question_answer

In the reaction \[{{H}_{2}}{{O}_{2}}\xrightarrow{{}}S+2{{H}_{2}}O\]

A)  \[{{H}_{2}}S\] is an acid and\[{{H}_{2}}{{O}_{2}}\]is a base

B)  \[{{H}_{2}}S\]is a base and\[{{H}_{2}}{{O}_{2}}\]is an acid

C) \[{{H}_{2}}S\]is an oxidizing agent and\[{{H}_{2}}{{O}_{2}}\]is a reducing agent

D)  \[{{H}_{2}}S\] is a reducing agent and\[{{H}_{2}}{{O}_{2}}\]is an oxidizing agent

E) \[{{H}_{2}}S\]is hydrolyzed to S

Correct Answer: D

Solution :

Here, the oxidation number of S increases from -2 in\[{{H}_{2}}S\]to 0 in elemental sulphur, while that of O decreases from -1 in\[{{H}_{2}}O\]to -2 in\[{{H}_{2}}O\],therefore\[{{H}_{2}}S\]is a reducing agent and \[{{H}_{2}}{{O}_{2}}\]is an oxidizing agent.