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) \[2HI+{{H}_{2}}{{O}_{2}}\,\to \,\,2{{H}_{2}}O+{{I}_{2}}\]
D) \[{{H}_{2}}S{{O}_{3}}+{{H}_{2}}{{O}_{2}}\,\to \,\,{{H}_{2}}S{{O}_{4}}+{{H}_{2}}O\]
Correct Answer: B
Solution :
\[C\overset{o}{\mathop{{{l}_{2}}}}\,+{{H}_{2}}{{O}_{2}}\to \overset{\,\,\,\,\,\,\,\,\,-1}{\mathop{2HCl}}\,+{{O}_{2}}\]. In this reaction chlorine reduced from zero to ? 1 oxidation state.
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