Question

In the reaction, $HAS{O}_2+S{n}^{2+}\to As+A{n}^{4+}+H_{2}O$ Oxidising agent is:

• A. $S{n}^{2+}$
• B. $S{n}^{4+}$
• C. $As$
• D. $HAs{O}_2$

Answer 1

Answer: (D)

Oxidising agent itself undergoes reduction during a redox reaction.
Oxidation state of $A$s in $HAs{O}_2$ is $=x$
$\therefore$ $1+x+(2\times -2)=0$
or $x=3$ Oxidation state of $As=0$
Oxidation state of $Sn$ in $S{n}^{2+}=+2$
Oxidation state of $Sn$ in $S{n}^{4+}=+4$
$\therefore$ Increase in oxidation number is oxidation.
$\therefore$ $S{n}^{2+}$ is oxidised to $S{n}^{4+}$
during reaction and it is reducing agent.
$\because$ Decrease in oxidation number is reduction.
$\therefore$ $HAs{O}_2$ is reduced to $As$ and it is oxidising agent.