Question

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. ${{H}_{2}}S{{O}_{4}}$

Answer 1

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.