question_answer

In the reaction, \[HAS{{O}_{2}}+S{{n}^{2+}}\xrightarrow{{}}As+S{{n}^{4+}}+{{H}_{2}}O\] Oxidising agent is:

A)  \[S{{n}^{2+}}\]         

B)  \[S{{n}^{4+}}\]

C)  As           

D)  \[SAs{{O}_{2}}\]

Correct Answer: D

Solution :

Key Idea: Oxidising agent itself undergoes reduction during a redox reaction.  Oxidation state of As 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.