Chapter 9.5 - Hydrogen Peroxide

In the previous section, we saw hard water and soft water. In this section, we will see hydrogen peroxide.

Some basics about peroxides can be written in 4 steps:
1. Peroxides are a group of compounds having O-O single bond.
• The O-O group in a peroxide is called peroxide group.
2. Hydrogen peroxide will have the formula H-O-O-H
• In short form, it is written as H₂O₂
• All the three bonds in H₂O₂ are covalent bonds.
3. Barium peroxide will have the formula Ba²⁺ O^--O^-
• It is an ionic compound. The ions are Ba²⁺and O^--O^-.     
• In short form, it is written as: BaO₂.     
4. Sodium peroxide will have the formula 2Na⁺ O^--O^-
• It is an ionic compound. The ions are Na⁺, Na⁺ and O^--O^-.
• In short form, it is written as: Na₂O₂.    

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• Hydrogen peroxide is the most common peroxide.
• In pure form, hydrogen peroxide is a very pale blue liquid.
• When it is made into a solution with water, it becomes almost color less.
• Hydrogen peroxide has many practical uses. It is used as an oxidizer, bleaching agent and antiseptic. It is also used as a rocket propellant. 

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First we will see the preparation of hydrogen peroxide.

Method I: From sodium peroxide and sulfuric acid.
This can be written in 3 steps:
1. Calculated quantities of sodium peroxide is gradually added to an ice-cold solution of 20% H₂SO₄.
Na₂O₂ + H₂SO₄ ⟶ Na₂SO₄ + H₂O₂
2. This solution is kept undisturbed for some time.
• Crystals of Na₂SO₄.10H₂O separates out. Those crystals can be removed by filtration.
• The resulting solution will contain 30% H₂O₂.
3. This method is known as Merck’s method for preparing hydrogen peroxide.

Method II: From barium peroxide and sulfuric acid.
This can be written in 3 steps:
1. Solid crystals of hydrated barium peroxide is made into a thin paste using ice-cold water.
• This paste is then added slowly to an ice-cold solution of 20% H₂SO₄.
BaO₂.8H₂O (s) + H₂SO₄ (aq) ⟶ BaSO₄ (s) + H₂O₂ (aq) + 8H₂O (l)
• The white precipitate of BaSO₄ is removed by filtration. After filtration, we get a 5% solution of H₂O₂.
2. Note that, we use hydrated barium peroxide (BaO₂.8H₂O) for this reaction. We do not use anhydrous barium peroxide (BaO₂). The reason can be written in 2 steps:
(i) Suppose that, we use BaO₂. Then the newly forming BaSO₄ will be able to stick to the crystals of BaO₂.
• Soon, a coating of BaSO₄ will be formed around the BaO₂ crystals.
• This coating will prevent the further reaction of BaO₂ with H₂SO₄.
(ii) In hydrated barium peroxide, the BaO₂ molecules are surrounded by water molecules.
• So the BaSO₄ will not be able to stick to the crystal surface. So the coating will not form.
3. The solution obtained after filtration will contain Ba²⁺ ions and H₂SO₄ molecules. The presence of these ions and molecules will increase the rate of decomposition of hydrogen peroxide. So this method is not preferred if the hydrogen peroxide is to be stored for a longer period of time.         

Method III: From barium peroxide and carbon dioxide.
This can be written in 3 steps:
1. A rapid stream of CO₂ is bubbled through a thin paste of BaO₂.
BaO₂ + H₂O + CO₂ ⟶ BaCO3q + H₂O2.
2. The BaCO₃ (barium carbonate) is insoluble. It is filtered out.
3. After filtration, we get a dilute solution of H₂O₂.

Method IV: From barium peroxide and phosphoric acid.
This can be written in 3 steps:
1. Action of phosphoric acid on barium peroxide ⟶ hydrogen peroxide.
3BaO₂ + H₃PO₄ ⟶ Ba₃(PO₄)₂ + 3H₂O₂
2. All the heavy metal impurities present in BaO₂ are converted into insoluble phosphates.
3. The heavy metal impurities are known to catalyze the decomposition of H₂O₂. So the H₂O2 obtained by this method has greater shelf life.

Method V: Electrolysis of 50% H₂SO₄
This can be written in 4 steps:
1. Electrolysis of cold 50% solution of H₂SO₄ is carried out with platinum anode and graphite cathode.
2. The H₂SO₄ first dissociates as shown below:
H₂SO₄ ⟶ H⁺ + HSO₄^-
• Reaction at cathode is:
2H⁺ + 2e^- ⟶ H₂
• Reaction at anode is:
2HSO₄^- ⟶ H₂S₂O₈ + 2e^-
3. H₂S₂O₈ (peroxodisulphuric acid) formed at the anode is taken out and water is added to it.
• This solution is distilled under low pressure. The following reactions take place during distillation:
    ♦ H₂S₂O₈ + H₂O ⟶ H₂SO₅ + H₂SO₄
    ♦ H₂SO₅ + H₂O ⟶ H₂SO₄ + H₂O₂
4. The hydrogen peroxide has a low boiling point when compared with H₂SO₄. So during distillation, it evaporates leaving the H₂SO₄ behind.
• The H₂O will also evaporate. So after condensation, we will be getting a dilute solution of H₂O₂.

Method VI: Electrolysis of equimolar mixture of  H₂SO₄ and ammonium sulphate
This can be written in 5 steps:
1. The ammonium sulphate first reacts with H₂SO₄ to give ammonium bisulphate.
(NH₄)₂SO₄ + H₂SO₄ ⟶ 2NH₄HSO₄
2. The ammonium bisulphate dissociates as shown below:
NH₄HSO₄ ⟶ H⁺ + NH₄SO₄^-
• Reaction at cathode is:
2H⁺ + 2e^- ⟶ H₂
• Reaction at anode is:
2NH₄SO₄^- ⟶ (NH₄)₂S₂O₈ + 2e^-
3. (NH₄)₂S₂O₈ (ammonium persulphate) formed at the anode is taken out and water is added to it.
• This solution is distilled under low pressure.
(NH₄)₂S₂O₈ + H₂O ⟶ 2NH₄HSO₄ + H₂O₂
4. This method gives more concentrated solution of H₂O₂.
5. This method is now used to prepare D₂O₂ in the laboratory.
K₂S₂O₈ + 2D₂O ⟶ 2KDSO₄ + D₂O₂

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Industrial preparation of hydrogen peroxide

Auto-oxidation of 2-ethylanthraquinol gives H₂O₂. It can be written in 4 steps:
1. Auto-oxidation refers to the reaction with oxygen at normal temperatures. Flame or electric spark is not required for such a reaction.
2. O₂ is bubbled through a 10% solution of 2-ethylanthraquinol.
• 2-ethylanthraquinol undergoes auto-oxidation. 
• During the oxidation process, the ‘ol’ changes to ‘one’.
• That is., the alcohol changes to ketone. H₂O₂ is also produced. This is shown in the fig.9.5 below:

Fig.9.5

• Note that, the -OH group (alcohol group) changes to =O group (keto group)
3. So the resulting mixture contains the ketone and H₂O₂  
• To this mixture, water is added. The H₂O₂ is miscible with water.
• A layer of H₂O₂-water combination is formed. The ketone forms another layer.
• The two layers are separated using a separating funnel.
• H₂O₂-water combination is distilled under reduced pressure to obtain a 30% solution of H₂O₂.
4.  The ketone is recycled to obtain the same original alcohol.
• This is done using hydrogen gas as shown in the fig.9.6 below:

Fig.9.6

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Expressing the strength of hydrogen peroxide solution

This can be written in 6 steps:
1. H₂O₂ decomposes slowly on exposure to light.
• The equation is: 2H₂O₂ ⟶ 2H₂O + O₂
• So the strength of a H₂O₂ solution can be determined based on the amount of oxygen gas which can be produced from that solution.
2. Based on the equation in (1), we can write:
Two moles of H₂O₂ will give one mole of O₂
• Two moles of H₂O₂ is (2 × 34) = 68 grams of H₂O₂.
• So we can write:
68 grams of H₂O₂ will give 22.7 litres of O₂ gas at STP.
• That means:
to make 1 litre of O₂, we will need $\frac{68}{22.7}$ = 2.99 = 3 grams of H₂O₂.
• In other words:
to make 1000 mL of O₂, we will need 3 grams of H₂O₂.  
• It follows that:
to make 1 mL of O₂, we will need $\frac{3}{1000}$ = 0.003 grams of H₂O₂.
3. At this stage, we need to become familiar with an internationally accepted rule.
• This rule gives us the method for writing the strength of a H₂O₂ solution. It can be explained in 4 steps:
(i) Consider a H₂O₂ solution marked as 1 V.
• It means that, 1 mL of that solution is capable to produce 1 mL of O₂.
• It is a 1:1 ratio.
• So 10 mL of that solution is capable to produce 10 mL of O₂.     
(1 V is usually written simply as V)
(ii) Consider a H₂O₂ solution marked as 10 V.
• It means that, 1 mL of that solution is capable to produce 10 mL of O₂.
• It is a 1:10 ratio.
• So 10 mL of that solution is capable to produce 100 mL of O₂.     
(iii) Consider a H₂O₂ solution marked as 100 V.
• It means that, 1 mL of that solution is capable to produce 100 mL of O₂.
• It is a 1:100 ratio.
• So 10 mL of that solution is capable to produce 1000 mL of O₂.
(iv) In this way, strength of the H₂O₂ available in the market is expressed in terms of ‘V’.
• Here 'V stands for 'volume'.
4. Now we will see how much grams of H₂O₂ is present in each type. It can be written in 3 steps:
(i) Consider a 1 V solution of H₂O₂. Take out 1 mL from that solution.
• We will get 1 mL of O₂ from that 1 mL
• From (2), we know that:
to make 1 mL of O₂, we will need 0.003 grams of H₂O₂.
• So it is clear that:
1 mL of 1 V solution will contain 0.003 grams of H₂O₂.
(ii) Consider a 10 V solution of H₂O₂. Take out 1 mL from that solution.
• We will get 10 mL of O₂ from that 1 mL
• From (2), we know that:
to make 1 mL of O₂, we will need 0.003 grams of H₂O₂.
• So it is clear that:
1 mL of 10 V solution will contain 0.03 grams of H₂O₂.
(iii) Consider a 100 V solution of H₂O₂. Take out 1 mL from that solution.
• We will get 100 mL of O₂ from that 1 mL
• From (2), we know that:
to make 1 mL of O₂, we will need 0.003 grams of H₂O₂.
• So it is clear that:
1 mL of 100 V solution will contain 0.3 grams of H₂O₂.
5. At this stage, we need to become familiar with another internationally accepted rule.
• This rule gives us the method for writing the strength of any solution in percentage format. It can be explained in 5 steps:
(i) To calculate the percentage, we multiply a fraction by 100.
(ii) Numerator of the fraction should be:
Mass of the solute. It should be in grams.
(iii) Denominator of the fraction should be:
Volume of the solution. It should be in mL.
(iv) So the expression will be: $\frac{\rm{Mass~in~grams}}{\rm{Volume~in~mL} }~\times ~ 100$
(v) ‘grams’ in the numerator and ’mL’ in the denominator do not match. So during calculation of the percentage, units do not cancel each other.
• We are inclined to write the unit as g/mL. But g/mL indicates ‘density’.
• ‘Density’ is different from ‘strength of solution’. So we cannot use g/mL here.
• To solve this problem, we write m/V instead of g/mL. We also write the chemical equation of the solute.
6. Let us see some examples:
(i) In 4 (i), we saw that:
• 1 mL of 1 V solution will contain 0.003 grams of H₂O₂.
   ♦ So the numerator will be 0.003 grams
   ♦ The denominator will be 1 mL
• Thus the percentage is: $\frac{0.003}{1} × 100$ = 0.3%
• We can write:
1 V H₂O₂ solution is a 0.3% m/V H₂O₂ solution.
(ii) In 4 (ii), we saw that:
• 1 mL of 10 V solution will contain 0.03 grams of H₂O₂.
   ♦ So the numerator will be 0.03 grams
   ♦ The denominator will be 1 mL
• Thus the percentage is: $\frac{0.03}{1} × 100$ = 3%
• We can write:
10 V H₂O₂ solution is a 3% m/V H₂O₂ solution.
(iii) In 4 (iii), we saw that:
• 1 mL of 100 V solution will contain 0.3 grams of H₂O₂.
   ♦ So the numerator will be 0.3 grams
   ♦ The denominator will be 1 mL
• Thus the percentage is: $\frac{0.3}{1} × 100$ = 30%
• We can write:
100 V H₂O₂ solution is a 30% m/V H₂O₂ solution.

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Physical properties of hydrogen peroxide

• In pure form, H₂O₂ has a very pale blue color.
   ♦ When water is added to make a H₂O₂ solution, it becomes colorless.
• H₂O₂ is miscible with water in all proportions. It exists as a hydrate H₂O₂.H₂O
• The physical properties of H₂O₂ can be obtained from standard tables. Let us see some interesting points from the table. It can be written in 5 steps:
1. Melting point of H₂O₂ is 272.4 K
• We know that, the normal room temperature is 298 K. At this temperature, H₂O₂ will be in liquid form.
• So if we start cooling this liquid, it will solidify when the temperature drops to 272.4 K
• Recall that, water solidifies at nearly the same temperature, which is 273 K
2. When in liquid form, the density of H₂O₂ is 1.44 g/cm³
(This is at the room temperature of 298 K)
• Density of liquid water is nearly the same, which is 1 g/cm³
3. When in solid form, the density of H₂O₂ is 1.64 g/cm³
(This is at a temperature of 268.5 K)
• Density of solid ice is 0.9 g/cm³
4. Note that:
   ♦ When water solidifies, density decreases.
   ♦ When H₂O₂ solidifies, density increases.
5. H₂O₂ is slightly more viscous than water.

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Solved example 9.5
On a bottle of hydrogen peroxide solution available in the market, the strength is marked as 40 V. How many grams of hydrogen peroxide is present in each 1 mL of that solution?
Solution:
1. The strength is marked as 40 V.
• So each 1 mL in that solution is capable to produce 40 mL of oxygen at STP.
2. We have the equation:
2H₂O₂ ⟶ 2H₂O + O₂
• So 68 grams of H₂O₂ will give 22.7 litres of O₂ at STP
• So to produce 1 litre of O₂, we must have $\frac{68}{22.7}$ = 2.99 = 3 grams of H₂O₂
⇒ To produce 1000 mL of O₂, we must have 3 grams of H₂O₂
⇒ To produce 1 mL of O₂, we must have $\frac{3}{1000}$ = 0.003 grams of H₂O₂
⇒ To produce 40 mL of O₂, we must have (40 × 0.003) = 0.12  grams of H₂O₂
3. So it is clear that:
0.12 grams of hydrogen peroxide is present in each 1 mL of that solution.
4. Strength of the solution in percentage format is: $\frac{0.12}{1}~\times ~100$ = 12%

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In the next section, we will see structure of hydrogen peroxide.

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