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 :
[b] \[{{\overset{0}{\mathop{Cl}}\,}_{2}}+{{H}_{2}}{{O}_{2}}\to +2\overset{-1}{\mathop{HCl}}\,+{{O}_{2.}}\]in this reaction chlorine is reduced from zero to -1 oxidation state.
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