The correct option is
C H2O2A species can act as oxidizing as well as reducing species, if it can get reduced and oxidized further, respectively.
A. H2O2:
oxidation number of O is: 2×1+2x=0
x=−1
Since O can have oxidation number as -2 and 0 when reduced and oxidized respectively as shown below in the reaction.
Reduction of peroxide:
-1 -2
H2O2+2H++2e−→2H2O
Oxidation of peroxide:
-1 0
H2O2→O2+2H++2e−
B. SO3: oxidation number of S is:
x+3×(−2)=0
x=+6
maximum oxidation state of S is +6 and can not be reduced further, thus can act as only reducing agent and not as oxidizing agent.
C. H2SO4: oxidation number of S is:
2×1+x+4×(−2)=0
x=+6
maximum oxidation state of S is +6 and can not be reduced further, thus can act as only reducing agent and not as oxidizing agent.
D. HNO3: oxidation number of N is:
1+x+3×(−2)=0
x=+5
maximum oxidation state of N is +5 and can not be reduced further, thus can act as only reducing agent and not as oxidizing agent.
therefore H2O2 can act as oxidising as well as reducing agent.