Question
Which of the following chemical reactions depict the oxidizing beahviour of $${H_2}S{O_4}$$ ?
A.
$$NaCl + {H_2}S{O_4} \to NaHS{O_4} + HCl$$
B.
$$2PC{l_5} + {H_2}S{O_4} \to 2POC{l_3} + 2HCl + S{O_2}C{l_2}$$
C.
$$2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O$$
D.
$$Ca{\left( {OH} \right)_2} + {H_2}S{O_4} \to CaS{O_4} + 2{H_2}O$$
Answer :
$$2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O$$
Solution :
$$2H{I^{ - 1}} + {H_2}\mathop {S{O_4}}\limits^{ + 6} \to I_2^0 + \mathop {S{O_2}}\limits^{ + 4} + 2{H_2}O$$ in this reaction oxidation number of $$S$$ is decreasing from $$+ 6$$ to $$+4$$ hence undergoing reduction and for $$HI$$ oxidation Number of $$I$$ is increasing from $$-1$$ to $$0$$ hence underegoing oxidation therefore $${H_2}S{O_4}$$ is acting as oxidising agent.