Question

i) $H_2{O}_2\to H_{2}O+\left(o\right)$

ii) $H_2{O}_2+2I^{-}\to I_2+2O{H}^{-},H_2{O}_2$ acts as:

• A. oxidising agent in both (i) and (ii)
• B. reducing agent in both (i) and (ii)
• C. oxidising agent in (i) and reducing agent in (ii)
• D. reducing agent in (i) and oxidising agent in (ii)

Answer 1

The correct option is A oxidising agent in both (i) and (ii)

GIven:

(i) $H_2{O}_2\to H_{2}O+\left[O\right]$

Here, the oxidation state of $O$ changes from $-1$ to $-2$ and $0$.

(ii) $H_2{O}_2+2I^{-}\to I_2+2O{H}^{-}$

Here, the oxidation state of $I$ changes from $-1$ to $0$. Hence, $H_2{O}_2$ oxidises iodine while itself gets reduced.

Hence, $H_2{O}_2$ acts as oxidising agent in both the above reactions.