In which of the following reaction(s) $H_{2} O_{2}$ acts as a reducing agent?

H_{2} O_{2} + 2 H^{+} + 2 e^{-} ⟶ 2 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} O_{2} ⟶ - 2 e^{-} O_{2} + 2 H^{+}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} O_{2} + 2 e^{-} ⟶ 2 O H^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O$
Losing or giving away electrons reduces the oxidation number/state of the species with which $H_{2} O_{2}$ reacts. Hence, in those cases, $H_{2} O_{2}$ will be acting as a reducing agent.

In reactions:
$H_{2} O_{2} ⟶ - 2 e^{-} O_{2} + 2 H^{+}$

$H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O$

It is clear that $H_{2} O_{2}$ is losing electrons. Hence $H_{2} O_{2}$ acts as reducing agent in the above reactions.

Suggest Corrections

Similar questions

H_{2} O_{2}

H_{2} O_{2}

H_{2} O_{2}

H_{2} O_{2}

(a)

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

(b)

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

(c)

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

(d)

H_{2} O_{2} + 2 O H^{-} - 2 e^{-} \to O_{2} + 2 H_{2} O

H_{2} O_{2}

H_{2} O_{2} + 2 H^{\oplus} + 2 e^{⊖} \to 2 H_{2} O

H_{2} O_{2} - 2 e^{⊖} \to O_{2} + 2 H^{\oplus}

H_{2} O_{2} + 2 e^{⊖} \to 2 O H^{⊖}

H_{2} O_{2} + 2 O H^{⊖} - 2 e^{⊖} \to 2 H_{2} O

Join BYJU'S Learning Program