A) \[2HI+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{I}_{2}}+S{{O}_{2}}+2{{H}_{2}}O\]
B) (b)\[Ca(OH)+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}CaS{{O}_{4}}+2{{H}_{2}}O\]
C) \[NaCl+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}NaHS{{O}_{4}}+HCl\]
D) (d)\[2PC{{l}_{5}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}2POC{{l}_{3}}+2HCl+S{{O}_{2}}C{{l}_{2}}\]
Correct Answer: A
Solution :
In the reaction, Ist half reaction: \[2\underset{+6}{\mathop{HI}}\,\xrightarrow{{}}\underset{0}{\mathop{{{I}_{2}}}}\,\] In this reaction oxidation number of 1 increases by one, thus this is an oxidation reaction and HI behaves as a reducing agent. IInd half reaction: \[\underset{+6}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\xrightarrow{{}}\underset{+4}{\mathop{S{{O}_{2}}}}\,\] On the other hand, in this reaction, oxidation number of S decreases by two, thus this is a reduction reaction and \[{{H}_{2}}S{{O}_{4}}\] behaves as oxidizing agent.You need to login to perform this action.
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