A) \[{{H}_{2}}{{O}_{2}}+S{{O}_{2}}\to {{H}_{2}}S{{O}_{4}}\]
B) \[2KI+{{H}_{2}}{{O}_{2}}\to 2KOH+{{I}_{2}}\]
C) \[PbS+4{{H}_{2}}{{O}_{2}}\to PbS{{O}_{4}}+4{{H}_{2}}O\]
D) \[A{{g}_{2}}O+{{H}_{2}}{{O}_{2}}\to 2Ag+{{H}_{2}}O+{{O}_{2}}\]
Correct Answer: D
Solution :
[d] \[S{{O}_{2}}\] changes to \[{{H}_{2}}S{{O}_{4}}\] (O.N. changes from +4 to +6 oxidation) \[2KI\to {{I}_{2}}\] (O.S. changes from \[-1\] to 0 oxidation) \[PbS\to PbS{{O}_{4}}\] (O.S. changes from \[-2\] to +6 oxidation) \[A{{g}_{2}}O\to 2Ag\] (O.S. changes from +1 to 0 reduction)
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