Solution:
Option (ii) is the answer.
(A) $H_2O_2 + 2HI → I_2 + 2H_2O$
Iodine undergoes oxidation, transitioning from the -1 oxidation state to the 0 oxidation state. As a result, H2O2 works as an oxidising agent and undergoes a reduction process.
(B) $HOCl + H_2O_2 → H_3O^+ + Cl^–+ O_2$
The oxidation state of chlorine is changed from +1 to -1 during the reduction process. As a result, H2O2 serves as a reducing agent while also becoming oxidised.