Question

Which of the following chemical reactions depicts the oxidising behaviour of $H_{2}S{O}_4$?

• A. $2HI+H_{2}S{O}_4\to I_2+S{O}_2+2H_{2}O$
• B. $Ca(OH{)}_2+H_{2}S{O}_4\to CaS{O}_4+2H_{2}O$
• C. $NaHS{O}_4+HCl\to NaCl+H_{2}S{O}_4$
• D. $2PC{l}_5+H_{2}S{O}_4\to 2POC{l}_3+2HCl+S{O}_{2}C{l}_2$

Answer 1

The correct option is A $2HI+H_{2}S{O}_4\to I_2+S{O}_2+2H_{2}O$

(a) In the reaction,
$2\stackrel{+1}{H}\stackrel{-1}{I}+{\stackrel{-+1}{H}}_2\stackrel{+6}{S}{\stackrel{-2}{O}}_4\to {\stackrel{0}{I}}_2+\stackrel{+4}{S}{\stackrel{-2}{O}}_2+2{\stackrel{+1}{H}}_2\stackrel{-2}{O}$
1st half reaction: $2H\stackrel{-1}{I}\to \stackrel{0}{I_2}$
In this reaction oxidation number of $I$ increases by one, thus this is an oxidation reaction and $HI$ behaves as a reducing agent.
IInd half reaction : $H_2\stackrel{+6}{S}{O}_4\to \stackrel{+4}{S}{O}_2$
In this reaction oxidation number of S decreases by two, thus this is a reduction reaction and $H_{2}S{O}_4$ behaves as oxidising agent.
(b) $H_{2}S{O}_4\to CaS{O}_4$
Oxidation state of Sulphur changes from $+6\to +6$
Hence, it will not show oxidising behaviour.
(c) $NaHS{O}_4\to H_{2}S{O}_4$
Oxidation state of Sulphur changes from $+6\to +6$, hence $H_{2}S{O}_4$ will not show oxidising or reducing behavouir.
(d) $H_{2}S{O}_4\to S{O}_{2}C{l}_2$
Oxidation state of sulphur changes from $+6\to +6$, hence it will not show oxidising or reducing behavouir.