Question

In which of the following reaction $H_2{O}_2$ acts as a reducing agent?
(a) $H_2{O}_2+2H^{+}+2e^{-}\to 2H_{2}O$
(b) $H_2{O}_2-2e^{-}\to O_2+2H^{+}$
(c) $H_2{O}_2+2e^{-}\to 2O{H}^{-}$
(d) $H_2{O}_2+2O{H}^{-}-2e^{-}\to O_2+2H_{2}O$.

• A. (a), (c)
• B. (b), (d)
• C. (a), (b)
• D. (c), (d)

Answer 1

The correct option is B (b), (d)

$H_2{O}_2\to O_2+2H^{+}+2e^{-}$

and $H_2{O}_2+2O{H}^{-}\to O_2+2H_{2}O+2e^{-}$

Oxidation state of $O$ changes from $-1$ in $H_2{O}_2$ to $0$ in $O_2$

As it is getting oxidized , it acts reducing agent.

Hence, the correct option is $B$