Hydrogen Peroxide

Q: Why is the structure of hydrogen peroxide called an open book structure?

Because the two O-H bonds lie in two different planes that meet along the O-O bond, like two pages of a half-open book with the O-O bond as the spine. The angle between the planes is 111.5° in the gas phase and 90.2° in the solid.

ChemistryHydrogenFor JEE aspirants

Hydrogen peroxide, $H_{2} O_{2}$ , is the second hydride of oxygen, and the oxygen in its peroxide link sits at $- 1$ , half way between $0$ and $- 2$ . That single fact explains almost all the properties of hydrogen peroxide: it decomposes on its own, and it acts as an oxidising agent or a reducing agent in acidic and basic medium. This page covers its preparation, open book structure, reactions and volume strength. The topic is in the JEE Advanced syllabus; JEE Main and NEET have dropped it.

On this page1Preparation2Manufacture3Structure4Decomposition5Oxidant or reductant6Volume strength7Flowchart8Examples

Key Formulas - Quick Reference

Laboratory preparation: $Ba O_{2} \cdot 8 H_{2} O +$ $H_{2} S O_{4}$ gives $BaS O_{4} + H_{2} O_{2} + 8 H_{2} O$ , with the $BaS O_{4}$ filtered off.
Merck's method: $N a_{2} O_{2} + H_{2} S O_{4}$ gives $N a_{2} S O_{4} + H_{2} O_{2}$ in ice-cold $20%$ acid.
★ Must learn Decomposition: $2 H_{2} O_{2}$ gives $2 H_{2} O + O_{2}$ . This is a disproportionation, and it is what volume strength measures.
Weak acid: $K_{a} = 1.55 \times 1 0^{- 12}$ at $298 K$ ; $H_{2} O_{2}$ gives $H^{+} + H O_{2}^{-}$ .
★ Must learn Oxidising in acid: $H_{2} O_{2} + 2 H^{+} + 2 e^{-}$ gives $2 H_{2} O$ . In base: $H_{2} O_{2} + 2 e^{-}$ gives $2O H^{-}$ .
★ Must learn Reducing in acid: $H_{2} O_{2}$ gives $O_{2} + 2 H^{+} + 2 e^{-}$ . In base: $H_{2} O_{2} + 2O H^{-}$ gives $O_{2} + 2 H_{2} O + 2 e^{-}$ .
★ Must learn Structure: non-planar open book. Gas phase dihedral $111. 5^{\circ}$ , $O - O$ $147.5 pm$ , $O - H$ $95 pm$ , angle $94. 8^{\circ}$ .
★ Must learn Volume strength $= 11.2 \times$ molarity $= 5.6 \times$ normality.
Strength in $g L^{- 1} = 34 \times$ molarity, and normality $= 2 \times$ molarity.
Bleaching works through nascent oxygen: $H_{2} O_{2}$ gives $H_{2} O + [O$ ].

1. What Hydrogen Peroxide Is

Hydrogen peroxide is another hydride of oxygen. Unlike water it is highly unstable, and so it does not exist in nature as such. Its importance has grown in recent years because of its use as a rocket fuel and as an industrial oxidising agent.

The $O - O$ single bond in $H_{2} O_{2}$ is called the peroxide linkage. Each oxygen atom in it has an oxidation state of $- 1$ . It can be pushed down to $- 2$ , as in water, or up to $0$ , as in dioxygen, which is why hydrogen peroxide can behave as an oxidising agent and as a reducing agent.

2. Preparation of Hydrogen Peroxide

Figure 1: Every laboratory route starts from a peroxide; both industrial routes build the peroxide link from scratch.

2.1 From sodium peroxide: Merck's method

A calculated amount of sodium peroxide is added in small lots, with constant stirring, to an ice-cold $20%$ solution of sulphuric acid.

N a_{2} O_{2} + H_{2} S O_{4} N a_{2} S O_{4} + H_{2} O_{2}

On cooling, crystals of $N a_{2} S O_{4} \cdot 10 H_{2} O$ separate out, and the solution left behind contains about $30%$ $H_{2} O_{2}$ . It still holds some dissolved $N a_{2} S O_{4}$ , which does not interfere with the ordinary reactions of $H_{2} O_{2}$ . A pure sample is obtained by vacuum distillation.

2.2 From barium peroxide: the laboratory method

(a) With dilute sulphuric acid. A paste of hydrated barium peroxide is prepared in ice-cold water and added slowly to an ice-cold solution of $20%$ sulphuric acid.

Ba O_{2} \cdot 8 H_{2} O + H_{2} S O_{4} ⟶ BaS O_{4} + H_{2} O_{2} + 8 H_{2} O

The white precipitate of $BaS O_{4}$ is filtered off, leaving a dilute solution of about $5%$ $H_{2} O_{2}$ .

Exam Trick

Two "why" questions come from this one equation. Why hydrated $Ba O_{2}$ and not the anhydrous solid? Because $BaS O_{4}$ forms a protective layer around unreacted anhydrous $Ba O_{2}$ and stops the reaction. Why does the product keep badly? Because traces of $B a^{2 +}$ and other metal ions left in solution catalyse the decomposition of $H_{2} O_{2}$ . The phosphoric acid route below removes them as insoluble phosphates.

(b) With carbon dioxide. A rapid stream of $C O_{2}$ is bubbled through a thin paste of $Ba O_{2}$ in ice-cold water.

Ba O_{2} + H_{2} O + C O_{2} BaC O_{3} + H_{2} O_{2}

The insoluble barium carbonate is filtered off, leaving a dilute solution of $H_{2} O_{2}$ .

(c) With phosphoric acid. Hydrogen peroxide can also be made by the action of phosphoric acid on barium peroxide.

3Ba O_{2} + 2 H_{3} P O_{4} B a_{3} (P O_{4})_{2} + 3 H_{2} O_{2}

This method has an advantage over the $Ba O_{2}$ and $H_{2} S O_{4}$ route: almost all the heavy metal impurities present in $Ba O_{2}$ , such as lead, which would otherwise catalyse the decomposition of $H_{2} O_{2}$ , are removed as insoluble phosphates. The resulting solution therefore keeps well.

3. Manufacture of Hydrogen Peroxide

3.1 By electrolysis of 50 per cent sulphuric acid

A cold $50%$ solution of sulphuric acid is electrolysed using platinum as the anode and graphite as the cathode.

2 H_{2} S O_{4} 2 H^{+} + 2HS O_{4}^{-}

2 H^{+} + 2 e^{-} cathode H_{2}

2HS O_{4}^{-} anode H_{2} S_{2} O_{8} + 2 e^{-}

The peroxydisulphuric acid formed at the anode is withdrawn and distilled with water under reduced pressure. It hydrolyses in two steps, through peroxymonosulphuric acid, and the low boiling $H_{2} O_{2}$ distils over with the water, leaving high boiling $H_{2} S O_{4}$ behind to be recovered and recycled.

H_{2} S_{2} O_{8} + H_{2} O H_{2} S O_{5} + H_{2} S O_{4}

H_{2} S O_{5} + H_{2} O H_{2} S O_{4} + H_{2} O_{2}

Adding the two steps gives the overall hydrolysis:

H_{2} S_{2} O_{8} + 2 H_{2} O H_{2} O_{2} + 2 H_{2} S O_{4}

Figure 2: The anode builds the

O - O

link inside

H_{2} S_{2} O_{8}

; two hydrolyses hand it over to water as

H_{2} O_{2}

. The sulphuric acid is recovered, so the net change is

2 H_{2} O \to H_{2} O_{2} + H_{2}

Modification. If an equimolar mixture of $H_{2} S O_{4}$ and ammonium sulphate is electrolysed instead of $50%$ $H_{2} S O_{4}$ , a more concentrated solution of $H_{2} O_{2}$ is obtained through ammonium persulphate. The same idea gives deuteroperoxide: $K_{2} S_{2} O_{8} + 2 D_{2} O$ gives $2KDS O_{4} + D_{2} O_{2}$ .

3.2 By autoxidation of 2-ethylanthraquinol

This is the modern method and the one used industrially today. Air is bubbled through a $10%$ solution of 2-ethylanthraquinol in benzene and cyclohexane. The quinol is oxidised to 2-ethylanthraquinone and hydrogen peroxide is formed.

Figure 3: The anthraquinol is a carrier, not a reactant. It is regenerated every cycle, so the raw materials are only

H_{2}

and

O_{2}

The $H_{2} O_{2}$ produced, about $1%$ , is extracted with water and the aqueous solution is concentrated by distillation under reduced pressure to give a $30%$ solution by weight. The 2-ethylanthraquinone is then reduced back to 2-ethylanthraquinol with hydrogen over a palladium catalyst, and the cycle starts again. The only raw materials actually consumed are $H_{2}$ and $O_{2}$ .

3.3 Concentrating the solution

Hydrogen peroxide cannot be concentrated by simple heating, because it decomposes well below its boiling point.

2 H_{2} O_{2} 2 H_{2} O + O_{2}

Slow evaporation on a water bath. The dilute solution is warmed carefully so that water molecules escape, giving about a $50%$ solution.
Evaporation in a vacuum desiccator. The $50%$ solution is kept over concentrated sulphuric acid, which absorbs the water vapour, leaving about $90%$ $H_{2} O_{2}$ .
Distillation under reduced pressure. At $10$ to $15 mm$ the water distils over at $303$ to $313 K$ and about $99%$ pure hydrogen peroxide is left behind.
Removing the last traces of water. The $99%$ solution is cooled in a freezing mixture of solid carbon dioxide and ether. Crystals of $H_{2} O_{2}$ separate out and are removed, dried and remelted to give completely pure hydrogen peroxide.

Figure 4: Every step removes water gently, because heat would destroy the product: from about

30%

50

90

99

and finally

100%

H_{2} O_{2}

3.4 Storage

The decomposition of $H_{2} O_{2}$ is catalysed by traces of metal impurities, by strong bases and by exposure to light, so a concentrated solution can decompose fast enough to explode.

It is stored in wax-lined coloured bottles, because a rough glass surface also triggers decomposition and light accelerates it.
A small amount of a stabiliser such as phosphoric acid, glycerol or acetanilide is added to retard the decomposition.

Key idea

Laboratory routes start from a peroxide (

N a_{2} O_{2}

Ba O_{2}

); industrial routes build the

O - O

link, electrolytically or by autoxidation, and concentrate the product without heating it.

Quick Recall: tap to check

Why is hydrated

Ba O_{2}

used rather than the anhydrous solid?

Anhydrous

Ba O_{2}

gets coated with

BaS O_{4}

, which stops the reaction.

What is actually consumed in the anthraquinol process?

Only

H_{2}

and

O_{2}

; the anthraquinol is regenerated every cycle.

Why can

H_{2} O_{2}

not be concentrated by boiling?

It decomposes to water and oxygen well below its boiling point.

4. Structure of Hydrogen Peroxide

X-ray studies show that the hydrogen peroxide molecule is non-planar. The two $O - H$ bonds lie in two different planes that meet along the $O - O$ bond, which gives the molecule its familiar open book shape: the $O - O$ bond is the spine and the two $O - H$ bonds are on separate pages.

Figure 5: Hydrogen peroxide is non-planar. Hydrogen bonding in the crystal pulls the dihedral angle down from

111. 5^{\circ}

90. 2^{\circ}

Dimension	Gas phase	Solid phase at $110 K$
$O - H$ bond length	$95 pm$	$98.8 pm$
$O - O$ bond length	$147.5 pm$	$145.8 pm$
$O - O - H$ bond angle	$94. 8^{\circ}$	$101. 9^{\circ}$
Dihedral angle	$111. 5^{\circ}$	$90. 2^{\circ}$

In the crystal the dihedral angle falls from $111. 5^{\circ}$ to $90. 2^{\circ}$ on account of hydrogen bonding between neighbouring molecules. The two oxygen atoms are joined by a single electron pair bond, which is the peroxide linkage.

5. Properties of Hydrogen Peroxide

5.1 Physical properties

Pure $H_{2} O_{2}$ is a colourless, syrupy liquid.
Its odour resembles that of nitric acid, and its aqueous solution has a bitter taste.
It is soluble in water, alcohol and ether in all proportions.
Its density is $1.44 g cm^{- 3}$ , higher than water, because the molecules are strongly associated by intermolecular hydrogen bonds.
Its melting point is $272 K$ . Its boiling point is $423 K$ at $670 mm$ and $358 K$ at $68 mm$ .

5.2 Decomposition

Hydrogen peroxide is an unstable liquid and decomposes readily on heating or on long standing to give water and dioxygen. It is an example of disproportionation: the same element is both oxidised and reduced.

2 H_{2} O_{2} 2 H_{2} O + O_{2}

In the peroxide, oxygen is at $- 1$ . In water it has gone down to $- 2$ and in $O_{2}$ it has gone up to $0$ . The decomposition is suppressed by adding glycerol, acetanilide or phosphoric acid.

5.3 Acidic behaviour

Pure hydrogen peroxide is a weak acid, with $K_{a} = 1.55 \times 1 0^{- 12}$ at $298 K$ . It ionises in water in two steps.

H_{2} O_{2} ⇌ H^{+} + H O_{2}^{-}

H O_{2}^{-} ⇌ H^{+} + O_{2}^{2 -}

The ions formed are the hydroperoxide ion and the peroxide ion. Its acidic character shows in its ability to neutralise bases and form the corresponding peroxides.

2NaOH + H_{2} O_{2} N a_{2} O_{2} + 2 H_{2} O

N a_{2} C O_{3} + H_{2} O_{2} N a_{2} O_{2} + H_{2} O + C O_{2}

5.4 Oxidising and reducing behaviour

Hydrogen peroxide can act as an oxidising agent as well as a reducing agent, in acidic and in basic solution alike. This is the most heavily examined part of the topic.

Figure 6: In the peroxide link oxygen is at

- 1

, half way between

0

and

- 2

, so hydrogen peroxide can move either way.

As an oxidising agent. Some standard examples:

Reagent	Product	What is seen
$PbS$ (black)	$PbS O_{4}$ (white)	old oil paintings are restored
$FeS O_{4}$ in $H_{2} S O_{4}$	$F e_{2} (S O_{4})_{3}$	green solution turns yellow
$KI$ with $HCl$	$KCl + I_{2}$	iodine is liberated
$N a_{2} S O_{3}$	$N a_{2} S O_{4}$	sulphite to sulphate
$NaN O_{2}$	$NaN O_{3}$	nitrite to nitrate
$N a_{3} As O_{3}$	$N a_{3} As O_{4}$	arsenite to arsenate
$K_{2} C r_{2} O_{7}$ in $H_{2} S O_{4}$	$Cr O_{5}$	blue colour, extracted into ether
$MnS O_{4}$ with $NaOH$	$Mn O_{2}$	brown precipitate

As a reducing agent. When it meets a stronger oxidising agent, hydrogen peroxide gives up its own oxygen instead:

Reagent	Product	What is seen
$KMn O_{4}$ in $H_{2} S O_{4}$	$MnS O_{4} + O_{2}$	the pink colour is discharged
$O_{3}$	$H_{2} O + O_{2}$	ozone is destroyed
$A g_{2} O$	$2Ag + O_{2}$	silver is deposited
$K_{3} [Fe (CN)_{6}$ ] with $KOH$	$K_{4} [Fe (CN)_{6}$ ]	ferricyanide to ferrocyanide

As an oxidising agentO goes from

- 1

- 2

product: water (or

O H^{-}

)
no gas given off

PbS

F e^{2 +}

I^{-}

Cr (III)

As a reducing agentO goes from

- 1

0

product:

O_{2}

gas
needs a stronger oxidant

Mn O_{4}^{-}

O_{3}

A g_{2} O

OC l^{-}

5.5 Bleaching action

The bleaching action of hydrogen peroxide is due to the nascent oxygen it liberates on decomposition.

H_{2} O_{2} H_{2} O + [O]

The nascent oxygen combines with the colouring matter, which is oxidised to a colourless product. The action is mild and leaves only water behind, so hydrogen peroxide is used for bleaching delicate materials such as ivory, feathers, silk and wool, which chlorine would damage. Because the colour is destroyed by oxidation, not by reduction as with sulphur dioxide, the bleaching is permanent.

5.6 Addition to alkenes

With a catalyst such as osmium tetroxide, hydrogen peroxide adds two $- OH$ groups across a carbon to carbon double bond to give a glycol. Ethene, for example, gives ethane-1,2-diol. Without a catalyst the reaction is too slow to be useful.

Key idea

Oxygen at

- 1

can go down to

- 2

(oxidising agent, water formed) or up to

0

(reducing agent,

O_{2}

evolved); decomposition does both at once.

Quick Recall: tap to check

What is the dihedral angle of

H_{2} O_{2}

in the gas phase and in the solid?

111. 5^{\circ}

in the gas,

90. 2^{\circ}

in the crystal at

110 K

Which gas is always given off when

H_{2} O_{2}

acts as a reducing agent?

Dioxygen,

O_{2}

How are old oil paintings restored?

H_{2} O_{2}

oxidises black

PbS

to white

PbS O_{4}

6. Strength of a Hydrogen Peroxide Solution

6.1 As a percentage

The concentration is sometimes expressed as the percentage of $H_{2} O_{2}$ in solution. A $40%$ solution of hydrogen peroxide means $40$ grams of hydrogen peroxide in $100$ grams of solution.

6.2 As volume strength

Volume strength is the volume of dioxygen, at STP, that one volume of the hydrogen peroxide solution gives on complete decomposition. A " $10$ volume" solution gives $10$ litres of oxygen per litre of solution.

The link between the units comes straight from the decomposition equation:

2 H_{2} O_{2} 2 H_{2} O + O_{2}

Two moles of $H_{2} O_{2}$ , that is $68 g$ , give one mole of $O_{2}$ , that is $22.4 L$ at STP. So one mole of $H_{2} O_{2}$ gives $11.2 L$ of oxygen, and every conversion follows from that.

Figure 7: Convert everything through molarity: every factor on the arrows multiplies

M

. One mole of

H_{2} O_{2}

releases

11.2 L

O_{2}

at STP, which is where the

11.2

comes from.

volume strength = 11.2 \times M = 5.6 \times N, strength in g L^{- 1} = 34 \times M

Exam Trick

Learn one number, 11.2: $1 M$ gives $11.2$ volume, so $1 N$ gives $5.6$ . And the chemist's bottle labelled 10 volume is $10/11.2 = 0.893 M$ , about $3%$ (w/v).

JEE Advanced

In a redox titration $H_{2} O_{2}$ has an $n$ factor of $2$ , whether it is being oxidised or reduced, because two electrons are involved in each half reaction. That is why normality is twice the molarity, and why the standard titration against $KMn O_{4}$ uses the ratio $2 KMn O_{4} : 5 H_{2} O_{2}$ .

Figure 8: Volume strength is simply

11.2 \times M

. A

10

volume solution is

0.893 M

, that is

30.4 g L^{- 1}

or about

3%

: the bottle sold in chemists' shops.

Key idea

Every strength converts through molarity:

g L^{- 1} = 34 M

N = 2 M

, volume strength

= 11.2 M

7. Uses of Hydrogen Peroxide

As a bleaching agent for ivory, feathers, silk, wool and hair, which chlorine would attack.
As an antiseptic. A dilute solution, about $3%$ , is used to wash wounds and as a mouthwash and gargle. The $30%$ solution is sold as perhydrol.
For restoring old oil paintings. The white lead in the paint darkens to $PbS$ over the years; hydrogen peroxide oxidises it back to white $PbS O_{4}$ .
As a rocket fuel and as a high energy oxidiser.
In industry, for making sodium perborate and sodium percarbonate, the bleaches used in detergents, and in the manufacture of antichlor, dyes and pharmaceuticals.
In pollution control, to oxidise cyanides in industrial effluent and to restore oxygen to polluted water.

8. Oxidant or Reductant: Flowchart and Mind Map

Most reaction questions on hydrogen peroxide ask one thing: is it the oxidising or the reducing agent here? The flowchart decides; the mind map after it puts the whole page on one screen.

Figure 9: Flowchart: decide first whether the partner is a stronger oxidising agent. If it is,

O_{2}

is evolved and

H_{2} O_{2}

is the reducing agent; otherwise it oxidises.

Figure 10: Mind map: the whole page on one screen. Oxygen at

- 1

is the thread through every branch: it can drop to

- 2

, rise to

0

, or do both at once.

9. Solved Examples

Solved Example 1

A certain sample of hydrogen peroxide is

1.5 M

. It is to be labelled as X volumes. Calculate the value of X.

Solution:

Molarity tells us the mass of $H_{2} O_{2}$ in one litre:

mass per litre = 1.5 \times 34 = 51 g

From the decomposition equation, $68 g$ of $H_{2} O_{2}$ gives $22.4 L$ of $O_{2}$ at STP:

2 H_{2} O_{2} 2 H_{2} O + O_{2}

V_{O_{2}} = \frac{22.4 \times 51}{68} = 16.8 L

Answer: $X = 16.8$ , so it is a $16.8$ volume solution. The short cut is volume strength $= 11.2 \times M = 11.2 \times 1.5 = 16.8$ .

Solved Example 2

Hydrogen peroxide is a strong oxidising agent in both acid and alkaline medium. Justify this with suitable half reactions.

Solution:

In both media the peroxide oxygen at $- 1$ is pulled down to $- 2$ , taking two electrons with it.

H_{2} O_{2} + 2 H^{+} + 2 e^{-} acidic medium 2 H_{2} O

H_{2} O_{2} + 2 e^{-} basic medium 2O H^{-}

It can also act as a reducing agent, when the oxygen is pushed up from $- 1$ to $0$ :

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

Answer: because oxygen in the peroxide link is at an intermediate oxidation state, hydrogen peroxide can move in either direction. Which way it goes depends entirely on what it is mixed with.

Solved Example 3

What happens when

H_{2} O_{2}

is treated with (a) acidified potassium permanganate (b) lead sulphide (c) alkaline potassium ferricyanide (d) acidified ferrous sulphate (e) sulphurous acid?

Solution:

(a) It reduces the acidified $KMn O_{4}$ and the pink colour is discharged, oxygen being released.

2KMn O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 5 O_{2}

(b) It oxidises black lead sulphide to white lead sulphate.

PbS + 4 H_{2} O_{2} PbS O_{4} + 4 H_{2} O

(c) It reduces alkaline potassium ferricyanide to potassium ferrocyanide.

2 K_{3} [Fe (CN)_{6}] + 2KOH + H_{2} O_{2} 2 K_{4} [Fe (CN)_{6}] + 2 H_{2} O + O_{2}

(d) It oxidises acidified ferrous sulphate to ferric sulphate.

2FeS O_{4} + H_{2} S O_{4} + H_{2} O_{2} F e_{2} (S O_{4})_{3} + 2 H_{2} O

(e) It oxidises sulphurous acid to sulphuric acid.

H_{2} S O_{3} + H_{2} O_{2} H_{2} S O_{4} + H_{2} O

Answer: in (a) and (c) hydrogen peroxide is the reducing agent; in (b), (d) and (e) it is the oxidising agent.

Solved Example 4

What happens when (i) chromium hydroxide is treated with

H_{2} O_{2}

in the presence of

NaOH

(ii) hydrazine reacts with

H_{2} O_{2}

(iii)

H_{2} O_{2}

reacts with

Ba (OH)_{2}

(iv) sodium hypochlorite reacts with

H_{2} O_{2}

Solution:

(i) Green chromium hydroxide is oxidised to yellow sodium chromate, which is the standard test for $H_{2} O_{2}$ in alkaline medium.

(ii) Hydrazine is oxidised to dinitrogen and water.

N_{2} H_{4} + 2 H_{2} O_{2} N_{2} + 4 H_{2} O

(iii) Acting as an acid, $H_{2} O_{2}$ neutralises the base and gives barium peroxide.

Ba (OH)_{2} + H_{2} O_{2} Ba O_{2} + 2 H_{2} O

(iv) Hypochlorite is reduced to chloride while $H_{2} O_{2}$ is oxidised to oxygen.

NaOCl + H_{2} O_{2} NaCl + H_{2} O + O_{2}

Answer: hydrogen peroxide is the oxidising agent in (i) and (ii), the acid in (iii) and the reducing agent in (iv).

Solved Example 5

Calculate the strength in

g L^{- 1}

, the molarity and the normality of a

40

volume solution of hydrogen peroxide.

Solution:

A $40$ volume solution means one litre gives $40 L$ of $O_{2}$ at STP.

2 H_{2} O_{2} 2 H_{2} O + O_{2}

$22.4 L$ of oxygen comes from $68 g$ of $H_{2} O_{2}$ , so:

mass per litre = \frac{68 \times 40}{22.4} = 121.43 g

M = \frac{121.43}{34 \times 1} = 3.57 M

N = \frac{121.43}{17 \times 1} = 7.143 N

Answer: $121.43 g L^{- 1}$ , $3.57 M$ and $7.143 N$ . The equivalent mass is $17$ because the $n$ factor of $H_{2} O_{2}$ is $2$ .

Solved Example 6

An aqueous solution of an inorganic compound (x) shows the following reactions. (i) It decolourises an acidified

KMn O_{4}

solution accompanied by the evolution of oxygen. (ii) It liberates iodine from an acidified

KI

solution. (iii) It gives a brown precipitate with alkaline

KMn O_{4}

solution with evolution of oxygen. (iv) It removes black stains from old oil paintings. Identify (x) and give the equations.

Solution:

Only one substance both reduces permanganate and oxidises iodide, so (x) is hydrogen peroxide.

2KMn O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 5 O_{2}

2KI + H_{2} S O_{4} + H_{2} O_{2} K_{2} S O_{4} + I_{2} + 2 H_{2} O

2KMn O_{4} + 3 H_{2} O_{2} 2KOH + 2Mn O_{2} + 2 H_{2} O + 3 O_{2}

PbS + 4 H_{2} O_{2} PbS O_{4} + 4 H_{2} O

Answer: (x) is $H_{2} O_{2}$ . Reaction (iv) is the restoration of old paintings: black lead sulphide is oxidised to white lead sulphate.

Solved Example 7

Why is hydrogen peroxide stored in wax-lined coloured bottles, and why is a stabiliser added?

Solution:

The decomposition $2 H_{2} O_{2}$ gives $2 H_{2} O + O_{2}$ is catalysed by traces of metal impurities, by strong bases and by light. A rough glass surface provides sites where the decomposition starts, and dissolved alkali leached from ordinary glass makes it worse.

Answer: the wax lining gives a smooth, inert, alkali-free surface, and the coloured glass keeps light out. A stabiliser such as phosphoric acid, glycerol or acetanilide is added because the reaction is otherwise fast enough in concentrated solution to be explosive.

Solved Example 8

20 mL

of a hydrogen peroxide solution reacts quantitatively with

20 mL

KMn O_{4}

solution in dilute

H_{2} S O_{4}

. The same volume of that

KMn O_{4}

solution is just decolourised by

10 mL

MnS O_{4}

solution in neutral medium, forming a dark brown precipitate of hydrated

Mn O_{2}

. The precipitate is dissolved in

10 mL

0.2 M

sodium oxalate on boiling with dilute

H_{2} S O_{4}

. Calculate the molarity of the

H_{2} O_{2}

solution.

Solution:

Step 1: find the $Mn O_{2}$ from the oxalate. Oxalate contains $10 \times 0.2 = 2 mmol$ . Both oxalate and $Mn O_{2}$ exchange two electrons each, so the $Mn O_{2}$ formed is also $2 mmol$ .

Mn O_{2} + N a_{2} C_{2} O_{4} + 2 H_{2} S O_{4} MnS O_{4} + N a_{2} S O_{4} + 2C O_{2} + 2 H_{2} O

Step 2: split that $Mn O_{2}$ between the two sources. In neutral medium all the manganese from both reactants ends up as $Mn O_{2}$ :

2KMn O_{4} + 3MnS O_{4} + 2 H_{2} O 5Mn O_{2} + K_{2} S O_{4} + 2 H_{2} S O_{4}

So the $2 mmol$ of $Mn O_{2}$ came from $KMn O_{4}$ and $MnS O_{4}$ in the ratio $2 : 3$ , which gives $0.8 mmol$ of $KMn O_{4}$ in the $20 mL$ used.

Step 3: use that $KMn O_{4}$ against the peroxide. In acidic medium the ratio is two permanganate to five peroxide:

2KMn O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 5 O_{2}

n_{H_{2} O_{2}} = 0.8 \times \frac{5}{2} = 2 mmol \Rightarrow M = \frac{2}{20} = 0.1

Answer: the hydrogen peroxide solution is $0.1 M$ , that is a $1.12$ volume solution.

Solved Example 9

The volume strength of

1.5 N

H_{2} O_{2}

solution is
(A)

4.8

(B)

8.4

(C)

3.0

(D)

8.0

Solution:

Answer: (B). Volume strength $= 5.6 \times N = 5.6 \times 1.5 = 8.4$ . Check through molarity: $M = N /2 = 0.75$ , and $11.2 \times 0.75 = 8.4$ .

Solved Example 10

In which reaction does

H_{2} O_{2}

act as a reducing agent?
(A)

PbS + 4 H_{2} O_{2}

(B)

2FeS O_{4} + H_{2} S O_{4} + H_{2} O_{2}

(C)

A g_{2} O + H_{2} O_{2}

(D)

2KI + H_{2} S O_{4} + H_{2} O_{2}

Solution:

Answer: (C). Silver oxide is reduced to silver and the peroxide oxygen rises from $- 1$ to $0$ , so dioxygen is given off:

A g_{2} O + H_{2} O_{2} ⟶ 2Ag + H_{2} O + O_{2}

In (A), (B) and (D) hydrogen peroxide is the oxidising agent, and water is formed with no gas.

Practice Questions

In basic medium $H_{2} O_{2}$ acts as an oxidising agent in its reaction with
(A) $C r_{2} (S O_{4})_{3}$ (B) $A g_{2} O$ (C) $K_{3} [Fe (CN)_{6}$ ] (D) $K_{2} C r_{2} O_{7}$ Answer: (A). Chromium goes from $+ 3$ to $+ 6$ as yellow $N a_{2} Cr O_{4}$ is formed, so it has been oxidised. With $A g_{2} O$ and $K_{3} [Fe (CN)_{6}$ ] the peroxide is the reducing agent.
Calculate the volume strength of a $0.5 M$ solution of hydrogen peroxide.Answer: $5.6$ volume, from $11.2 \times 0.5$ .
Why can a dilute solution of $H_{2} O_{2}$ not be concentrated by boiling?Answer: It decomposes into water and oxygen well below its boiling point, so boiling destroys the product instead of concentrating it.
Why is anhydrous barium peroxide not used in the laboratory preparation?Answer: The $BaS O_{4}$ formed makes a protective layer around the unreacted $Ba O_{2}$ and stops the reaction. Hydrated $Ba O_{2}$ avoids this.
Name the reaction type in $2 H_{2} O_{2}$ giving $2 H_{2} O + O_{2}$ and explain the name.Answer: Disproportionation. Oxygen at $- 1$ in the peroxide is reduced to $- 2$ in water and oxidised to $0$ in $O_{2}$ in the same reaction.
What is seen when $H_{2} O_{2}$ is added to an acidified solution of potassium dichromate, and the mixture is shaken with ether?Answer: A deep blue colour, due to $Cr O_{5}$ , which passes into the ether layer. This is the standard test for $H_{2} O_{2}$ .

Common Mistakes to Avoid

Watch out

Assuming $H_{2} O_{2}$ is always an oxidising agent. Against $KMn O_{4}$ , $O_{3}$ and $A g_{2} O$ it is the reducing agent and gives off oxygen.
Writing the peroxide oxygen as $- 2$ . In the $O - O$ link it is $- 1$ , which is the whole reason the molecule can go either way.
Using concentrated or anhydrous reagents in the laboratory preparation. The acid must be dilute and ice-cold, and the $Ba O_{2}$ must be hydrated.
Calling the structure planar. Hydrogen peroxide is non-planar, with an open book shape and a dihedral angle of $111. 5^{\circ}$ in the gas phase.
Trying to concentrate $H_{2} O_{2}$ by boiling. It decomposes first; distillation must be carried out under reduced pressure.
Forgetting the $n$ factor of $2$ in titration problems, which makes normality twice the molarity.
Thinking the bleaching is a reduction. $H_{2} O_{2}$ bleaches by oxidising the colouring matter with nascent oxygen, which is why the bleaching is permanent.
Storing $H_{2} O_{2}$ in a clear, ordinary glass bottle. Light, a rough surface and alkali from the glass all speed up its decomposition.

Frequently Asked Questions

How is hydrogen peroxide prepared in the laboratory?

A paste of hydrated barium peroxide in ice-cold water is added slowly to ice-cold dilute sulphuric acid. Insoluble $BaS O_{4}$ is filtered off, leaving about a $5%$ solution of $H_{2} O_{2}$ . Hydrated $Ba O_{2}$ is used because the sulphate would coat the anhydrous solid and stop the reaction.

Why is the structure of hydrogen peroxide called an open book structure?

Because the two $O - H$ bonds lie in two different planes that meet along the $O - O$ bond, like two pages of a half-open book with the $O - O$ bond as the spine. The angle between the planes is $111. 5^{\circ}$ in the gas phase and $90. 2^{\circ}$ in the solid.

Why can hydrogen peroxide act as both an oxidising and a reducing agent?

Because oxygen in the peroxide link has an oxidation state of $- 1$ , half way between $0$ in $O_{2}$ and $- 2$ in water. It can therefore be reduced to water, acting as an oxidising agent, or oxidised to dioxygen, acting as a reducing agent.

What is the volume strength of hydrogen peroxide?

It is the volume of oxygen at STP that one volume of the solution releases on complete decomposition. A $10$ volume solution gives $10 L$ of oxygen per litre. Volume strength is $11.2$ times the molarity and $5.6$ times the normality.

Why is hydrogen peroxide stored in coloured wax-lined bottles?

Because light, rough glass surfaces, alkali leached from glass and traces of metal impurities all catalyse its decomposition into water and oxygen. The wax lining gives a smooth inert surface, the dark glass blocks light, and a stabiliser such as phosphoric acid or acetanilide is added as well.

How does hydrogen peroxide restore old oil paintings?

White lead in the old paint reacts with traces of hydrogen sulphide in the air over the years and turns into black $PbS$ . Hydrogen peroxide oxidises that black sulphide to white lead sulphate, so the original colours reappear without the paint being touched.

Is hydrogen peroxide asked in NEET?

Not as a topic, because NMC removed the Hydrogen chapter from the NEET syllabus. Its reactions are still good practice for redox: balancing the permanganate and iodide reactions and working out the oxidation state of peroxide oxygen are skills NEET tests in the redox chapter.

Is hydrogen peroxide in the JEE syllabus?

Yes for JEE Advanced: the syllabus lists hydrogen peroxide, its preparation, reactions, use and structure, under the Hydrogen unit. JEE Main removed the Hydrogen chapter, although volume strength and the redox reactions of $H_{2} O_{2}$ are useful practice for its mole concept and redox questions.

Previous year questions on Hydrogen Peroxide

1 question from past papers, each with a step-by-step solution.

JEE Advanced 2022 Paper 1, Chemistry Section 3 Q2

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