Test Index

Redox Reactions

Section: Chemistry

Question : 73 of 73

Marks: +1, -0

\mathrm{H}_2\mathrm{SO}_4

$\mathrm{NaCl} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4 + \mathrm{HCl}$
$2\mathrm{PCl}_5 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{POCl}_3 + 2\mathrm{HCl} + \mathrm{SO}_2\mathrm{Cl}_2$
$2\mathrm{HI} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{I}_2 + \mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}$
$\mathrm{Ca}(\mathrm{OH})_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + 2\mathrm{H}_2\mathrm{O}$

Solution:

2\mathrm{HI}^{-1} + \mathrm{H}_2\overset{+6}{\mathrm{SO}_4} \rightarrow \mathrm{I}_2^0 + \overset{+4}{\mathrm{SO}_2} + 2\mathrm{H}_2\mathrm{O}

in this reaction oxidation number of

\mathrm{S}

is decreasing from

+6

+4

hence undergoing reduction and for

\mathrm{H} \mathrm{I}

oxidation Number of

\mathrm{I}

is increasing from

-1

to 0 hence undergoing oxidation therefore

\mathrm{H}_2\mathrm{SO}_4

is acting as oxidising agent.

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