Question

In which of the following reactions of $H_{2}S{O}_4$ sulphur retains its oxidation state?

• A. Reaction of dilute $H_{2}S{O}_4$ with sulphur
• B. Reaction of conc. $H_{2}S{O}_4$ with sulphur
• C. Reaction of conc. $H_{2}S{O}_4$ with glucose
• D. Reaction of dilute $H_{2}S{O}_4$ with zinc

Answer 1

Answer: (C), (D)

Reaction of conc. $H_{2}S{O}_4$ with glucose
Reaction of dilute $H_{2}S{O}_4$ with zinc

$C_{12}{H}_{22}{O}_{11}\stackrel{H_{2}S{O}_4}{\to }12C+11H_{2}O$

Here, $H_{2}S{O}_4$ indirectly participates and obtains as mixture of water and acid. Its oxidation state remains unchanged in this mixture.

$Zn\left(s\right)+H_{2}S{O}_{4}aq\to ZnS{O}_4\left(aq\right)+H_2$

In $H_{2}S{O}_4$, $S$ oxidation state is $+6$

In $ZnS{O}_4$, $Zn$ is in $+2$ state and $+2+x+4(-2)=0$

$2+x-8=0$

$x=6$

So, oxidation state of $S$ remains unchanged in this reaction also.