Question

Which of the following chemical reactions depicts the oxidizing behaviour of ${H_2}S{O_4}$?
A. \[Ca{(OH)_2} + {\text{ }}{{\text{H}}_2}S{O_4} \to CaS{O_4} + 2{H_2}O\]
B. $NaCl + {H_2}S{O_4} \to NaHS{O_4} + HCl$
C. $2PC{l_5} + {H_2}S{O_4} \to 2POC{l_3} + 2HCl + S{O_2}C{l_2}$
D. $2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O$

Answer 1

Hint: As we all know that oxidizing agent is basically a substance that can oxidize other substances by accepting their electrons and increases the oxidation state of that substance or we can say that an oxidizing agent undergoes a chemical reaction where it accepts one or more than one electrons and itself gets reduced, oxidizing the other substance.

Complete step by step answer:
Oxidising behaviour of a substance can be depicted by observing the oxidation state of the substance which is oxidized. So, we can calculate the oxidation states in the following reactions.
a. In the first reaction, the oxidation state of Calcium is $ + 2$, oxygen is $ - 2$, hydrogen is $ + 1$ and the oxidation state of sulfur remains the same that is $ + 6$ in ${H_2}S{O_4}$ and $CaS{O_4}$. Hence, in this reaction ${H_2}S{O_4}$ does not show any oxidizing behaviour.

b. In the second reaction, the oxidation state of sodium is $ + 1$, chlorine is $ - 1$, oxygen is $ - 2$, hydrogen is $ + 1$ and sulfur is same that is $ + 6$ in ${H_2}S{O_4}$ and $NaHS{O_4}$. Hence again the oxidizing behaviour is not seen in this reaction as well.

c. In the third reaction, the oxidation state of Phosphorus is $ + 5$, chlorine is $ - 1$, oxygen is $ - 2$, hydrogen is $ + 1$ and sulfur is same that is $ + 6$ in ${H_2}S{O_4}$ and $S{O_2}C{l_2}$. Thus, in this reaction also the oxidation state remains the same depicting that there is no oxidizing behaviour shown by ${H_2}S{O_4}$.

d. In the fourth reaction, the oxidation state of Iodine is$ - 1$which is reduced to $0$ and the oxidation state of Sulfur changes from $ + 6$ to $ + 4$. Hence, this is the only reaction where Sulfur gets reduced itself and oxidises Iodine depicting its oxidizing behaviour.

Therefore, the correct answer is (D).

Note:
The simplest way to describe the oxidizing and reducing behaviour of a substance is by calculating the oxidation states of the species involved in the chemical reaction. Remember the oxidation states of the common ones and you will get the answer for those you do not know.