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Chemical Properties of Hydrogen Peroxide (H2O2)

Chemical Properties of Hydrogen Peroxide (H2O2) | chemca
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Chemical Properties of Hydrogen Peroxide

The Redox Dual-Nature, Bleaching Action, and Disproportionation.

By chemca Team • Updated Sep 2026

Hydrogen Peroxide ($H_2O_2$) is a chemical chameleon. Because the oxygen atom in $H_2O_2$ is in an unstable, intermediate oxidation state, the molecule can act as an oxidizing agent, a reducing agent, or it can simply decompose (disproportionate) into water and oxygen. Mastering these specific reactions is a guaranteed way to score high in inorganic chemistry.

1. The Dual Nature: Oxidation States

In $H_2O_2$, the presence of the peroxide linkage ($-O-O-$) means each Oxygen atom is in the $-1$ oxidation state.

H₂O₂ O = -1 H₂O O = -2 O₂ O = 0 Reduction (Acts as Oxidant) +1 e⁻ Oxidation (Acts as Reductant) -1 e⁻

Figure 1: The intermediate oxidation state of Oxygen in H₂O₂ allows it to undergo both oxidation and reduction.

  • As an Oxidizing Agent: It accepts electrons, dropping its oxidation state from $-1$ down to $-2$ (forming $H_2O$ or $OH^-$).
  • As a Reducing Agent: It loses electrons, raising its oxidation state from $-1$ up to $0$ (evolving $O_2$ gas).
The Golden Rule for Exams:
Whenever $H_2O_2$ acts as a Reducing Agent, it is oxidized. The product of this oxidation is ALWAYS Oxygen Gas ($O_2$). If you see a reaction where $H_2O_2$ produces $O_2$ gas, it is acting as a reducing agent!

2. Oxidizing Action (Acidic & Basic Medium)

$H_2O_2$ is a powerful oxidizing agent in both acidic and basic mediums, but it is generally stronger in acidic medium.

A. In Acidic Medium

Half Reaction: $H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$

  • The PbS Trap (Qualitative Analysis): $H_2O_2$ oxidizes black Lead(II) Sulfide ($PbS$) into white Lead(II) Sulfate ($PbSO_4$). This reaction is used to restore the white color of old oil paintings blackened by atmospheric $H_2S$.
    $\underbrace{PbS}_{\text{Black}} + 4H_2O_2 \rightarrow \underbrace{PbSO_4}_{\text{White}} + 4H_2O$
  • Oxidizing Ferrous to Ferric:
    $2Fe^{2+} + 2H^+ + H_2O_2 \rightarrow 2Fe^{3+} + 2H_2O$
  • Liberating Iodine from KI:
    $2I^- + 2H^+ + H_2O_2 \rightarrow I_2 + 2H_2O$

B. In Basic Medium

Half Reaction: $H_2O_2 + 2e^- \rightarrow 2OH^-$

  • Oxidizing Manganese:
    $Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^-$

3. Reducing Action (Acidic & Basic Medium)

In the presence of stronger oxidizing agents (like $KMnO_4$ or $Cl_2$), $H_2O_2$ is forced to act as a reducing agent. Notice that $O_2$ gas is produced in every single reaction below.

A. In Acidic Medium

Half Reaction: $H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$

  • Decolorization of Potassium Permanganate ($KMnO_4$): This is a standard volumetric titration reaction. $H_2O_2$ reduces the purple $MnO_4^-$ to the colorless $Mn^{2+}$.
    $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + \mathbf{5O_2 \uparrow}$
  • Reaction with Chlorine (Halogens): Reduces $Cl_2$ to $HCl$.
    $Cl_2 + H_2O_2 \rightarrow 2HCl + \mathbf{O_2 \uparrow}$

B. In Basic Medium

Half Reaction: $H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$

  • Reaction with Iodine: (Compare this with the acidic medium where it oxidizes $I^-$. Here, it reduces $I_2$).
    $I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + \mathbf{O_2 \uparrow}$
  • Reduction of Metal Oxides:
    $Ag_2O + H_2O_2 \rightarrow 2Ag + H_2O + \mathbf{O_2 \uparrow}$

4. Bleaching Action & Disproportionation

A. Permanent Bleaching Action

$H_2O_2$ is heavily used as a mild bleaching agent for delicate materials like silk, wool, hair, and ivory. It bleaches due to its oxidizing property.

It decomposes to release nascent oxygen ($[O]$), which oxidizes the colored pigments into colorless matter. Because it is an oxidation process, the bleaching effect is permanent (unlike $SO_2$ bleaching, which is temporary reduction).

$H_2O_2 \rightarrow H_2O + [O]$
$\text{Colored substance} + [O] \rightarrow \text{Colorless oxidized substance}$

B. Auto-Oxidation-Reduction (Disproportionation)

As discussed in the physical properties section, pure $H_2O_2$ is unstable. Even upon standing at room temperature (and especially when exposed to light, heat, or heavy metals), it undergoes disproportionation.

$\overset{\mathbf{-1}}{2H_2O_2}_{(l)} \rightarrow \overset{\mathbf{-2}}{2H_2O}_{(l)} + \overset{\mathbf{0}}{O_2}_{(g)}$

In this single reaction, the Oxygen in $H_2O_2$ (at $-1$) is simultaneously oxidized to $O_2$ (at $0$) and reduced to $H_2O$ (at $-2$).

Mastery Check: H₂O₂ Reactions

15 High-Yield Questions to test your JEE/NEET Preparation

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