
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
498.3k+ views
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.
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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
