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 :
\[\underset{0}{\mathop{C{{l}_{2}}}}\,+{{H}_{2}}{{O}_{2}}\to \underset{-1}{\mathop{2HCl}}\,+{{O}_{2}}\] In this reaction \[{{H}_{2}}{{O}_{2}}\] works as reducing agent
You need to login to perform this action.
You will be redirected in
3 sec
