Hydrogen - Practically Study Material

HYDROGEN

Introduction:

In 1766, Sir Henry Cavendish prepared hydrogen in pure state by the action of disulphuric acid on zinc. It was named as inflammable air. Lavoisier a French chemist called it hydrogen mean water producer i.e., Greek hydro-water; gene-producer. Hydrogen is the lightest element of periodic table. Atomic form of hydrogen exists only at high temperature. In the normal elemental form, it exists as a diatomic molecule. Hence it is also called dihydrogen.

Occurrence:

Hydrogen is the most abundant element in the universe about 76% eg, Jupiter & Saturn contain mostly hydrogen. It is third most abundant element on the surface of the globe. In our atmosphere it occurs in a minute extent (i.e. H₂ is not found) but in solar atmosphere, H₂ is the main constituent as nuclear fuel. In the combined state, it constituents 15.4% of the earth crust and the oceans. In free state, it occurs in volcanic gases, some times in natural gases.

Unique position of hydrogen in the periodic table: It is a first element of the periodic table. The position of hydrogen in the periodic table is controversial or anomalous i.e. not fixed due to its resemblance both with alkali metals and halogens.

Resemblance with alkali metals:

(i) Like alkali metals, it has only one electron in the outer shell.

(ii) Like alkali metal, it loses its only electron to form hydrogen ion i.e. H⁺.

(iii) Like alkali metals, it exhibits an oxidation state of +1 in its compounds.

(iv) Like alkali metal, it has a strong affinity for non-metals.

(v) Like alkali metals, it also acts as strong reducing agent.

Resemblance with halogens:

(i) Like halogens, it has required one electron to gain the stable electronic configuration.

(ii) Like halogen, it can gain an electron and form a negative ion.

(iii) The ionization energy value of hydrogen is comparable to that of halogens i.e. ${IE}_{H} = 1312 kJ {mol}^{- 1}$ .

(iv) Like halogens, it exhibits oxidation number of -1 in some of its compounds.

(v) Like halogens, it can combine with alkali metals to form salts with similar formulae.

(vi) Like halogens, it also exists as a diatomic molecule.

On the basis of some resemblance it can be placed either in group 1 of periodic table among alkali metals or group 17 of periodic table among halogens.

Isotopes of hydrogen: Hydrogen has three isotopes.

Relative atomic mass 1.007825 2.014102 3.016049

Relative abundance 99.985 0.0156 10^–¹⁵

Radioactive stability Non radioactive Non radioactive radioactive

(Stable) (Stable) t_1/2 = 12.33 yrs

Discovery → Prestley Urey Bleakney Gould

Isotopic effect:

In general, chemical properties of isotopes are same but quantitative differences are noticed amongst them e.g. the reaction between H₂ and Cl₂ is 13.4 times faster than between D₂ and Cl₂ under similar conditions. Such differences in chemical properties which are due to difference in atomic masses is called isotopic effect.

Preparation of Hydrogen:

(i) From cold water: By the action of very active metals like Na, K, Ca etc.

$2 Na + 2 H_{2} O ⟶ 2 NaOH + H_{2} (highly exothermic and H_{2} catches fire )$

Zn-Cu and Al-Hg couple decompose water to give nascent hydrogen.

$Zn + H_{2} O ⟶ Zn (OH)_{2} + 2 H$

(ii) From boiling water: Less active metals like Zn, Mg, Al etc decompose boiling water liberating dihydrogen.

$2 Al + 3 H_{2} O \overset{Δ}{⟶} {Al}_{2} O_{3} + 3 H_{2}$

(iii) From steams: Less reactive metals like Fe, Sn, Ni etc, decompose steam at high temperature.

$3 Fe + 4 H_{2} O ⟶ {Fe}_{3} O_{4} + 4 H_{2}$

(iv) By electrolysis of water: Pure water is bad conductor of electricity.

$2 H_{2} O (ℓ) \frac{electrolysis}{acid/base} \to 2 H_{2} (g) + O_{2} (g)$

(v) From alkalies: By the action of Be, Zn, Al, Sn etc.

$Zn + \underset{(b o i l)}{2 NaOH} ⟶ {Na}_{2} {ZnO}_{2} + H_{2}$

(vi) From dil. Acids (Lab): By the action of Sn, Zn, Fe, Mg etc.

$Sn + 2 HCl ⟶ {SnCl}_{2} + H_{2} \underset{(granulated)}{Zn} + H_{2} {SO}_{4} ⟶ {ZnSO}_{4} + \underset{collected by down displacement of water}{H_{2}}$

Preparation of pure hydrogen (>99.95%):

(i) $\underset{(Ribbbon)}{Mg} + \underset{(Pure oil.)}{H_{2} {SO}_{4}} ⟶ {MgSO}_{4} + H_{2}$

(ii) $2 NaOH + 2 Al + 2 H_{2} O \frac{U y e n o ‘ s}{method} ⟶ 2 {NaAlO}_{2} + 3 H_{2}$

(iii) $NaH + H_{2} O ⟶ NaOH + H_{2}$

$(i) \underset{(R i b b o n)}{Mg} + \underset{(P u r e d i l)}{H_{2} {SO}_{4}} ⟶ {MgSO}_{4} + H_{2}$

Manufacture of H_{2 $:$}

(i) From water gas (Bosch process): By passing a mixture of water gas and steam over heated catalyst, Fe₃O₄ and Cr₂O₃ or FeCrO₄, at high temperature, H₂ is obtained.

$\underset{(c o a l o r c o k e)}{C (s)} + \underset{(S t e a m)}{H_{2} O (g)} \frac{1270 K, Ni}{coal gasficafion} \to \underset{Water gas (1.1) o r s y n g a s}{\underset{⏟}{CO (g) + H_{2} (g)}} - 121.3 k$

$(CO + H_{2}) + \underset{(steam)}{H_{2} O} \frac{{Fe}_{2} O_{3} \cdot {Cr}_{2} O_{3} or {FeCrO}_{4}}{\underset{(water gas shift reaction)}{450 - 500^{\circ} C}} {CO}_{2} + 2 H_{2}$

The CO₂ gas is removed by dissolving gaseous mixture in water under high pressure (25 atm)

(ii) By lane’s process: The super heated steam is passed over heated iron at 600-800°C.

$3 Fe + 4 H_{2} O \overset{oxidation}{⇌} {Fe}_{3} O_{4} + 4 H_{2} ↑ m\} gassin g$

Iron is regenerated by passing water gas (H₂O + CO). This reaction is called vivifaction and time allotted for this reaction is about 20 minutes.

$\begin{array}{l} {Fe}_{3} O_{4} + 4 CO ⟶ 3 Fe + 4 {CO}_{2} \\ {Fe}_{3} O_{4} + 4 H_{2} ⟶ 3 Fe + 4 H_{2} O \end{array}\} vivifaction$

This iron is again used for decomposition of steam. In order to make the process continuous, the above two reactions are carried out alternatively using two or more furnaces.

(iii) From natural gas: Natural gas containing CH₄ is also a good source of hydrogen. Natural gas is mixed with steams and then passed over ‘Ni’ catalyst heated to 900°C.

$\underset{Natural gas}{{CH}_{4}} (g) + \underset{Steam}{H_{2} O (g)} \to_{900^{0} C}^{N i} \underset{Syngas}{\underset{⏟}{CO (g) + 3 H_{2} (g)}}$

CO from the mixture is removed as in Bosch’s process. This process of obtaining H₂ from natural gas is called steam reforming process.

(iv) Electrolytic method: By the electrolysis of acidified or alkaline water.

$\underset{(alk)}{H_{2} O} ⇌ H^{+} + {OH}^{-}$

At cathode $2 H^{+} + 2 e^{-} ⟶ H_{2}$

At anode $4 {OH}^{-} - 4 e^{-} ⟶ 4 OH$

$4 OH ⟶ 2 H_{2} O + O_{2}$

(v) As by product:

When NaOH is manufactured by electrolysis of NaCl in Nelson or castner kellner cell, hydrogen is obtained as by product.

Reaction: $NaCl {⇌ Na}^{+} + {Cl}^{-}$

At anode: $2 {Cl}^{-} + 2 e^{-} ⟶ {Cl}_{2}$

At cathode: $2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2 {OH}^{-}$

$2 {Na}^{+} + 2 {OH}^{-} ⟶ 2 NaOH$

Properties of Hydrogen:

Physical properties:

It is a colourless, odourless and tasteless gas. It is nonpoisonous but in presence of AsH₃ is poisonous. It is slightly soluble in water. Its critical temperature is -234.5°C, which makes its liquification difficult. It is lightest of all element (density = 0.08987 gm L^–¹). B.P = -252.5 °C (20.5 K)

M.P. = -259.0 °C (14.0 K). Ionization potential = 13.54 eV/atm. Electronegativity = 2.1

Chemical properties:

(i) Combustion: It burns with blue flame in oxygen atmosphere. It is not a supporter of combustion.

$2 H_{2} + O_{2} \overset{exo}{⟶} 2 H_{2} O; Δ H = - 485 {KJmol}^{- 1}$

Care should be taken with these gases since mixture of H₂ and O₂ close to a 2:1 ratio is often explosive.

ii) Reaction with non-metals:

$H_{2} + X_{2} \to AHX (X = F, Cl, Br, I)$

$3 H_{2} + N_{2} \to 2 {NH}_{3}$

iii) Reaction with metals:

A number of metals react with H₂ forming hydrides.

The reactions are not violent and usually require a high temperature.

$\begin{array}{l} 2 Li + H_{2} \to 2 LiH \\ 2 Na + H_{2} \to 2 NaH \end{array}$

Metals like Fe, Ni and Pd form interstitial or metallic hydrides.

iv) Reducing property: The oxides of less electropositive metals such as copper, tin, iron, lead, etc, are reduced to the metals when heated in hydrogen.

$\begin{array}{l} PbO + H_{2} \to Pb + H_{2} O {Ag}_{2} O + H_{2} \to 2 Ag + H_{2} O \\ CuQ + H_{2} \to Cu + H_{2} O {Fe}_{3} O_{4} + 4 H_{2} \to 3 Fe + 4 H_{2} O \end{array}$

Applications of Hydrogen:

i) Manufacture of methyl alcohol in presence of ZnO, Cr₂O₃ catalyst.

$CO + 2 H_{2} \frac{{ZnO}_{2}, {Cr}_{2} O_{S}}{300^{\circ} C, 200 stm} {CH}_{3} OH$

ii) Manufacture of Ammonia – Haber’s process.

$N_{2} + 3 H_{2} \frac{Fectisyst}{450 - 500 ° C} 2 {NH}_{3}$

iii) Manufacture of hydrogen chloride (absorbed in water).

$H_{2} + {Cl}_{2} \to 2 HCl$

iv) Hydrogenation of oils

$Unsaturated oil + H_{2} \overset{Ni}{⟶} saturatedfat or ghee$

v) Synthetic petrol – Fisher – Tropsch process

$a) n (CO + H_{2}) + {nH}_{2} \to C_{n} H_{2 n} + {nH}_{2} O$

$b) nCO + (2 n + 1) H_{2} \to \underset{paraffin}{C_{n} H_{2 n + 2}} + {nH}_{2} O$

vi) When mixed with helium: It is used for filling balloons.

vii) Oxy – hydrogen flame: It produces temperature of 2800 – 3000⁰C and oxyatomic hydrogen flame produces a temperature of 4000⁰. It is used for welding purposes and for melting platinum and quarts. The atomic hydrogen welding torch is also used for the same purpose.

viii) As cryogenic fluid – To produce low temperature.

ix) As fuel:

a) In rocket fuel: Because of its low mass and high enthalpy of combustion. Saturn V, the rocket that took Neil Armstrong to the moon was powered by liq. hydrogen fuel. Oxygen (need for combustion) as well as hydrogen were carried in rocket in separate tanks. Inspite of its high heat of combustion (242 kJ mol^–¹) its use as a fuel is limited due to lack of availability of free hydrogen in nature & the difficulty of its safe storage and distribution.

b) In coal gas:

Composition: H₂ = 45 – 55%, N₂ = 2 – 12%, CH₄ = 25 – 35%

CO₂ = 0 – 3%, CO = 4 – 11% , O₂ = 1 – 1.5%

Ethylene, acetylene, benzene, etc = 2.5 – 5%.

It’s calorific value is about 20,00kJ/m³.

Uses:

a) As illuminant

b) As a fuel

c) To provide a inert atmosphere in metallurgical processes.

c) In water gas: (H₂ + CO)

Preparation :

$\underset{\underset{coke or coal}{hotbed of}}{C} + \underset{steam}{H_{2} o} \overset{1000^{0} C}{\to} CO \underset{water}{+} H_{2} (endothermic)$

This reaction is endothermic; hence, a blast of steam brings down the temperature of the bed of coke in a few minutes. To maintain the temperature above 1000⁰C alternate blast of air and steam are passed over the coke bed.

Uses of water gases:

i) It burns with blue flame, calorific value is 13000kJ/m³.

ii) As industrial source of hydrogen (Bosch process)

iii) In manufacture of methyl alcohol (Patart Process) and synthetic petrol (Fisher – Tropsch process).

d) In carburetted water gas: It is a enriched mixture of water gas and gaseous hydrocarbons obtained from cracking of petroleum oils.

The gaseous hydrocarbons are used to increase the calorific value of water gas.

e) In semi water gas: It is a mixture of water gas and producer gas.

Composition: CO = 25 – 28% ; H = 10 – 12%; CH₄ = 1 – 2%; CO₂ = 4 – 5%; N₂ = 50 – 55%

Its calorific value is 160 – 180 BTU per cubic foot.

Uses:

i) As fuel in steel industry

ii) For production of power in internal combustion engine.

f) In fuel Cell (H₂–O₂ fuel cell): Electrical cells that are designed to convert the energy from the combustion of fuels like hydrogen and carbon monoxide or methane directly into electrical energy are called fuel cells. This cell was used to power the Apollo space program and to form the drinking water for the astronauts. In this cell, hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide or potassium hydroxide solution (electrolyte). Hydrogen is fed into the anode compartment where it is oxidized and the oxygen is fed into cathode compartment where it is reduced. Each electrode is made of porous compressed carbon containing a small amount of catalyst like ${Pt}_{1} Ag or C_{0} O$

The electrode reactions are given below

Anode: $[H_{2 (g)} + 2 O {\bar{H}}_{(sq)} \to 2 H_{2} O (ℓ) + 2 e^{-}] 2$

Cathode : $[O_{2 (g)} + 2 H_{2} O (ℓ) + 4 e^{-} \to 4 {OH}^{-} (aq)]$

—————————————————————-

Overall cell reaction: $2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O (ℓ)$

Advantages:

i. High efficiency (60 – 70%)

ii. Continuous source of energy

iii. Pollution free working.

xii) In metallurgy: Hydrogen is used in metallurgy of molybdenum and tungsten for obtaining pure metals from their oxides at high temperature.

${WO}_{3} + 3 H_{2} \to W + 3 H_{2} O$

xiii) In nuclear fusion: (Thermo nuclear reaction):

$\begin{array}{l} _{1}^{3} H +_{1}^{2} H \to_{2}^{4} He +_{0}^{1} n + energy \\ Tritium deuterium \end{array}$

Special forms of Hydrogen:

i) Nascent hydrogen: Freshly prepared hydrogen is known as nascent hydrogen and is more reactive than ordinary hydrogen.

ii) Atomic hydrogen: When hydrogen is passed through an electric arc established between two tungsten filaments, hydrogen is dissociated into atoms. This form of hydrogen is known as atomic hydrogen. Reducing power of atomic hydrogen is more than that of nascent hydrogen.

iii) Active hydrogen: It is obtained by subjecting molecular hydrogen to silent electric discharge at ordinary temperature and 0.5 mm pressure. It has great chemical activity and its half life period is similar to that of atomic hydrogen.

iv) Occluded hydrogen: It is more active than ordinary hydrogen. Hydrogen adsorbed by certain metals e.g., Pd, Pt, Fe, Co, Ni etc., is known as occluded hydrogen. One volume of finely divided metals adsorb the following volumes of hydrogen. Pd black = 870, Pt = 49.5; Au = 3, Fe = 15.7, Cu = – 4.5, Al = 2.7. Colloidal Pd takes up 2050 volumes of H₂ than itself. Occlusion decreases with rise in temperature. Order of occlusion Colloidal Pd > Pt > Au > Ni

v) Ortho and para hydrogen (Allotropes): When the spins of the two protons (nuclei) are parallel in the hydrogen molecule, it is known as ortho hydrogen (s = 1). When the spins of the two protons (nuclei) are antiparallel, it is known as para hydrogen (s = 0). Orthoform is more stable than para and para form always tends to revert in orthoform, still orthoform has not been isolated in the pure form. When ordinary H₂ is passed over activated charcoal at 20K in quartz vessel. The charcoal adsorb only para H₂. The charcoal consists of about 99.8%, para form and may be kept for a week in glass vessel at room temperature without any appreciable change to orthoform. Thus at room temperature ordinary H₂ is an equilibrium mixture of 75% ortho and 25% para hydrogen. At low temperature, the percentage of para form increases.

Difference in ortho & para forms:

The two forms are differ in their physical properties but their chemical properties are same

i) Ortho is more stable than para.

ii) Vapour pressure of ortho is more than para.

iii) Specific heat of ortho is more than para.

iv) Conductivity of ortho is less than para.

v) Magnetic moment of para is zero while ortho has twice that of proton.

Transportation of H₂:

It is transported in the form of hydrolith (CaH₂) or ammonia (NH₃). Ammonia is cracked by passing over heated catalysts yielding a mixture of hydrogen (75%) & nitrogen (25%).

$2 {NH}_{3} \to \underset{(75 %)}{3 H_{2}} + \underset{(25 %)}{N_{2}}$

Hydrogen economy:

Hydrogen economy is basically a use of liquid hydrogen as an alternate source of energy. The technology involves the production, transportation and storage of energy in the form of liquid hydrogen in bulk quantities. One major advantage of using hydrogen as a fuel is that it is environmentally clean, giving only water as a combustion product i.e., cleaner technology. The second advantage with hydrogen is that the heat of courbustion of hydrogen per gram is higher than any other fuel. e.g., $H_{2} + \frac{1}{2} O_{2} \to H_{2} O; {ΔH}^{0} = - 284 kJ {Mol}^{- 1}$

Illustration 1: How do isotopes of hydrogen differ in their nuclear composition?

Solution: Isotopes of hydrogen have only one proton but contain nil, 1 and 2 neutrons in the nucleus of protium, deuterium and tritium respectively.

Illustration 2: Name the material which shows a great promise for holding hydrogen during the use of hydrogen in vehicles.

Solution: Carbon nanotubes and metal hydrides (of magnesium, magnesium-nickel alloys and iron- titanium alloys) can hold large quantities of hydrogen. Therefore, these materials are very potential for storing (holding) hydrogen gas.

Illustration 3: In the reaction, Zn(s) + H₂SO₄(dil) → ZnSO₄(aq) + H₂(g)

a) Which is more electropositive – hydrogen or zinc?

b) Name another metal that gives similar reaction with dilute H₂SO₄ but faster.

Solution: a) In this reaction, zinc (Zn) displaces hydrogen (H) from H₂SO₄. Therefore, zinc is more electropositive than hydrogen.

b) Magnesium, because magnesium (Mg) is more electropositive (reactive) than zinc (Zn).

Hydrides: Binary compounds of hydrogen and other elements are called hydrides. The types of hydrides depend upon electro negativity of an element and hence on the type of bond formed. Hydrides are classified into the following four classes.

1) Ionic or saline or salt like hydrides: These are formed by metals of low electronegativity i.e., elements of IA, IIA (except Be & Mg) and some highly positive lanthanides by heating the metal in hydrogen at high temperature. BeH₂ & MgH₂ have covalent polymeric structures. These are white colourless crystalline solids having high mp, bp, easily decomposed by water, alcohol, CO₂ or SO₂. These are used as strong reducing agent. Alkali metal hydrides are used for making etc and for removing last traces of water from organic compounds.

2) Molecular or covalent hydrides:

These are formed by element of group IIIA, IVA, VA, VIA, VIIA by sharing electrons with hydrogen atoms e.g., NH₃, HCl, B₂H₆, AsH₃. These are usually gases or liquids having low mp & bp (volatile).

The hydrides of IIIA group elements like B₂H₆, (AlH₃)_x act as lewis acid because they are electron deficient molecules. These have been classified as electron precise, electron deficient & electron rich hydrides.

i) Electron precise hydrides: All the valence electrons of the central atom are involved in bond formation. eg., CH₄ (Methane); C₂H₆ (Ethane).

All these compounds have simple covalent bonds between the atoms.

ii) Electron deficient hydrides: The available number of valence electrons is less than the number required for normal covalent bond formation or for writing the Lewis structure for the molecule. e.g., B₂H₆ ; (AIH₃)_n;

iii) Electron rich hydrides: The valence electrons on the central atom are more than that are needed for bond formation. i.e., lone pairs are present on the central atom. e.g.,

Nomenclature of covalent hydrides: The systematic or IUPAC names of these hydrides are deduced from the name of the element and the suffix ‘ – ane’.

Table – 1. Some covalent or molecular hydrides and their names

S.No.	Element	Group no. in the PT		Formula of the hydride	Name of the hydride
S.No.	Element	Short	Long	Formula of the hydride	IUPAC	Common
1.	B	III	13	B₂H₆	Diborane (6)	Diborane
2.	C	IV	14	CH₄	Methane	Methane
3.	N	V	15	NH₃	Azane	Ammonia
4.	P	V	15	PH₃	Phosphane	Phosphine
5.	As	V	15	AsH₃	Arsane	Arsine
6.	Sb	V	15	SbH₃	Stibane	Stibine
7.	O	VI	16	H₂O	Oxidane	Water
8.	S	VI	16	H₂S	Sulfane	Hydrogen sulfide
9.	F	VII	17	HF	Hydrogen fluoride	Hydrogen fluoride
10.	Cl	VII	17	HCl	Hydrogen chloride	Hydrogen chloride

3. Metallic or Interstitial hydrides: These are formed by many transition and inner transition elements at elevated temperature, by absorbing hydrogen into the interstices of their lattices. They exhibit metallic properties and are powerful reducing agents. They are non-stoichiometric compounds and their composition varies with temperature and pressure. e.g., TiH_1.73, CeH_2.7, LaH_2.8, PdH_0.60, ZrH_1.92. Metals of groups 7, 8, 9 do not form hydrides. This gap is referred as hydride gap.

4. Polymeric hydrides: These are amorphous solids containing molecules linked together in two or three dimensions by hydrogen bridge bonds.

e.g.,. ${({BeH}_{2})}_{n}, {({MgH}_{2})}_{n} & {({AlH}_{3})}_{n} {({InH}_{3})}_{n}, {({GaH}_{3})}_{n}, {({SiH}_{3})}_{n} etc.$

Illustration 4: Metallic hydrides are called nonstoichiometric hydrides. Why are these so called?

Solution: Many metals form hydrides having nonstoichiometric compositions. That is why these hydrides are called nonstoichiometric hydrides.

Illustration 5: What is the hydride gap?

Solution: The elements in the middle of the d-block do not form hydrides. This absence of

hydrides in this part of the periodic table is called the hydride gap.

Illustration 6: Draw the structure of borohydride (BH₄^–) ion.

Solution: B in BH₄^– shows sp³ hybridisation. Each hybrid orbital forms bonds with the s orbitals of H atoms. Therefore, BH₄^– has a tetrahedral shape.

Water (H₂O):

Water is one of the most abundant available substance or compound of hydrogen in nature.

About three fourth of the earth surface is covered by water. Water is known as universal solvent because.

i) It can dissolve maximum number of compounds.

ii) It is cheaply available

iii) It has high dielectric constant (82D)

Oxygen in H₂O molecule is sp³ hybridized and the structure of water molecule is an angular or bent structure.

Fig. a) Structure of H₂O in the gas phase

b) Polar nature of H₂O molecule

c) Resultant dipole moment of water

Fig: Hydrogen bonding in liquid water – structure of water molecule in the liquid state

Fig. a)Hexagonal honey comb structure of ice

b) Tetrahedral arrangement of oxygen in ice

Most common physical state of water is liquid, however, it exist as solid (ice) below 0⁰C and gas (steam) above 100⁰C.

Physical properties: of ordinary water and heavy water:

Physical property	H₂O	D₂O
Maximum density (g/ml)	1.00 at 4⁰C	1.1073 at 11.6⁰C
Melting point	0⁰C	3.82⁰C
Boiling point	100⁰C	101.42⁰C
Specific heat at 20⁰C	1.00 Cal/gK	1.01 cal/gK
Latent heat of fusion	79.7 cal/g	75.5 cal/g
Latent heat of vaporization	225.3 J/g	232.9 J/g
Viscosity at 20⁰C	10.9 millipoise	12.6 millipoise
Dielectric constant at 20⁰C	82.0D	80.5D
Solubility of NaCl (per lit) at 25⁰C	359 g	305 g
Molecular mass	18.016	20.03

Chemical properties of water:

i) Reaction with metals: Metals of group IA & IIA decompose water to liberate H₂ along with evolution of energy (Exothermic reaction). Hot metals like Zn, Mg, Fe etc, decompose water to liberate H₂. Pb and Cu decompose water only on heating. Ag, Au, Hg & Pt metals donot decompose water.

ii) Reaction with Non-metals: Fluorine decomposes cold water forming ozonized oxygen. Chlorine decomposes cold water forming HCl & HClO, However, in presence of sunlight HCl and O₂ is formed. Red hot coke on reaction with steam produce water gas.

iii) Reaction with non metal oxide: Acidic oxides combine with water to form oxo-acid like carbonic acid, sulphurous acid, sulphuric acid, orthophosphoric acid, nitric acid, perchloric acid etc.,

iv) Reaction with Metal oxides: Basic oxides combine with water to form alkalies.

v) Reaction with hydrides, carbides, nitrides & phosphides: Water decomposes these compounds with liberation of hydrogen, acetylene, ammonia, phosphine respectively.

vi) Hydrolysis: Salt of strong bases with weak acids, weak bases with strong acids and weak bases with weak acids undergo hydrolysis in water. Some slats like BiCl₃, SbCl₃ on hydrolysis form oxy compounds. Halides of non-metals like PCl₃, PCl₅, SiCl₄ etc are decomposed by water.

vii) Water of crystallization: It forms hydrates like CuSO₄. 5H₂O, MgSO₄. 7H₂O, FeSO₄.7H₂O, BaCl₂ . 2H₂O etc., on combination with water during crystallization. The water present in the hydrates is called water of crystallization.

These are of 3 – types

(i) cationic (ii) Anion (iii) Lattice.

viii) Water as catalyst: Perfectly dry gases generally do not react but ammonia & HCl gas combine only in presence of moisture.

Soft and Hard water: The water which produces large amount of lather with soap readily is known as softwater e.g., rain water. The water which does not produce lather with soap readily is known as hard water e.g., sea water, river water, lake water, spring water & well water.

Hard water is of two types :

i) Temporary hard water

ii) Permanent hard water.

Cause of hardness of water: Presence of bicarbonates, chlorides and sulphates of calcium and magnesium in it, causes hardness of water.

Types of Hardness of water: Hardness of water is of two types:

i) Temporary hardness: It causes due to presence of soluble bicarbonates of calcium or magnesium or both. It is also called as carbonate hardness. It can be easily removed by simply boiling & filtering the water.

ii) Permanent hardness: It causes due to presence of soluble chlorides and sulphates, nitrates of calcium and magnesium. It is also called as non-carbonate hardness. It can not be removed simply by boiling the water.

Removal of temporary hardness (methods):

i) By boiling: The soluble bicarbonates are decomposed into insoluble carbonates, which are removed by filtration.

$M {({HCO}_{3})}_{2 (aq)} \overset{boiling}{\to} {MCO}_{3 (9)} ↓ + H_{2} O + {CO}_{2} (M = Mg, Ca)$

ii) By Clark’s process (Commercial scale): By adding lime water or milk of lime, soluble bicarbonates are converted into insoluble carbonates (easily filtered off). e.g.,

$\begin{array}{l} Ca {({HCO}_{3})}_{2} + Ca (OH)_{2} \to 2 {CaCO}_{3} ↓ + 2 H_{2} O \\ M \underset{(m g)}{{({HCO}_{3})}_{2}} + 2 Ca (OH)_{2} \to 2 {CaCO}_{3} ↓ + M (OH)_{2} ↓ + 2 H_{2} O \end{array}$

Excess of lime causes temporary hardness.

Removal of permanent hardness (methods):

i) By adding washing soda:

Both types of hardness are precipitated as insoluble carbonates which are filtered off

$\begin{array}{l} Mg {({HCO}_{3})}_{2} + {Na}_{2} {CO}_{3} \to {MgCO}_{3} ↓ + 2 {NaHCO}_{3} \\ {MgCl}_{2} + {Na}_{2} {CO}_{3} \to {MgCO}_{3} ↓ + 2 NaCl \\ Ca {({HCO}_{3})}_{2} + {Na}_{2} {CO}_{3} \to {CaCO}_{3} ↓ + 2 {NaHCO}_{3} \\ {CaCl}_{2} + {Na}_{2} {CO}_{3} \to {CaCO}_{3} ↓ + 2 NaCl \end{array}$

ii) By adding caustic soda:

The temporary and permanent hardness of water can be removed by adding caustic soda (NaOH). e.g.,

$\begin{array}{l} M {({HCO}_{3})}_{2} + 2 Na (OH) \to M (OH)_{2} ↓ + 2 {NaHCO}_{3} (M = Mg, Ca) \\ {MCl}_{2} + 2 NaOH \to M (OH)_{2} ↓ + 2 NaCl \\ {MSO}_{4} + 2 NaOH \to M (OH)_{2} ↓ + {Na}_{2} {SO}_{4} \end{array}$

iii) By adding sodium phosphate (Na₃PO₄):

Temporary & permanent hardness of water are precipitated as phosphate of calcium & magnesium.

$\begin{array}{l} 3 M {({HCO}_{3})}_{2} + 2 {Na}_{3} {PO}_{4} \to M_{3} {({PO}_{4})}_{2} ↓ + 6 {NaHCO}_{3} (M = Mg, Ca) \\ {3 MCl}_{2} + 2 {Na}_{3} {PO}_{4} \to M_{3} {({PO}_{4})}_{2} ↓ + 6 NCl \\ 3 {MSO}_{4} + 2 {Na}_{3} {PO}_{4} \to M_{3} {({PO}_{4})}_{2} ↓ + 3 {Na}_{2} {SO}_{4} \end{array}$

iv) Calgone process (sequestration) :

Sodium hexameta phosphate [Na₂[Na₄(PO₃)₆]] is called calgone (means calcium gone).

It forms soluble complex with Ca²⁺ or Mg⁺ present in hard water. e.g. $2 {CaSO}_{4} + {Na}_{2} [{Na}_{4} {({PO}_{3})}_{6}] \to 2 {Na}_{2} {SO}_{4} + {Na}_{2} [{Ca}_{2} {({PO}_{3})}_{6}]$

The complex donot form any precipitate with soap & readily produce lather with soap, Ca²⁺ & Mg²⁺ ions are rendered ineffective (sequestration). Water becomes free from Ca²⁺ and Mg²⁺ ions.

v) Permutit process: Inorganic ion exchanges: Hydrated sodium aluminium silicate [Na₂Al₂ Si₂O₈.x H₂O] is called permutit or zeolite (Na₂Z). It exchanges its sodium ions with Ca²⁺ and Mg²⁺ ions present in permanent hard water.

$\begin{array}{l} {Na}_{2} Z + {CaCl}_{2} \to CaZ + 2 NaCl \\ {Na}_{2} Z + {MgCl}_{2} \to MgZ + 2 NaCl \end{array}$

After some time permutit when fully exhausted, can be regenerated by passing a 10% solution of NaCl through it.

$\begin{array}{l} CaZ + 2 NaCl \to {Na}_{2} Z + {CaCl}_{2} \\ MgZ + 2 NaCl \to {Na}_{2} Z + {MgCl}_{2} \\ (regenerated zeolite) \end{array}$

The soluble calcium & magnesium slats thus formed are washed away by water and the regenerated permutit can be used again.

Advantages of the permutit process:

i) It is most efficient method to get water with zero degree hardness.

ii) It can be used to remove both temporary and permanent hardness completely.

iii) This is a very cheap process, since during the process only NaCl is consumed which is quite cheap.

iv) By synthetic Resins: (Done in two connecting tanks). In this process two types of resins are used.

a) Cation exchange resins: (H⁺ – resin): These are giant molecules containing carboxyl ( – COOH) or sulphonic acid group ( – SO₃H) and are capable of exchanging H⁺ ions with Ca ²⁺ & Mg²⁺

${Ca}^{2 +} + 2 H^{+} – resin \to Ca (resin)_{2} + 2 H^{+}$

The exhausted resins like permutit can be regenerated by passing conc H₂SO₄ or HCl acid.

$Ca (resin)_{2} + 2 HCl \to 2 H^{+} - resin + {CaCl}_{2}$

b) Anion exchange resins (OH^– – resin):

These are also giant molecules containing basic group (NH₃OH) and are capable of exchanging OH^– ions with anions such as Cl^– & SO₄² ions.

${Cl}^{-} + {OH}^{-} - resin \to {Cl}^{-} - resin + {OH}^{-}$

Exhausted anion resin can be regenerated by adding moderately strong NaOH solution.

${Cl}^{-} - resin + NaOH \to {OH}^{-} - resin + NaCl$

Thus, the water is first passed through cation resins and then through anion resins and pure distilled or deionized water is obtained from second tank i.e.,

$H^{+} + {OH}^{-} \to H_{2} O (deionized water)$

Disadvantage of hard water:

Hard water, when used in boilers in industries causes formation of scales in the boiler. This eats away the metal layer and also causes wastage of heat energy. Hence only soft water is used in industries. Besides hard water causes wastage of soap.

Degree of hardness: The hardness of water is expressed in terms of ppm of calcium carbonate.

$\begin{array}{l} 1 {CaCO}_{3} \equiv 1 {CaCl}_{2} \equiv 1 {CaSO}_{4} \equiv 1 {MgCl}_{2} \equiv 1 {MgSO}_{4} \\ 100 ppm 11 ppm 136 ppm 95 ppm 120 ppm \end{array}$

Illustration 7: A water sample contains 136mg of CaSO₄ per litre calculate the hardness in term of CaCO₃ equivalent

Solution: CaSO₄ = CaCO₃

136 100 or

136mg L^– of CaSO₄ = 100mg L^– of CaCO₃

Hence, hardness = 100ppm

Illustration 8: A one litre water sample contains 1 mg of CaCl₂ and 1mg of MgCl₂. Find out the total hardness in terms of CaCO₃ in ppm.

Solution:

$\begin{matrix} 1 {CaCO}_{3} \equiv 1 {CaCl}_{2} \equiv 1 {MgCl}_{2} \\ 100 g 111 g 95 g \end{matrix}$

i) $111 g of {CaCl}_{2} \equiv 100 g of {CaCO}_{3}$

$∴ 1 mg of {CaCl}_{2} \equiv \frac{100 g \times 1 mg of {CaCO}_{3}}{11 g} = 0.9 mgof {CaCO}_{3}$

ii) $95 g of {MgCl}_{2} \equiv 100 g of {CaCO}_{3}$

$1 mg of {MgCl}_{2} \equiv \frac{100 gof {CaCO}_{3} x 1 mg of {CaCO}_{3}}{95 g} = 1.05 mg of {CaCO}_{3}$

Thus, degree of hardness in 1 litre = 0.9 + 1.05 = 1.95 ppm

Illustration 9: Why do lakes freeze from top towards bottom?

Solution: Ice is lighter than water. When the lake water cools, it starts freezing at its surface. The ice top does not sink to the bottom. Being lighter than water, it remains at the top. This top layer of ice acts as a thermal insulation for the lower layers of water and does not allow the lower layers of water to freeze. This phenomenon is of great ecological significance – the aquatic animals and plants can survive in lakes even when the lake from its top side is frozen.

Heavy water (D₂O):

Preparation: It is prepared by exhaustive electrolysis of water containing heavy water with Ni – electrode (perforated). The electrolysis is carried out in seven stages.

In the first stage, the electrolyte is reduced to ${\frac{1}{6}}^{t h}$ of its original volume and the alkali present is partly neutralized by passing carbon dioxide.

It is then distilled and taken to another cell for the second stage of electrolysis. The process is repeated and 99% D₂O is obtained after the seventh stage. About 30 litres of ordinary water gives 1 ml of heavy water. In India, heavy water is manufactured at Nangal in Punjab and at Bhabha Atomic Research Centre Trombay, Rourkela & Nayveli by electrolysis of ordinary water. Protium bonds are weaker than deuterium bonds and thus in electolysis of ordinary water, H₂ is liberated much faster than D₂ and remaining water becomes enriched in D₂O.

Properties of heavy water:

i) Physical properties:

It is a colourless, odourless and tasteless mobile liquid.

Nearly all the physical constant for D₂O are higher than ordinary water.

ii) Chemical properties:

Heavy water is chemically similar to ordinary water. However D₂O reacts more slowly than H₂O in chemical reactions.

i) Eelectrolysis : In presence of Na₂CO₃ or P₂O₅ (which makes it conductor)

$2 D_{2} O \overset{electrolysis}{\to} \underset{(Cathode)}{2 D_{2}} + \underset{(Anode)}{O_{2}}$

ii) Reaction with metals:

Metals of group IA & IIA liberated deuterium from D₂O.

$2 {Na}_{(s)} + 2 D_{2} O_{(ℓ)} \to 2 {NaOD}_{(aq)} + D_{2 (g)}$

Hot metals like Zn, Fe etc decompose the vapour of heavy water.

iii) Reaction with metal oxides: It form heavy alkalies with oxides like Na₂O, CaO etc.

$\begin{array}{l} {Na}_{2} O + D_{2} O \to 2 NaOD \\ CaO + D_{2} O \to \underset{C a l c i u m d e u t e r o x i d e}{Ca (OD)_{2}} \end{array}$

iv) Reaction with non metal oxides:

It forms deutero acids with acidic oxides like N₂O₅, P₂O₅, SO₃ etc.

$\begin{array}{l} N_{2} O_{5} + D_{2} O \to 2 DNO \\ Deuteronitric acid \\ {SO}_{3} + D_{2} O \to \underset{Deutero sulphuric acid}{D_{2} {SO}_{4}} \end{array}$

v) Reaction with metal carbides, phosphides, nitrides, arsenides.

Like H₂O, it forms corresponding deuterocompounds.

$\begin{array}{l} \underset{c a l c i u m c a r b i d e}{{CaC}_{2}} + 2 D_{2} O \to Ca (OD)_{2} + C_{2} D_{2} \\ \underset{aluminium carbide}{{Al}_{4} C_{3} + 12 D_{2} O} \to 4 Al (OD)_{3} + 3 {CD}_{4} \end{array}$

$\begin{array}{l} {Ca}_{3} P_{2} + 6 D_{2} O \to 3 Ca (OD)_{2} + 2 {PD}_{3} \\ calcium phosphide \end{array}$

$\begin{array}{l} {Mg}_{3} N_{2} + 6 D_{2} O \to 3 Mg (OD)_{2} + 2 {ND}_{3} \\ magnecium nitride \end{array}$

$\begin{array}{l} AIN + 3 D_{2} O \to Al (OD)_{3} + {ND}_{3} \\ aluminium nitride \end{array}$

$\begin{array}{l} {Na}_{3} As + 3 D_{2} O \to 3 NaOD + {AsD}_{3} \\ sodium arsenide \end{array}$

v) Deuterolysis : Water brings hydrolysis of certain inorganic salts. It gives similar reactions which are termed as deuterolysis.

$\begin{array}{l} BaS + 2 D_{2} O \to Ba (OD)_{2} + D_{2} S \\ {AlCl}_{3} + 3 D_{2} O \to Al (OD)_{3} + 3 DCl \end{array}$

vi) Exchange reactions :

It can exchange labile hydrogen of a compound by deuterium partially or completely.

$\begin{array}{ll} NaOH + D_{2} O \to NaOD + HDO \\ {HCl}_{(aq)} + D_{2} O \to DCl + HDO \\ {NH}_{4} Cl + {4 D}_{2} O \to {ND}_{4} Cl + 4 HDO \end{array}$

viii) Formation of deutero hydrates: Like ordinary water, it may be associated with salts as water of crystallisation, giving deutero hydrates.

e.g., ${NaSO}_{4} .10 D_{2} O, {CuSO}_{4} .5 D_{2} O, {MgSO}_{4} .7 D_{2} O etc.,$

Uses of D₂O:

i) As a tracer compound.

ii) As a neutron moderator and coolent in nuclear reactors.

iii) In production of heavy hydrogen.

iv) In deciding the basicity of acids.

$2 {CH}_{3} COOH + D_{2} O \to {CH}_{3} COOD + H_{2} O (Basicity = 1)$

$H_{3} {PO}_{3} + D_{2} O \to {HD}_{2} {PO}_{3} + H_{2} O (Basicity = 2)$

$H_{3} {PO}_{4} + 3 D_{2} O \to D_{3} {PO}_{4} + 3 H_{2} O (Basicity = 3)$

Hydrogen peroxide (H₂O₂):

Structure: H₂O₂ is best represented as the equilibrium mixture of two tautomeric forms in equilibrium.

In H₂O₂the two oxygen atoms are linked to each other by a single covalent bond (i.e., peroxide bond) and each oxygen is further linked to a hydrogen atom by a single covalent bond. The two O – H bonds are, however, in different planes due to repulsions between different bonding and anti bonding orbitals. The dihedral (inter planar) angle between the two planes being 111.5⁰ in gas phase but reduced to 90.2⁰ in the crystalline state due to hydrogen bonding. The other molecular dimensions of hydrogen peroxide in the gas phase and the crystalline state are show in figure.

Fig. Open book structure of H₂O₂ (a) in the gas phase dihedral angle is 111.5⁰ and (b) in the solid phase at 110K, dihedral angle is 90.2⁰.

Preparation of hydrogen Peroxide:

1) Laboratory method:

a) From sodium peroxide: By the action of ice cold dil H₂SO₄

${Na}_{2} O_{2} + H_{2} {SO}_{4} (20 %) \to {Na}_{2} {SO}_{4} + H_{2} O_{2} (30 %)$

b) From BaO₂: ${BaO}_{2} 8 H_{2} O + H_{2} {SO}_{4} (20 %) \to {BaSO}_{4} + H_{2} O_{2} + 8 H_{2} O (5 %)$

Anhydrous BaO₂ does not react readily with H₂SO₄ because a coating of insoluble BaSO₄ is formed on its surface which stop further action of the acid. Excess of BaO₂ should be avoided because it tends to decompose H₂O₂. H₂O₂ obtained by this method can not store for long.

The weaker acid such as H₂CO₃ (CO₂) and H₃PO₄ acid treatment is preferred over sulphuric acid because a coating of insoluble impurities like barium per sulphate tends to decompose H₂O₂.

H₃PO₄ also acts as a negative catalyst for H₂O₂ because H₃PO₄ removes all heavy metals (catalyst for H₂O₂ decomposition) in the form of insoluble phosphates.

3BaO₂ + 2H₃PO₄ $\to$ Ba₃ (PO₄)₂ + 3H₂O₂

Ba₃ (PO₄)₂ + 3H₂SO₄ $\to$ 3 BaSO₄ + 2H₃PO₄.

2) Industrial Methods:

i) By electrolysis of 50% ice cold H₂SO₄:

$2 H_{2} {SO}_{4} ⇌ 2 H^{+} + 2 {HSO}_{4}^{-}$

At anode: $2 {HSO}_{4}^{-} \to \underset{\underset{(Marshall ‘ s acid)}{Peroxodisulphuric acid}}{H_{2} S_{2} O_{8}} + 2 e^{-}$

$H_{2} S_{2} O_{8} + H_{2} O \to H_{2} {SO}_{4} + \underset{\underset{(Caro ‘ s acid)}{peroxomonosulphuric acid}}{H_{2} {SO}_{5}}$

$H_{2} {SO}_{5} + H_{2} O \to H_{2} {SO}_{4} + \underset{(20 - 30 %)}{H_{2} O_{2}}$

At cathode: $2 H^{+} + 2 e^{-} \to H_{2} ↑$

Modification:

In place of 50% H₂SO₄, an equimolar mixture of H₂SO₄ and ammonium sulphate is electrolysed.

${({NH}_{4})}_{2} {SO}_{4} + H_{2} {SO}_{4} \to \underset{Ammoniumhydrogen sulphate}{2 {NH}_{4} {HSO}_{4}}$

NH₄HSO₄ $⇌$ NH₄SO₄^– + H⁺

At anode: 2NH₄SO₄^– $\to$ (NH₄)₂S₂O₈ + 2e^–

Ammonium peroxy disulphate or

Ammonimum per sulphate

${({NH}_{4})}_{2} S_{2} O_{8} + 2 H_{2} O \frac{Δ}{43 mmpresure} \underset{(can be used again)}{\underset{↓}{{2 NH}_{4} {HSO}_{4}}} + \underset{(30 - 40 %)}{H_{2} O_{2}}$

At cathode: 2H⁺ + 2e^– $\to$ H₂ $↑$

ii) By auto – oxidation: By auto oxidation of 2 – ethyl anthraquinol

$+ H_{2} O_{2} \to (20 %)$

3) By oxidation of isopropyl alcohol: A little H₂O₂ is added to initiate the reaction

${CH}_{3} {CHOHCH}_{3} \frac{{Little H}_{2} O_{2} + [O]}{slight pressure, 70 ° C} ⟶ {CH}_{3} {COCH}_{3} + H_{2} O_{2}$

Concentration of H₂O₂ solution: It is very carefully concentrated since it decomposes on heating or on standing.

$2 H_{2} O_{2} \to 2 H_{2} O + O_{2} (auto – oxidation)$

Decomposition is catalysed by Cu, Ag, Pt, Fe, Co, MnO₂ etc. Following methods are employed for concentration of H₂O₂ :

i) Evaporation on water bath: 50% H₂O₂ is obtained by careful evaporation of solution on water bath.

ii) Dehydration in vaccum desicator: 50% H₂O₂ is dehydrated in a vaccum desicator in presence of conc H₂SO₄, then 90% H₂O₂ is obtained.

iii) Vaccum distillation: In this step 90% H₂O₂ is distilled under reduced pressure (10 – 15) to get 99% H₂O₂.

iv) Cooling: The last traces of water in H₂O₂ are removed by freezing it in a freezing mixture (dry ice + ether) when crystal of hydrogen peroxide separate out.

Storage of H₂O₂: It is stored in coloured paraffin wax coated plastic or teflon bottles in presence of trace of alcohol, acetanilide, glycerine or phosphoric acid (i.e., some stabilizer or negative catalyst), which slow down the rate of decomposition of hydrogen peroxide. Impurities like silica, MnO₂, Iron, Mn or Al₂O₃ causes rapid decomposition.

Physical properties:

i) Anhydrous H₂O₂ is colourless, viscous liquid soluble in ether, alcohol and water (sp. gravity 1.45g/ml at 0⁰C).

ii) It gives blue tinge in thick layers.

iii) It causes blisters on skin.

iv) It is bitter in taste.

v) Pure H₂O₂ is weak acidic in nature and exist as more associated liquid (due to hydrogen bonding) than water and turns blue litmas red (Ka = 1.57 ´ 10^–² at 20⁰c) & ionize in 2 steps H₂O₂ $⇌$ H⁺ + HO₂^– and HO₂^– $\to$ H⁺ + O₂

vi) It boil at 152⁰C and freeze at – 0.89⁰

vii) It begins to decompose at bp and thus, distilled under reduced pressure.

viii) It is diamagnetic in nature.

ix) In pure state, its dielectric constant is 93.7, which increases with dilution. Due to high dielectric constant, H₂O₂ and its aqueous solutions are good solvents. However not used commonly on account of its oxidant nature.

Chemical properties:

i) Stability: Pure H₂O₂ decomposes on long standing or heating into water and O₂.

2H₂O₂ $⇌$ 2H₂O + O₂

It is an example of disproportionation reaction.

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 $\to$ Na₂O₂ + 2H₂O

H₂O₂ + Na₂CO₃ $\to$ Na₂O₂ + CO₂+ H₂O

3) Oxidising properties (Powerful): It accepts electrons and acts as powerful oxidant in acidic as well as in alkaline (basic) medium.

In acidic solution: $H_{2} O_{2} + 2 H^{+} + 2 e^{-} \overset{Fast}{⟶} 2 H_{2} O; E_{RP}^{0} = 1.77 V$

In basic solution: $H_{2} O_{2} + {OH}^{-} + 2 e^{-} \overset{Slow}{⟶} 3 {OH}^{-}; E_{RP}^{0} = 0.87 V$

Thus, H₂O₂ is more powerful oxidant in acidic medium.

The simple interpretation of H₂O₂ as oxidant can be shown by the equation.

$H_{2} O_{2} \to H_{2} O + [O]$

Some other oxidizing properties of H₂O₂ are given below

i) It oxidizes black lead sulphide to white lead sulphate.

PbS + 4H₂O₂ $\to$ PbSO₄ + 4H₂O

ii) It oxidizes ferrous sulphate to ferric sulphate in presence of H₂SO₄.

2FeSO₄ + H₂SO₄ + H₂O₂ $\to$ Fe₂ (SO₄)₃ + 2H₂O

iii) It oxidizes nitrites to nitrates

NaNO₂ + H₂O₂ $\to$ NaNO₃ + H₂O

iv) It oxidizes benzene to phenol in presence of FeSO₄

C₆H₆ + H₂O₂ $\to$ C₆H₅OH + H₂O

v) It oxidizes sulphites to sulphates

Na₂SO₃ + H₂O₂ $\to$ Na₂SO₄ + H₂O

vi) It oxidizes ferrocyanide to ferricyanide in acidic medium

2K₄ [Fe (CN)₆ ]+ H₂O₂ + H₂SO₄ $\to$ 2K₃ [Fe (CN)₆ ]+ K₂SO₄ + 2H₂O

vii) Sulphurous acid to sulphuric acid

H₂SO₃ + H₂O₂ $\to$ H₂SO₄ + H₂O

viii) Chromic hydroxide (ppt) to chromate in alkaline solution

2 Cr (OH)₃ + 4OH^– + 3H₂O₂ $\to$ 2CrO₄^2- + 8H₂O

Chromate(yellow)

x) In basic medium, formaldehyde to formic acid, in presence of pyrogallol & itself reduced to H₂.

2HCHO + H₂O₂ $\to$ 2HCOOH + H₂

xi) Acidified KI to I₂

2KI + H₂SO₄ + H₂O₂ $\to$ K₂SO₄ + 2H₂O + I₂

xi) Acidified chromic acid or acidified K₂Cr₂O₇ (ether) to blue peroxide of chromium.

$H_{2} {Cr}_{2} O_{7} + 4 H_{2} O_{2} \overset{H^{+}}{⟶} 2 {CrO}_{5} + 5 H_{2} O$

$K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} + 4 H_{2} O_{2} \to \underset{↓ ether}{2 {CrO}_{5}} + K_{2} {SO}_{4} + 5 H_{2} O \underset{(blue)}{{CrO}_{5}}$

4) Reducing properties: (Poor):

It also acts as a reducing agent towards powerful oxidizing agent in acidic as well as basic medium.

in acidic medium:

$H_{2} O_{2} \to O_{2} + 2 H^{+} + 2 e^{-}; E_{op}^{0} = - 0.08 V$

in basic medium:

$H_{2} O_{2} + 2 {OH}^{-} \to O_{2} + 2 H_{2} O + 2 e^{-}; E_{op}^{0} = - 0.08 V$ .

From E⁰_op and E⁰_Rp values, it is apparent that H₂O₂ is powerful oxidizing agent and poor reducing agent.

Reducing property of H₂O₂ can be expressed as [H₂O₂ + O $\to$ H₂O + O₂]

Some other reducing properties of H₂O₂ are given below

It reduces:

i) Moist silver oxide to silver.

Ag₂O + H₂O₂ $\to$ 2Ag + H₂O + O₂.

ii) Ozone to oxygen :

O₃ + H₂O₂ $\to$ 2O₂ + H₂O

iii) Manganese dioxide to mangnous salt in acidic medium

MnO₂ + H₂SO₄ + H₂O₂ $\to$ MnSO₄ + 2H₂O + O₂

iv) Chlorine & bromine to corresponding hydracids. This property is called antichlor.

Cl₂ + H₂O₂ $\to$ 2HCl + O₂

v) Red lead to plumbous acid in presence of HNO₃

${Pb}_{3} O_{4} + 6 {HNO}_{3} + H_{2} O_{2} \to 3 Pb {({NO}_{3})}_{2} + 4 H_{2} O + O_{2}$

vi) Decolorises acidified KMnO₄

$2 {KMnO}_{4} + 3 {HSO}_{4} + 5 H_{2} O_{2} \to 2 {MnSO}_{4} + K_{2} {SO}_{4} + 8 H_{2} O + 5 O_{2}$

vii) Potassium ferricyanide to potassium ferro cyanide in alkaline medium

$2 {[Fe (CN)_{6}]}^{3 -} + H_{2} O_{2} + 2 {OH}^{-} \to 2 {[Fe (CN)_{8}]}^{4 -} + 2 H_{2} O + O_{2}$

viii) KMnO₄ to MnO₂:

$2 {KMnO}_{4} + H_{2} O_{2} \to {MnO}_{2} + 2 KOH + 2 O_{2}$

ix) Hypohalites to halides:

$NaOCl + H_{2} O_{2} \to NaCl + H_{2} O + O_{2}$

5) Addition reaction:

$\begin{matrix} {CH}_{2} & H - O & {CH}_{2} OH \\ | | & + & | & \to & | \\ \underset{ethylene}{{CH}_{2}} & H - O & \underset{ethyleneglycol}{{CH}_{2} OH} \end{matrix}$

6) Reaction with N₂O₅:

$N_{2} O_{5} + H_{2} O_{2} \to {HNO}_{3} + {HOONO}_{2}$ (Pernitric acid)

7) Reaction with NO₂:

2NO₂ + H₂O₂ $\to$ 2HNO₃

8) Bleaching properties:

It acts as a bleaching agent.

Coloured material + [O] $\to$ colourless material.

Strength of H₂O₂ solution:

i) Percentage strength:

It expresses the amount of H₂O₂ by weight present in 100 ml of the solution.

For example : 30% aq solution of H₂O₂ implies that 30gm of H₂O₂ are present in 10ml of the solution.

ii) Volume strength:

The strength of aq solution of H₂O₂ is expressed in terms of volume strength. For example 10 volume, 20 volume and so on H₂O₂ solution. 10 volume H₂O₂ implies that 1ml H₂O₂ produces 10ml of O₂ at NTP or 1 litre H₂O₂ produces 10 litre of O₂ at NTP and so on. Let us now calculate the % strength of a 10 volume H₂O₂ solution.

Table: strength of H₂O₂:

Sample of H₂O₂	% strength (w/v)	Molarity (M)	Normality (N)
10 vol H₂O₂	3.036	0.893	1.786
20 vol H₂O₂	6.072	1.786	3.572
30 vol H₂O₂	9.108	2.679	5.358

Some important relations:

Concentration or strength of 10 volume H₂O₂ solution = 30.35 g/L

Volume strength = 5.6 x Normality = $5.6 \times \frac{% strength}{eq . wt of H_{2} O_{2}} \times 10$

Volume strength = 11.2 x Molarity.

Note:

i) 30% H₂O₂ is called perhydrol.

ii) During photosynthesis H₂O₂ is oxidize to O₂.

Illustration 10: What is meant by “30 volume H₂O₂”?

Solution: The statement “30 volume H₂O₂” mean that 1 volume of this sample of H₂O₂ gives 30 volumes of oxygen at NTP or STP.

Illustration 11: Hydrogen peroxide is used to restore the colour of old oil-paintings containing lead oxide. Write a balanced equation for the reaction that takes place in this process.

Solution: In old paintings, the original white lead paint over the period of time gets converted to black lead sulphide (PbS) due to the presence of H₂S in the atmospheric air. Hydrogen peroxide oxidizes the black PbS into PbSO₄.

Illustration 12: Explain why hydrogen peroxide is stored in coloured / plastic bottles?

Solution: Hydrogen peroxide decomposes when exposed to light. It is due to this that it is stored in wax-lined coloured glass or plastic bottles in the presence of a stabilizer such as urea.

Illustration 13: Calculate the percentage strength and normality of 20 volumes H₂O₂.

Solution: (1) 20 volumes H₂O₂ solution means 1ml of H₂O₂ on decomposition gives 20ml of O₂ at STP i.e.., 1 litre of H₂O₂ on decomposition gives 20 lit of O₂.

Further, let us consider decomposition of H₂O₂

$\underset{2 mole}{2 H_{2} O_{2}} \to 2 H_{2} O + \underset{1 mole}{O_{2}}$

Since 68g of H₂O₂ gives 22.4 lit of O₂

∴ 20 lit of O₂ is given by = $68 \times \frac{20}{22.4}$ g of H₂O₂ = 60.17g

∴ strength of H₂O₂ = 6.071% $(\frac{w}{v})$

Now, Normality of 20 volume H₂O₂ solution is

$N = \frac{Wt of H_{2} O_{2}}{Equivalent weight of H_{2} O_{2}} \times \frac{1}{Volume of solution in lit} = \frac{60.71}{17 \times 1} = 3.57 N$

Illustration 14: The lable on a bottle of hydrogen peroxide solution reads as 10 volumes. Report the concentration of hydrogen peroxide in percentage.

Solution: We know that hydrogen peroxide decomposes as follows

$\underset{2 mole}{2 H_{2} O_{2}} \to 2 H_{2} + \underset{1 mole}{O_{2}}$

Since 22400 ml of oxygen is obtained from 68g of H₂O₂ at STP

10ml of oxygen is obtained from = $\frac{68}{22400} \times 10 = 0.3035 gm$ of H₂O₂

1 ml of H₂O₂ solution contains 0.3035 gm

100 ml of H₂O₂ – 0.3035 ⨯ 100 = 3.035 grams of H₂O₂

or the solution is 30.35%

Concentration of 10 volumes H₂O₂ solution = 3.035 ⨯ 10 = 30.35 gm/litre

Illustration 15: A sample of water contains 12g of MgSO₄ per 100 kg of water. Calculate its degree of hardness.

Solution: Wt. of of MgSO₄ = 120

Mol. wt. of MgSO₄ = 120

Weight of water = 100 kg = 100 ⨯ 10³ = 10⁵ 9

Degree of hardness = $\frac{mol . wt . {CaCO}_{3}}{Mol . wt . of {MgSO}_{4}} \times \frac{Wt . of {MgSO}_{4} in g}{Wt . of water in g} \times 10^{5}$

$\frac{100}{200} \times \frac{12}{10^{5}} \times 10^{6}$ = 100 ppm.

Exercise 1:

(i) The volume strength of 1.6 N H₂O₂ solution is

(A) 8.96 (B) 7.96 (C) 3.56 (D) 10

(ii) The normality of 20 volume hydrogen peroxide solution is

(A) 4.57N (B) 3.57 N (C) 3.0 N (D) None of these

(iii) The volume strength of 3% solution of H₂O₂ is

(A) 6.5 volume (B) 7.5 volume (C) 9.88 volume (D) 8.5 volume

Answer to Exercises

Exercise 1:

(i) A

(ii) B

(iii) C

SOLVED PROBLEMS

Prob 1: Ortho and para hydrogen differ in

(A) atomic number

(B) mass number

(C) electron spin in two atoms

(D) nuclear spin in two atoms

Sol: (A) Ortho & para hydrogen differ in nuclear spin in two atom.

Prob 2: The volume strength of 1.5N H₂O₂ solution is

(A) 4.8 (B) 8.4 (C) 3.0 (D) 8.0

Sol: (B) Volume strength = 5.6 ⨯ normality = 5.6 ⨯ 1.5 = 8.4

Prob 3: The oxide which gives H₂O₂ on treatment with dilute acid is

(A) PbO₂ (B) Na₂O₂ (C) MnO₂ (D) TiO₂

Sol: (B) Na₂O₂ peroxide gives H₂O₂ on treatment with dilute acid.

Prob 4: One mole of calcium phosphide on reaction with excess of water gives

(A) one mole of phosphine

(B) two moles of phosphoric acid

(C) two moles of phosphine

(D) one mole of phosphorus pentoxide

Sol: (C) Two moles of phosphine.

Prob 5: The structure of H₂O₂ is

(A) planar (B) non-planar (C) spherical (D) linear

Sol: (B) The structure of H₂O₂ is non-linear (open book)

Prob 6: Action of water or dilute mineral acids on metals can give

(A) monohydrogen (B) tritium (C) dihydrogen (D) trihydrogen

Sol: (C) H₂O + Mg $\to$ MgO + H₂; 2HCl + Zn $\to$ ZnCl₂ + H₂

Prob 7: The H^– ions is stronger base than hydroxide ion. Which of the following reactions will occur if sodium hydride (NaH) is dissolved in water?

(A) $H^{-} (aq) + H_{2} O (I) \to H_{3} O^{+} (aq)$

(B) $H^{-} (aq) + H_{2} O (I) \to {OH}^{-} (aq) + H_{2} (g)$

(D) None of these

Sol: (C) H^– + H₂O $\to$ No reaction.

Prob 8: Which of the following pairs of substances on reaction will not evolve H₂ gas?

(A) Fe and H₂SO₄ (aqueous)

(B) Copper and HCl (aqueous)

(C) Sodium and ethyl alcohol

(D) Iron and steam

Sol: (B) Cu will not evolve H₂ gas on reaction with aqueous HCl.

Prob 9: Heavy water is obtained by

(A) boiling water

(B) fractional distillation of H₂O

(C) prolonged electrolysis of H₂O

(D) heating H₂O₂

Sol: (C) Heavy water is prepared by prolonged electrolysis of H₂O.

Prob 10: Polyphosphates are used as water softening agents because they

(A) form soluble complexes with anionic species

(B) precipitate anionic species

(C) form soluble complexes with cationic species

(D) precipitate cationic species

Sol: (C) Polyphosphates are used as water softening agent because they form soluble complexes with cationic species.

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