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Hydrogen peroxide
Laboratory Preparation:
(i) From Sodium
peroxide: It is prepared in the laboratory by adding small
amounts of sodium peroxide to ice cold water.
Na2O2 + 2H2O ¾® 2NaOH + H2O2
In place of cold
water, dilute sulphuric acid (20%) can be used. Sodium peroxide in small
amounts at a time is added to cold 20% sulphuric acid.
Na2O2 + H2SO4 ¾®
Na2SO4 + H2O2
The crystals of
hydrated sodium sulphate (Na2SO4·10H2O) are removed by cooling the solution.
The solution of hydrogen peroxide obtained always consists some dissolved
sodium sulphate. The concentration of H2O2 is about 3%.
BaO2·8H2O + H2SO4 = BaSO4 + H2O2 + 8H2O
The barium sulphate formed is filtered off and a
solution of H2O2 is obtained.
The use of sulphuric
acid has a disadvantage as it catalyses the decomposition of hydrogen peroxide
formed. In place of sulphuric acid, weak acids like orthophosphoric acid,
carbonic acid are preferred. Barium peroxide on treatment with orthophosphoric
acid gives a precipitate of barium phosphate and H2O2 which goes into the solution.
3BaO2 + 2H3PO4 ¾®
Ba3(PO4)2 + 3H2O2
ppt.
Insoluble barium
phosphate is filtered off. The filterate consists of hydrogen peroxide.
The barium phosphate
is decomposed by dilute sulphuric acid.
Ba3(PO4)2 + 3H2SO4
¾®
3BaSO4 + 2H3PO4
BaSO4 is removed by filtration and orthophosphoric
acid is used again.
Merck’s process: H2O2 can be obtained by passing a current of CO2 through a cold pasty solution of barium
peroxide in water.
BaO2 + CO2 + H2O ¾® BaCO3 + H2O2
Barium carbonate being
insoluble is filtered off.
Manufacture of Hydrogen Peroxide
(i) Auto oxidation
of 2-butyl anthraquinol
(ii) By Oxidation of
isopropyl alcohol
(iii) Electrolytic
process: In this process, the electrolysis of 50% sulphuric
acid is carried out at low temperature using platinum electrodes and a current
of high density. Proxy disulphuric acid is formed.
2H2SO4 2H+ + 2HSO4–
2HSO4– ¾®
H2S2O8 + 2e (anode)
peroxy
disulphuric acid
Peroxy disulphuric
acid is distilled. Hydrolysis occurs and a distillate containing about 30% H2O2 is obtained.
H2S2O8 + 2H2O ¾® 2H2SO4 + H2O2
In place of 50% H2SO4, ammonium hydrogen sulphate dissolved
in excess of sulphuric acid can be used for
electrolysis.
NH4HSO4 NH4SO4– + H+
At anode 2NH4SO4– ¾®
(NH4)S2O8 + 2e
Ammonium peroxy
disulphate
At cathode 2H+ + 2e ¾®
H2(g)
The solution
containing ammonium peroxy disulphate is heated at 43 mm pressure when its
hydrolysis occurs forming hydrogen peroxide.
(NH4)2S2O8 + 2H2O ¾® 2NH4·HSO4 + H2O2
Hydrogen peroxide
along with water distils over. Ammonium bisulphate can be used again. This
method gives 30–40% aq. solution of hydrogen peroxide.
Properties of H2O2
Physical Properties:
(i) Pure anhydrous hydrogen peroxide is a syrupy liquid. It is colourless but
gives a bluish tinge in thick layers. It is odourless.
(ii) It is soluble in
water, alcohol and ether.
(iii) It has bitter
taste. It is injurious to skin (blisters are formed).
Chemical Properties:
(i) Stability: It is unstable in nature. It decomposes on standing and
heating. It is an example of auto oxidation-reduction reaction.
2H2O2 = 2H2O + O2
(ii) Acidic nature:
The pure liquid has weak acidic nature but its aqueous solution is neutral
towards litmus. It reacts with alkalies and carbonates to give their
corresponding peroxides.
H2O2 + 2NaOH = Na2O2 + 2H2O
H2O2 + Ba(OH)2 = BaO2 + 2H2O
H2O2 + Na2CO3 = Na2O2 + CO2 + H2O
(iii) Oxidising
nature: It is a powerful oxidising agent. It is a electron
acceptor in acidic and alkaline solutions.
H2O2 + 2H+ + 2e = 2H2O (In
acidic solutions)
H2O2 + 2e = 2OH- (In alkaline solutions)
The reactions are
generally slow in acid solutions but fast in alkaline solution.
Oxidising nature of
hydrogen peroxide can be interpreted on account of the possession of label
oxygen atom. The potential equation for its oxidising nature can be written as,
H2O2 ¾® H2O + O
the following examples
show the oxidising nature of H2O2:
(a) It oxidises black
lead sulphide (PbS) to which lead sulphate (PbSO4).
PbS
+ 4H2O2 ¾® PbSO4 + 4H2O
(b) It oxidises
nitrites to nitrates.
NaNO2 + H2O2
¾®
NaNO3 + H2O
(c) It oxidises
sulphites into sulphates.
Na2SO3 + H2O2 ¾® Na2SO4 + H2O
(d) It oxidises
arsenites into arsenates
Na3AsO3 + H2O2 ¾®
Na3AsO4 + H2O
(e) It liberates
iodine from potassium iodide
2KI
+ H2O2 ¾® 2KOH + I2
(f) It oxidises H2S into sulphur
H2S + H2O2 ¾® 2H2O + S
(g) It oxidises
acidified ferrous sulphate to ferric sulphate
2FeSO4 + H2SO4 + H2O2 ¾® Fe2(SO4)3 + 2H2O
(h) It oxidises
acidified potassium ferrocyanide to potassium ferricyanide
2K4Fe(CN)6 + H2SO4 + H2O2 ¾®
2K3Fe(CN)6 + K2SO4 + 2H2O
(i) It oxidises
formaldehyde to formic acid. This oxidation occurs in presence of pyrogallol
and in alkaline medium.
HCHO
+ H2O2 ¾® HCOOH + H2O
or HCHO +
H2O2 ¾® 2HCOOH + H2
(j) Benzene in
presence of ferrous sulphate is oxidised to phenol.
(k) It dissolves the
chromic hydroxide precipitate present in NaOH solution forming a yellow
solution of sodium chromate.
2Cr(OH)3 + 4NaOH + 3H2O2 ¾® 2Na2CrO4 + 8H2O
(l) A solution of
chromic acid in sulphuric acid or acidified potassium dichromate is oxidised to
blue peroxide of chromium (CrO5) which is unstable, however, it is
soluble in ether and produces blue coloured solution.
K2Cr2O7 + H2SO4 + 4H2O2 ¾®
2CrO5 + K2SO4 + 5H2O
Peroxide of chromium
decomposes to form chromic sulphate and oxygen.
4CrO5 + 6H2SO4 ¾®
2Cr2(SO4)3 + 6H2O + 7O2
Peroxide of chromium
is represented as
(m) In acidic solution, mercury is oxidised to
mercuric oxide.
Hg
+ H2O2HgO + H2O
(n) Bleaching action: Due to its oxidising nature, it
acts as a bleaching agent.
Coloured
material + O ¾® Colourless
It bleaches materials like silk, hair, ivory, cotton,
wool, etc.
(iv) Reducing nature: It can also act as a reducing
agent towards powerful oxidising agents.
H2O2 ¾® 2H+ + O2 + 2e–
In alkaline solution, however, its reducing action is
more effective.
H2O2 + 2OH– ¾®
2H2O
+ O2 + 2e–
The potential equation when H2O2 acts as a reducing agent can be expressed as,
H2O2 + O ¾® H2O + O2
(a) Ag2O is reduced to silver
Ag2O + H2O2 ¾® 2Ag + H2O + O2
(b) It reduces ozone to oxygen.
H2O2 + O3 ¾®
H2O + 2O2
(c) Manganese dioxide in acidic medium is reduced to
manganous salt.
MnO2 + H2SO4 ¾®
PbO + H2O
+ O2
(e) Red lead in presence of HNO3 is reduced to plumbous salt.
Pb3O4 + 6HNO3 + H2O2 ¾®
3Pb(NO3)2 + 4H2O + O2
(f) Chlorine and bromine are reduced to corresponding
hydracids.
Cl2 + H2O2 ¾®
2HCl + O2
This reaction can be shown in following steps:
Cl2 + H2O2 ¾®
2HCl + O2
Similarly,
Br2 + 2O2 ¾®
2HBr + O2
(g) It reduces acidified KMnO4 solution, i.e., acidified KMnO4 is decolorised by H2O2.
2KMnO4 + 3H2SO4 + 5H2O 2 ¾®
K2SO4 + 2MnSO4 + 8H2O + 5O2
(h) Potassium ferricyanide (alk. solution) is reduced
to potassium ferrocyanide.
2K3Fe(CN)6 + 2KOH + H2O2 ¾®
2K4Fe(CN)6 + 2H2O + O2
(i) Hypohalites are reduced to halides
NaOCl
+ H2O2 ¾® NaCl + H2O + O2
(j) KIO4 is reduced to KIO3
KIO4 + H2O2 ¾®
KIO3 + H2O + O2
Uses:
(a) for restoring colour of old paintings.
(b) as ‘antichlor’ to remove traces of chlorine and
hypochlorite.
(c) Highly
concentrated solution (about 40%) of H2O2 is used to oxidise petrol, alcohol, and
hydrozine hydrate for the propelling of rockets and torpedoes.
NH2·NH2 +
2H2O2 ¾® N2 + 4H2O
Problem for practice
Q.1 H2O2 can be obtained when following reacts with H2SO4 except with -
(A) PbO2 (B)
BaO2
(C) Na2O2 (D) SrO2
Q.2 H2O2
¾® 2H+ + O2 + 2e¯,
E0 = – 0.68 V
The above equation represent which of the following behaviour
of H2O2 ?
(A) Reducing agent (B)
Oxidising agent
(C) Acidic (D)
Catalytic
Q.3 Which one
of the following reactions shows oxidising nature of H2O2
?
(A) H2O2 + 2KI ¾® 2KOH + I2
(B) Cl2 + H2O2 ¾® 2HCl + O2
(C) H2O2 + Ag2O ¾® 2Ag +
H2O + O2
(D) NaClO + H2O2 ¾® NaCl +
H2O + O2
Q.4 In which of
the following reaction, H2O2 is behaving as a reducing agent -
(A) C6H6 + H2O2 ¾® C6H5OH + H2O
(B) PbS + 4H2O2 ¾® PbSO4 + 4H2O
(C) 2I¯ + 2H+ + H2O2 ¾® I2 + 2H2O
(D) 2MnO4¯ + 6H+ + 5H2O2 ¾®
2Mn2+ + 5O2 + 8H2O
Q.5 The
reaction H2S + H2O2 ¾® S + 2H2O
manifests -
(A) Acidic nature of H2O2
(B) Alkaline nature of H2O2
(C) Oxidising action of H2O2
(D) Reducing nature of H2O2
Q.6 Which is
true about different forms of hydrogen -
(A) Ortho hydrogen
has same spins of two nuclei clockwise or anticlockwise
(B) Para hydrogen has different spins of two nuclei
(C) At absolute
zero, there is 100% para form and at high temperature, there is 75% ortho form
(D) All are correct
Q.7 H2O2
¾® H2O + O2
This represents -
(A) Oxidation of H2O2
(B) Reduction of H2O2
(C) Disproportionation of H2O2
(D) Acidic nature of H2O2
Q.8 10 Volumes of H2O2 has a
strength of approximately -
(A) 3% (B) 30%
(C) 10% (D) 5%
Q. 9 A given solution of H2O2
is 30 volumes. Its concentration in terms of molarity is -
(A) 9.1 M (B) 2.68 M
(C) 2.5 M (D) 26.8
M
Q. 10 When the same amount of zinc is treated separately
with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes
of hydrogen evolved is -
(A) 1 :
1 (B) 1
: 2
(C) 2 : 1 (D) 9:
4
Q.11 The volume of '10 Vol' of H2O2
required to liberate 500 mL O2 at NTP is -
(A) 50
mL (B) 25 mL
(C) 100
mL (D) 125 mL
Q.12 Decomposition of H2O2 is accelerated by -
(A) Traces of acids
(B) Acetanilide
(C) Finely divided metals
(D) Alcohol
Q.13 Hydrogen
peroxide does not -
(A) Liberate iodine from KI
(B) Turn the titanium salt yellow
(C) Give silver peroxide with moist silver oxide
(D) Decolourise the violet colour of KMnO4 solution
Q.14 Which of the
following statements about H2O2 is not true -
(A) H2O2
is used to clean oil paintings
(B) H2O2
acts as oxidising as well as reducing agent
(C) Two hydroxyl
groups in H2O2 lie in the same plane
(D) It retains same
structure in liquid and solid form
Q.15 For which
purpose H2O2 is used -
(A) In refreshing old oil painting
(B) In detection of Ti4+, Cr3+ ions
(C) As water softener
(D) As rocket fuel
Q.16 Hydrogen
peroxide is prepared by which of the following acid ?
(A) H2SO4 (B) HCl
(C) HNO3 (D)
H3PO4
Q.17 H2O2 is obtained when dilute H2SO4
reacts
with - [REE-1994]
with - [REE-1994]
(A) BaO2 (B)
MnO2
(C) Na2O2 (D) PbO2
Q.18 H2O2 is prepared by the action of -
[REE-1997]
(A) Cold H2SO4 on BaO2
(B) Cold H3PO4 on BaO2
(C) Cold H2SO4 on MgO2
(D) Dilute HCl on MnO2
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