of $Y$ is $-3$, then both will be oxidised.