A) \[2FeC{{l}_{2}}+2HCl+{{H}_{2}}{{O}_{2}}\to 2FeC{{l}_{3}}+2{{H}_{2}}O\]
B) \[C{{l}_{2}}+{{H}_{2}}{{O}_{2}}\to 2HCl+{{O}_{2}}\]
C) \[2Hl+{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{l}_{2}}\]
D) \[{{H}_{2}}S{{O}_{3}}+{{H}_{2}}{{O}_{2}}\to {{H}_{2}}S{{O}_{4}}+{{H}_{2}}O\]
Correct Answer: B
Solution :
\[C{{l}_{2}}+{{H}_{2}}{{\overset{-1}{\mathop{O}}\,}_{2}}\to 2HCl+{{\overset{0}{\mathop{O}}\,}_{2}}\]Oxidation state of oxygen changes from -1 \[({{H}_{2}}{{O}_{2}})\] to zero \[({{O}_{2}})\] i.e., \[{{H}_{2}}{{O}_{2}}\] getting oxidised and acting as reducing agent.
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