Question

In the reaction of $H_{2}S{O}_4$ with $H_{2}S$, the change in oxidation states of sulphur are:

• A. $+6$ to $+4$ and $-2$ to $0$
• B. $+4$ to $+2$ and $+2$ to $0$
• C. $+6$ to $-2$ and $+6$ to $4$
• D. $+2$ to $+6$ and $0$ to $-2$

Answer 1

Answer: (A)

$+6$ to $+4$ and $-2$ to $0$

The reaction between sulphuric acid and hydrogen sulphide is as follows:

$H_{2}S{O}_4+H_{2}S\to S{O}_2+S+2H_{2}O$

Oxidation state of sulphur in $H_{2}S{O}_4$ is +6. It changes to +4 in $S{O}_2$.

Oxidation state of sulphur in $H_{2}S$ is -2. It changes to 0 in $S$.