A) \[PbS+4{{H}_{2}}{{O}_{2}}\to PbS{{O}_{4}}+4{{H}_{2}}O\]
B) \[2KI+{{H}_{2}}{{O}_{2}}\to 2KOH+{{I}_{2}}\]
C) \[2FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2{{H}_{2}}O\]
D) \[A{{g}_{2}}O+{{H}_{2}}{{O}_{2}}\to 2Ag+{{H}_{2}}O+{{O}_{2}}\]
Correct Answer: D
Solution :
In \[A{{g}_{2}}O\] (O.N. of \[Ag+1\]) in \[Ag\] the O.N. is O. There is gain of electrons, hence \[{{H}_{2}}{{O}_{2}}\] is lasting as reducing agent.
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