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Hydrogen - Hydrogen Peroxide

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hydrogen Peroxide (H2O2H_2O_2) was discovered by J.L. Thenard in 1818 and is an important chemical used in pollution control and as a bleaching agent.

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Structure: H2O2H_2O_2 has a non-planar, 'open-book' structure. The O−OO-O bond length is 147.5147.5 pm. The dihedral angle is 111.5∘111.5^\circ in the gas phase and decreases to 90.2∘90.2^\circ in the crystalline (solid) phase due to hydrogen bonding.

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Preparation: In the laboratory, it is prepared by the action of dilute H2SO4H_2SO_4 or H3PO4H_3PO_4 on hydrated barium peroxide (BaO2⋅8H2OBaO_2 \cdot 8H_2O). Industrially, it is prepared by the auto-oxidation of 22-ethylanthraquinol.

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Chemical Properties: H2O2H_2O_2 acts as both an oxidizing agent and a reducing agent in both acidic and alkaline media. Its oxidizing nature is due to the liberation of nascent oxygen: H2O2→H2O+[O]H_2O_2 \rightarrow H_2O + [O].

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Bleaching Action: The bleaching action of H2O2H_2O_2 is due to oxidation. It oxidizes coloring matter to colorless products. Unlike SO2SO_2 (which bleaches by reduction), the bleaching by H2O2H_2O_2 is permanent.

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Storage: H2O2H_2O_2 decomposes slowly on exposure to light or in the presence of metal surfaces or traces of alkali. Reaction: 2H2O2→2H2O+O22H_2O_2 \rightarrow 2H_2O + O_2. It is stored in wax-lined glass or plastic bottles in the dark. Urea can be added as a stabilizer.

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Volume Strength: The concentration of H2O2H_2O_2 is often expressed as 'volume strength'. For example, '10 volume' H2O2H_2O_2 means 11 L of this solution will give 1010 L of oxygen at STP.

📐Formulae

2H2O2→2H2O+O2↑2H_2O_2 \rightarrow 2H_2O + O_2 \uparrow

Molarity (M)=Volume Strength11.2\text{Molarity (M)} = \frac{\text{Volume Strength}}{11.2}

Normality (N)=Volume Strength5.6\text{Normality (N)} = \frac{\text{Volume Strength}}{5.6}

Strength in g/L=Normality×17\text{Strength in g/L} = \text{Normality} \times 17

Strength in g/L=Molarity×34\text{Strength in g/L} = \text{Molarity} \times 34

Percentage Strength=Strength in g/L10\text{Percentage Strength} = \frac{\text{Strength in g/L}}{10}

💡Examples

Problem 1:

Calculate the normality and molarity of '10 volume' H2O2H_2O_2 solution.

Solution:

  1. Normality N=Volume Strength5.6=105.6≈1.786 NN = \frac{\text{Volume Strength}}{5.6} = \frac{10}{5.6} \approx 1.786\text{ N}.
  2. Molarity M=Volume Strength11.2=1011.2≈0.893 MM = \frac{\text{Volume Strength}}{11.2} = \frac{10}{11.2} \approx 0.893\text{ M}.

Explanation:

Using the standard conversion factors derived from the decomposition reaction where 22 moles of H2O2H_2O_2 (6868 g) produce 22.422.4 L of O2O_2 at STP.

Problem 2:

Complete and balance the following equation showing the oxidizing action of H2O2H_2O_2 in acidic medium: PbS+H2O2→PbS + H_2O_2 \rightarrow ?

Solution:

PbS(s)+4H2O2(aq)→PbSO4(s)+4H2O(l)PbS(s) + 4H_2O_2(aq) \rightarrow PbSO_4(s) + 4H_2O(l)

Explanation:

In this reaction, Black Lead Sulphide (PbSPbS) is oxidized to White Lead Sulphate (PbSO4PbSO_4) by Hydrogen Peroxide.

Problem 3:

Show the reducing action of H2O2H_2O_2 on Silver Oxide (Ag2OAg_2O).

Solution:

Ag2O(s)+H2O2(aq)→2Ag(s)+H2O(l)+O2(g)Ag_2O(s) + H_2O_2(aq) \rightarrow 2Ag(s) + H_2O(l) + O_2(g)

Explanation:

Here, H2O2H_2O_2 reduces Silver Oxide to metallic silver, while H2O2H_2O_2 itself is oxidized to O2O_2.

Hydrogen Peroxide - Revision Notes & Key Formulas | ICSE Class 11 Chemistry