Which of the following molecule can act as both an oxidizing agent as well as reducing agent:
A. \[{{H}_{2}}S\]
B. \[S{{O}_{3}}\]
C. \[{{H}_{2}}{{O}_{2}}\]
D. \[{{F}_{2}}\]
E. \[{{H}_{2}}S{{O}_{4}}\]

Hint: To find a substance that can act as both a reducing agent and an oxidizing agent it should possess a higher as well as a lower oxidation state which is stable and can be attained by undergoing a redox reaction with other compounds.


Complete answer:
A substance can act as oxidising as well as reducing agent when it is present in its intermediate oxidation state.

A. In \[{{H}_{2}}S\], sulphur is present in \[-2\] oxidation state, which means it cannot go on to any lower oxidation state because it has no stable oxidation state less than\[-2\] that is \[-2\] is its minimum oxidation state. So it cannot act as an oxidising agent because it cannot get reduced further.

B. In \[S{{O}_{3}}\]​, sulphur is present in \[+6\] oxidation state, which means it cannot go on to any higher oxidation state because it has no stable oxidation state more than \[+6\] that is \[+6\] is its maximum oxidation state. So it cannot act as a reducing agent because it cannot get oxidised further.

C. In \[{{H}_{2}}{{O}_{2}}\]​, oxygen is present in \[-1\] oxidation state, which means it can go on to a higher oxidation state because it has a stable oxidation state more than \[-2\] which is 0 in \[{{O}_{2}}\] molecule and it can go on to a lower oxidation state because it has a stable oxidation state less than \[-1\] which is \[-2\] in \[{{H}_{2}}O\] molecule. So it can act as oxidising as well as reducing agent.

D.\[{{F}_{2}}\]​ has fluorine present in 0 oxidation state, which means it cannot go on to any higher oxidation state because it has no stable oxidation state more than 0 that is 0 is its maximum oxidation state. So it cannot act as a reducing agent because it cannot get oxidised further.

E. In \[{{H}_{2}}S{{O}_{4}}\]​, sulphur is present in \[+6\] oxidation state, which means it cannot go on to any higher oxidation state because it has no stable oxidation state more than \[+6\] that is \[+6\] is its maximum oxidation state. So it cannot act as a reducing agent because it cannot get oxidised further.

So \[{{H}_{2}}{{O}_{2}}\]​ can act as both oxidising as well as reducing agent. Hence, the correct answer is \[{{H}_{2}}{{O}_{2}}\] which is option C.

Note: Oxidation number which is also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. It is a positive or negative number that represents the effective charge of an atom or element which further indicates the extent or possibility of its oxidation or reduction.