Hydrogen Class 11 Important Extra Questions Chemistry Chapter 9

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Class 11 Chemistry Chapter 9 Important Extra Questions Hydrogen

Hydrogen Important Extra Questions Very Short Answer Type

Question 1.
Which gaseous compound on treatment with dihydrogen produces methanol?
Answer:
Carbon monoxide (CO).

Question 2.
What are the constituents of water gas?
Answer:
Carbon monoxide and hydrogen.

Question 3.
Arrange H₂, D₂, T₂ in the decreasing order of their
(i) Boiling point
Answer:
T₂ > D₂ > H₂

(ii) Heat of fusion.
Answer:
T₂ > D₂ > H₂.

Question 4.
Which isotope of hydrogen
(i) does not contain neutron
Answer:
Protium

(ii) is radioactive?
Answer:
Tritium.

Question 5.
Out of the following metals which can be used to liberate H2 gas on reaction with dil. hydrochloric acid?
(i) Cu,
(ii) Zn,
(iii) Iron,
(iv) Silver,
(v) Magnesium.
Answer:
Only Zn, Fe, Mg.

Question 6.
Name one compound each of hydrogen in which it exists in:
(i) Positive oxidation state
Answer:
HCl

(ii) Negative oxidation state.
Answer:
NaH.

Question 7.
What is the importance of heavy water in nuclear power generation?
Answer:
It is used as a moderator in nuclear reactions to slow down the speed of fast-moving neutrons.

Question 8.
State two properties in which hydrogen resembles alkali metals.
Answer:

1. Both form unipositive ion
2. Both have one electron in their s orbital (ns1).

Question 9.
Give an example of each anionic and covalent hydride.
Answer:
Ionic Hydride NaH Covalent hydride NH₃.

Question 10.
Why is H₂O₂ concentrated at low pressure?
Answer:
Because it decomposes at ordinary pressure or on heating.

Question 11.
What is the mass of 1 mole of deuterium oxide and tritium oxide?
Answer:
D₂O = 20g, T₂O = 22g.

Question 12.
Arrange H₂O, H₂S, H₂Se, H₂Te in the decreasing order of boiling point.
Answer:
H₂O > H₂S > H₂Se > H₂Te.

Question 13.
Give one example of a zeolite used in softening hard water.
Answer:
Sodium aluminium silicate Na₂Al₂Si₂Og. xH₂O.

Question 14.
Name the compounds which retard the decomposition of H₂O₂.
Answer:
Acetanilide, glycerol.

Question 15.
Which is heavier out of ice and water?
Answer:
Water.

Question 16.
What is the trade name of hydrogen peroxide used as an antiseptic?
Answer:
Perhydrol.

Question 17.
What is the significance of H₂O₂ labelled as 30 volumes’?
Answer:
“30 volume” labelled hydrogen peroxide means that 1 mL of this sample of solution gives 30 mL of oxygen gas at STP.

Question 18.
What happens when water is added to calcium carbide.
Answer:
CaC₂ + O → Ca(OH)₂ + C₂H₂
Acetylene gas is produced.

Question 19.
What is the cause of the temporary hardness of water?
Answer:
Presence of Ca(HCO₃)₂ and Mg(HCO₃)₂ in water.

Question 20.
How is the temporary hardness of water removed?
Answer:
By boiling and filtering.

Question 21.
How is heavy water produced from ordinary water?
Answer:
By repeated electrolysis of ordinary water containing 3 % of NaOH in it.

Question 22.
Which of the two dihydrogen (H₂) or deuterium (D₂) undergoes reactions more readily?
Answer:
Dihydrogen (H₂).

Question 23.
Define hard water.
Answer:
Hard water is one that does not produce lather with soap solution readily.

Question 24.
Which isotope of hydrogen contains an equal number of protons and neutrons?
Answer:
Deuterium.

Question 25.
What happens when ethylene reacts with hydrogen peroxide?
Answer:
H₂C = CH₂ + H₂O₂ → HO-CH₂-CH₂-OH
Ethylene glycol

Question 26.
Name the phenomenon of absorption of hydrogen by palladium?
Answer:
Occlusion.

Question 27.
H₂O₂ is a better oxidant than H₂O. Explain.
Answer:
H₂O₂ is easily reduced to form H₂O and O.
H₂O₂ → H₂O + O.

Question 28.
Name one example of a reaction in which hydrogen acts as an oxidizing agent.
Answer:
The reaction of hydrogen with active metals
2Na + H₂ → 2NaH .

Question 29.
Concentrated H₂SO₄ cannot be used to dry moist H₂ gas. Why?
Answer:
Cone. H₂SO₄ on absorbing H₂O from moist H₂ gas produces so much heat that H₂ gas catches fire.

Question 30.
Give an example where water acts as an oxidizing agent.
Answer:
2Na + 2H₂O → 2NaOH + H₂

Question 31.
Name the element which when reacted with dil. H₂SO₄ gives pure hydrogen.
Answer:
Magnesium.

Question 32.
Does hydrogen support combustion?
Answer:
No.

Question 33.
Why is dihydrogen not preferred in balloons these days?
Answer:
Dihydrogen is a highly combustible gas and hence is likely to catch fire in presence of excess air.

Question 34.
Why is sodium less soluble in heavy water than in ordinary water?
Answer:
Due to the lower dielectric constant of D₂O over H₂O, NaCl is less soluble in heavy water.

Question 35.
Although D₂O resembles H₂O chemically, yet it is toxic. Explain.
Answer:
D₂O is toxic since D⁺ ions react at a much slower rate than H⁺ in enzyme catalysed reactions.

Question 36.
Is it correct to say that hydrogen can behave as a metal? State the conditions under which such behaviour can be possible.
Answer:
Yes. H2 can act as a metal under very high pressures.

Question 37.
Give two advantages of using hydrogen as a fuel over gasoline.
Answer:
The high heat of combustion and production of no pollutants like SO₂, NO₂, CO₂, etc.

Question 38.
What happens when chloroform is treated with heavy water in presence of an alkali?
Answer:

Question 39.
What is hydrolith? How is it prepared?
Answer:
Hydrolith is a calcium hydride (CaH₂). It can be prepared by

Question 40.
Write the structures of two complex metal hydrides which are used as reducing agent in organic synthesis.
Answer:
Lithium aluminium hydride (LiAlH₄) and sodium borohydride (NaBH₄).

Question 41.
What type of elements form interstitial hydrides.
Answer:
d and f-block elements.

Question 42.
Explain why beryllium forms a covalent hydride while calcium forms an ionic hydride?
Answer:
Because of higher electronegativity (= 1.5). Be forms covalent hydride while due to lower electronegativity (= 1.0), calcium forms ionic hydride.

Question 43.
Write two uses of interstitial hydrides.
Answer:

1. For storing Hg and
2. Catalysts of hydrogenation reactions.

Question 44.
Explain why electrolysis of ordinary water occurs faster than heavy water.
Answer:
Due to lower bond dissociation energy of H-O-H bonds in water than D-O-D bonds in D₂O, electrolysis of H₂O is much faster than of D₂O.

Question 45.
Can marine species live in distilled water?
Answer:
No, because distilled H₂O does not contain dissolved oxygen.

Question 46.
Can distilled water be called deionised water?
Answer:
Yes. Distilled water does not contain any cations and anions.

Question 47.
Which isotope of hydrogen is used as a tracer in organic reactions?
Answer:
Out of the three isotopes of hydrogen H, D, T both D and T can be used. But due to the radioactive nature of T, it is the only D that is used as a tracer in understanding the mechanism of organic reactions.

Question 48.
Which salts present in water make it permanent hard?
Answer:
Calcium and magnesium chlorides and sulphates.

Question 49.
Name a process that can remove both temporary and permanent hardness of the water.
Answer:
Pemiutit process.

Question 50.
Complete the reaction:
Fe(s) + H₂O(g) →?
Answer:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

Hydrogen Important Extra Questions Short Answer Type

Question 1.
Hydrogen forms three types of bonds in its compounds. Describe each type of bonding using suitable examples.
Answer:
Hydrogen forms compounds in three different ways:
1. By loss of electrons as in the reactions of H₂ with CuO

2. By gain of electrons as in reactions of H₂ with metals.

3. By sharing of electrons as in the reactions of H₂ with halogens

Question 2.
Name one example of a reaction in which dihydrogen acts as
(i) an oxidizing agent
Answer:
As an oxidizing agent

Here Na has been oxidized to Na while dihydrogen has been reduced to H⁺ ion.

(ii) a reducing agent.
Answer:

Here CuO has been reduced to copper and H₂ has been oxidized to H₂O.

Question 3.
The process \(\frac{1}{2}\) H₂(g) + e^– → H^– (g) is endothermic (DH = +151 kJ mol^-1), yet salt- like hydrides are known. How do you account for this?
Answer:
This is due to the reason that high lattice energy released (energy released during the formation of solid metal hydride from their corresponding gaseous ions, i.e., M⁺ and H⁺) more than compensates the energy, needed for the formation of H^– ions from H₂ gas.

Question 4.
Find the volume strength of 1.6 N H₂O₂ solution.
Answer:
Strength = Normality × EQuestion wt.
Eq. wt.of H₂O₂ = 17
∴ Strength of 1.6N H₂O₂ solution = 1.6 × 17g L^-1

Now 68g of H₂O₂ gives 22400 mL O₂ at NTP/STP
∴ 1.6 × 17g of H₂O₂ will give = \(\frac{22400}{68}\) × 1.6 × 17
= 8960 mL of O₂ at STP
But 1.6 × 17g of H₂O₂ are present in 1000 mL of H₂O₂ solution

Hence 1000 mL of H₂O₂ solution gives 8960 mL of O₂ at STP 1 mL of H₂O₂ will give = 8.96 mL of O₂ at STP.

Hence the volume strength of 1.6N H₂O₂ solution is = 8.96 volume

Question 5.
A sample of hard water is allowed to pass through an anion exchanger. Will it produce lather with soap easily?
Answer:
No. Ca²⁺ and Mg²⁺ ions are still present and these will react with soap to form curdy white ppt. Therefore it will not produce lather with soap solution easily.

Question 6.
Anhydrous Ba02 is not used for preparing H₂O₂ Why?
Answer:
BaSO₄ formed during the reaction of BaO₂ with H₂SO₄ forms a protective layer around unreacted BaO₂ and the reaction stops after some time.

Question 7.
Find the volume strength of the 2N H₂O₂ solution.
Answer:
Volume strength = 5.6 × Normality
Volume strength = 5.6 × 2
= 11.2 volumes.

Question 8.
Calculate the concentration in g. L-1 of a 20 volume H₂O₂ solution.
Answer:
1 L of 20 volume H₂O₂ solution on decomposition gives 20L of O₂ at STP.

Now 22.4 L of O₂ at STP is obtained from 68g of H₂O₂

20L of O₂ at STP is obtained from \(\frac{68}{22.4}\) × 20 = 60.7g

Thus the strength of 20 volume H₂O₂ solution = 60.7g L^-1.

Question 9.
Explain why an oxide ion is called a hard ion?
Answer:
Oxide ion (O^2-) is very small in size and thus cannot be easily polarised and hence it is called a hard oxide ion.

Question 10.
Statues coated with white lead on long exposure to the atmosphere turn black and the original colour can be restored on treatment with H₂O₂. Why?
Answer:
On long exposure to the atmosphere, white lead is converted into black PbS due to the action of H₂S present in the atmosphere.
PbO₂ + 2H₂S → PbS + 2H₂O

On the treatment of such blackened statues with H₂O_2, the black PbS gets oxidised to PbSO₄ and the colour is restored.
PbS + 4H₂O₂ → PbSO₄ + 4H₂O.

Question 11.
A mixture of hydrazine and H₂O₂ with Cu(II) catalyst is used as a rocket propellant. Why?
Answer:
The reaction between hydrazine (N₂H₄) and H₂O₂ is highly exothermic and is accompanied by a large increase in the volumes of the products and hence the mixture is used as a rocket propellant.

Question 12.
Calculate the volume of 10 volume H₂O₂ solution that will react with 200 ml. of 2N KMnO₄4 in an acidic medium.
Answer:
Normality 10 volume H₂O₂ = \(\frac{10 \times 68}{22.4 \times 17}=\frac{10}{5.6}\)N J
Applying normality equation

Question 13.
What is water gas? How is it prepared?
Answer:
An equimolar mixture of CO and H is called water gas. It is prepared by passing steam over a red hot iron.

Question 14.
The boiling point of H₂O is higher than that of H₂S. Explain.
Answer:
Due to intermolecular H-bonding in H₂O₂ extra energy in the form of heat is required to break these H-bonds due to which H₂O boils at a higher temperature than H₂S which does not have H-bonding.

Question 15.
Write two uses of interstitial hydrides.
Answer:
Two important uses to which interstitial hydrides are put are:

1. storing H₂ gas
2. Catalysts for hydrogenation reactions.

Question 16.
What is meant by auto protolysis of water?
Answer:
Auto-protolysis of water means that two molecules of water react with each other through proton transfer, i.e. one acts as the acid and the other acts as a base. The molecule which accepts a proton is converted into H₃O⁺ while the other which loses a proton is converted into OH^– ion

Question 17.
What is the difference between hydrolysis and hydration?
Answer:
Interaction of H⁺ ions and OH^– ions of H₂O with anions and cations of the salt respectively to give an acidic or a basic solution is called hydrolysis. For example:

Hydration,.on the other hand, means the addition of water to ions or molecules to form hydrated ions or hydrated salts. For examples:

Question 18.
Complete the following equations:
(i) Fe(s) + H₂O(g) →?
Answer:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

(ii) PbS(s) + H₂O₂(aq)→?
Answer:
PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O

(iii) MnO₄^–(aq) + H₂O(g) →?
Answer:
MnO₄-(aq) + 5H₂O₂(aq) + 6H+ → 2MnO⁺²+ 8H₂O + 5O₂

Question 19.
Discuss the importance of heavy water in a nuclear reactor.
Answer:
Heavy water is used as a moderator in nuclear reactors because it slows down the fast-moving neutrons & therefore helps in controlling the process of nuclear fission. It has also been used as a tracer compound to study the mechanism of many chemical reactions.

Question 20.
Distinguish clearly between:
(a) Hard & soft water.
Answer:

S.N. Hard water	Soft water
1. It does not give lather with soap freely.	1. It readily forms lather with soap.
2. Magnesium & calcium salts are present in it.	2. It is free from magnesium & calcium salts.

(b) Temporary hardness & permanent hardness of the water.
Answer:

S.No. Temporary hardness	Permanent hardness
1. This is due to calcium & magnesium bicarbonates in the water.	1. This is due to calcium & magnesium chlorides & sulphates.
2. It can be removed by boiling water.	2. It cannot be removed by boiling water.

Question 21.
Explain the correct context in which the following terms are used:
(a) diprotium
Answer:
Diprotium: It is used for the correct term for Hr

(b) dihydrogen
Answer:
Dihydrogen: The term dihydrogen is used for the H2 molecule while referring to the isotopic mixture with natural abundance for H & D.

(c) proton
Answer:
Proton: It is used for H⁺.

(d) hydrogen
Answer:
Hydrogen: It is used in relation to the isotope.

Question 22.
Hydrogen forms three types of bonds in its compound. Describe each type of bonding using suitable examples.
Answer:
(a) Ionic bond: The binary hydrides of alkali metals like (LiH, KH & NaH otc.) form an ionic bond. They on electrolysis give hydride ion (H^–)
NaH(s) + H₂O(aq) → H₂(g) + NaOH(aq)
Na⁺ + H^– → NaH

(b) Metallic bond: d & f block elements (Metals) form metallic bonds.
Example: SeH₂, YH₂ & LaH₃ etc.

(c) Covalent bond: With p-block elements hydrogen form a covalent bond. For example,

Question 23.
Why do lakes freeze from the top to the bottom?
Answer:
There are intermolecular hydrogen bonding in H₂O molecule. The density of water is greater than ice. It may be noted that at 4°C water has maximum density. In the severe cold, the upper layer of the seawater freezes & the heavier water (density more than that of ice) is present below the surface of the ice. Due to this sea animals can live safely in the water.

Question 24.
Why does elemental hydrogen react with other substances only slowly at room temperature?
Answer:
The high bond energy of di-hydrogen (436 kJ mol^-1) makes it a very stable molecule & therefore its reaction with other elements are slow at room temperature.

Question 25.
Why is water an excellent solvent for ionic or polar substances?
Answer:
Water molecules are highly polar. When water molecules interact with ions of an ionic compound, a large amount of hydration energy is released. The hydration energy is more than the energy needed to overcome the interionic attractions of the ionic compounds as well as the energy needed to break the hydrogen-bonded, association of water molecules. Therefore, the ionic compounds dissolve readily in water.

When water molecule interacts with strongly polar substances, the hydration energy released is sufficient to break the molecules of polar substances into ions as well as to break the hydrogen-bonded, association structure of water molecules. Therefore, water is an excellent solvent for polar compounds.

Question 26.
How is hydrogen obtained from:
(i) Nitric acid
Answer:
By action of Mg on cold water & dill. HNO₃
Mg + 2HNO₃(aq) → Mg(NO₃)₂ + H₂

(ii) Alcohol
Answer:
By action of Na or K on alcohol
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂

(iii) Ammonia?
Write a chemical equation in each case.
Answer:
By passing NH₃ over heated Na or K 2Na + 2NH₃ → 2NaNH₂ + H₂

Question 27.
Explain how is dil. solution of hydrogen peroxide concentrated?
Answer:
Hydrogen peroxide is concentrated by its distillation under reduced pressure at a temperature below 333K & the absence of heat metal impurities. 90% H₂O₂ is obtained by this method. For further concentration, the solution is cooled to about 263K when crystals enriched in hydrogen peroxide separate out. This process of fractional crystallisation is repeated to get 100% H₂O₂

Question 28.
Describe how is strength of hydrogen peroxide expressed?
Answer:
Strength of H₂O₂ as a percentage: H₂O₂ in solution (W/V), 70%. H₂O₂ means 70g of H₂O₂ is present in 100 g of solution.

Volume strength: Strength of hydrogen peroxide is expressed as the volume of oxygen liberated at S.T.P. given by ml. of a sample of hydrogen peroxide on decomposition into water & oxygen, for example, 30 volumes H₂O₂ means that 30 ml O₂ is obtained at S.T.P. by decomposing 1 ml of H₂O₂2 solution.

Question 29.
What happens when:
(i) Barium peroxide is treated with cold dilute H₂SO₄.
Answer:
H2O2 is obtained:
BaO₂. 8H₂O + H₂SO₄ → BaSO₄ ↑ + H₂O₂ + 8H₂O

(ii) Sodium peroxide is treated with cold dilute H₂SO₄ & the resulting mixture is cooled below 273 K.
Answer:
H₂O₂ is obtained:

(iii) Barium peroxide is treated with Phosphoric acid.
Answer:
H₂O₂ is obtained:
3BaO₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 3H₂O₂

Question 30.
Why is water an excellent solvent for ionic or polar substances?
Answer:
Water is a polar solvent with a high dielectric constant. Due to the high dielectric constant of water, the coelomic force of attraction between cations & anions gets weakened. Thus, a water molecule is able to remove ions from the lattice site using ion dipole forces easily. The dissolution of ionic/polar substances in water is further favoured by the hydrations of ions by the water molecule.

Hydrogen Important Extra Questions Long Answer Type

Question 1.
(a) Compare atomic hydrogen with nascent hydrogen.
Answer:
Comparison of atomic and nascent hydrogen

The main point of differences are:

1. Nascent hydrogen can be produced even at room temperature but atomic hydrogen is produced only at very high temperature.
2. Nascent hydrogen can never be isolated, but atomic hydrogen can be isolated.
3. The reducing power of atomic hydrogen is much greater than that of nascent hydrogen.

In general reactivity of the three forms of hydrogen increases in order.
Molecular hydrogen (H₂) < Nascent hydrogen < Atomic hydrogen.

(b) What is (i) active hydrogen
Answer:
Active Hydrogen: It is obtained by subjecting a stream of molecular hydrogen at ordinary temperature to silent electric discharge at about 30,000 volts. It is very reactive in nature (half-life = 0.33 second, and combines directly at ordinary temperatures with Pb and S forming their hydrides

(ii) heavy hydrogen? How are they formed?
Answer:
Heavy hydrogen: It is manufactured by the electrolysis of heavy water containing a little of IT SO, or NaOH to make the solution conducting.

In the laboratory, it can be prepared by the action of heavy water on sodium metal.
2D₂O(l) + 2Na(s) → 2NaOD(aq) + D₂(g).

Question 2.
How is the solution of H₂O₂ concentrated?
Answer:
The concentration of hydrogen peroxide: Hydrogen peroxide obtained by any method is always in the form of a dilute solution. Great care is to be taken for concentrating its solution because it is unstable and decomposes on heating.
2H₂O₂ → 2H₂O + O₂

The decomposition of H2O2 is catalysed by the ions of heavy metals present as impurities.

The solution of H2O2 is concentrated by the following methods.
1. By careful evaporation on a water bath: A dilute solution of H₂O₂ is taken in a shallow evaporating dish and is heated at 313K – 323 K. Water evaporates slowly and a hydrogen-peroxide solution of about 15 – 50% strength is obtained.

2. By dehydration in a vacuum desiccator: The dilute (50 %) solution of H₂O₂ obtained as above, is further concentrated by placing the same in a vacuum desiccator containing concentrated H₂SO₄ as a dehydrating agent. Here, water vapours are absorbed by concentrated sulphuric acid. This is shown in the diagram

(Concentration of H₂O₂ in vacuum desiccator)

3. By distillation under reduced pressure: The solution of hydrogen peroxide is further concentrated by subjecting it to distillation under reduced pressure. The solution is distilled at 308 – 313 K under a reduced pressure of 15 mm Hg. Water present in the solution distils over leaving behind about 98 – 99% concentrated solution of hydrogen peroxide.

4. By crystallisation: The last traces of water present in H₂O are removed by freezing it in a freezing mixture of solid CO₂ and others. The crystals of hydrogen peroxide separate out. These crystals are removed, dried and then remitted to obtain 100% pure hydrogen peroxide.

5. Storage of hydrogen peroxide: In order to check the decomposition of hydrogen peroxide, a small amount of acetanilide (i.e. negative catalyst) is added to it before storing the hydrogen peroxide.

Hydrogen peroxide cannot be concentrated by distillation at ordinary pressure because it undergoes decomposition into water and oxygen as it is a highly unstable liquid. It decomposes even on long-standing or on heating.

Question 3.
What are the different methods used for the softening of hard water? Explain the principle of each method.
Answer:
Hard water can be softened by the following methods depending upon the nature of hardness.
(a) Temporary hardness:
1. By boiling: It can be removed by merely boiling the water. Boiling decomposes the bicarbonates to give carbon dioxide and insoluble carbonates, which can be removed by filtration.

2. Clark’s process: Temporary hardness can be removed by the addition of a calculated amount of lime, whereupon magnesium and/or calcium carbonates is precipitated.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O
Mg(HCO₃)₂ + Ca(OH)₂ → CaCO3₃ + MgCO₃ + 2H₂O

(b) Permanent hardness:
1. With sodium carbonate: On treatment with washing soda, Ca²⁺ and Mg²⁺ in hard water are precipitated. The precipitate of the insoluble carbonates thus formed is removed by filtration.

2. Ion-exchange method: The common substance used for this process is zeolite which is hydrated sodium aluminium silicate, NaAl(SiO)₂, The exchange occurs when passing over the zeolite bed, sodium ions from zeolite are replaced by calcium and magnesium ions. Thus
Na(Ze) + Mg²⁺ → (Ze)₂Ca + 2Na⁺
2NaZe + Mg²⁺ → (Ze)₂Mg + 2Na⁺

when all the sodium ions of the zeolite have been replaced, the zeolite is said to be exhausted. It can be regenerated by treatment with a strong solution of sodium chloride.
2Na + (Ze)₂Ca → 2ZeNa + Ca²⁺.

Question 4.
Show how hydrogen peroxide can function both as an oxidising and a reducing agent.
Answer:
Oxidising properties: H₂O₂ has a tendency to accept electrons in chemical reactions and thus behaves as an oxidising agent in both acidic and alkaline medium.
H₂O₂ → H₂O + O
In acidic medium
H₂O₂ + 2H⁺ + 2e^– → 2H₂O

In alkaline medium
H₂O₂ + OH^– + 2e^– → 3OH^–
Example:
(a) In acidic medium:
2Fe²⁺ + 2H⁺ + H₂O₂ → 2Fe³⁺ + 2H₂O
(b) In alkaline medium:
3Cr3+ + 4H₂O₂ + 100H^– → 3CrO₄^2- + 8H₂O

Reducing properties: H₂O₂ can give electrons in a few reactions and thus behaves as a reducing agent.
In acidic medium
H₂O₂ → O₂ + 2H+ + 2e^–

In alkaline medium
H₂O₂ + 2OH^– → 2H₂O + O₂ + 2e^–

Reducing property in acidic medium:
2MnO₄^2- + 6H+ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂

Reducing property in basic medium:
2Fe³⁺ + H₂O₂ + 2OH^– → 2Fe²⁺ + O₂ + 2H₂O

Hydrogen Important Extra Questions Numerical Problems

Question 1.
Calculate the percentage strength & strength in g/L of 10 volume hydrogen peroxide solution.
Answer:
H₂O₂ decomposes on heating according to the equation:

From the equation
22.4L of O₂ at N.T.P are obtained from 68g of H202

∴ 10 ml of O₂ at N.T.P will be obtained from \(\frac{68}{22400}\) × 10g of H₂O₂
But 10 ml of O₂ at N.T.P are produced from 1 ml. of 10 volume H₂O₂ solution.

Thus 1 ml of 10 volume H₂O₂ solution contains \(\frac{68}{22400}\) × 10 g of H₂O₂
∴ 100 ml. of 10 volume H₂O₂ solution will contain
\(\frac{68}{22400}\) × \(\frac{10}{1}\) × 100 = 3.036g .

Thus a 10 volume H₂O₂ solution is approx. 3%
Alternatively, 1000 ml of 10 volume of H₂O₂ will contain H₂O₂
\(\frac{68}{22400}\) × 10 × 1000 = 30.36g

Therefore, strength of H₂O₂ in 10 volume H₂O₂ is 30.36 g/L

Question 2.
Calculate the normality of 20 volume hydrogen peroxide solution.
Answer:
Step-I: To calculate the strength in g/L of 20 volume H₂O₂ solution.
By definition, 1L of 20 volume H₂O₂ solution on decomposition gives 20 litres of oxygen at N.T.P. consider the chemical equation.

Now 22.4 litres of 02 at N.T.P. will be obtained from H₂O₂
\(\frac{68}{22400}\) × 20g = 60.7g

Thus, the strength of 20 volume H₂O₂ solution is 60.7 g/l

Step-II: To calculate the equivalent wt. of H₂O₂ consider the chemical equation,

From the above equation, 32 parts by wt. of oxygen are obtained from 68 parts by wt. of H₂O₂
∴ 8 parts by wt. of oxygen will be obtained from
\(\frac{68}{32}\) × 8 = 17 parts by wt. of H₂O₂
∴ Equivalent wt. of H₂O₂ = 17

Step-III: To calculate the normality = \(\frac{\text { Strength }}{\text { Eq. wt. }}=\frac{60.7}{17}\) = 3.57
Hence normality of 20 volume H₂O₂ solution = 3.57N

Question 3.
Find the volume strength of the 1.6N H₂O₂ solution.
Answer:
We know that strength = Normality × Eq. wt. of H₂O₂

∴ Strength of 1.6N H₂O₂ solution = 1.6N × 17
Now 68g of H₂O₂ gives 22400 ml O₂ at N.T.P.

∴ 1.6 × 17g of H₂O₂ will give
\(\frac{22400}{68}\) × 1.6 × 17 = 8960 ml of O₂ at N.T.P.

But 1.67 × 17g of H₂O₂ are present in 1000 ml of H₂O₂ solution.
Hence 1000 ml of H₂O₂ solution gives 8960 ml of O₂ at N.T.P.

∴ 1 ml of H₂O₂ solution will give = \(\frac{8960}{1000}\)
= 8.96 ml of O₂ at N.T.P.
Hence the volume strength of 1.6N H₂O₂ solution = 8.96 volume.

Question 4.
Calculate the amount of H₂O₂ present in 10 ml of 25 volume H₂O₂ solution.
Answer:
ml. of 25 volume H₂O₂ liberate O₂
10 × 25 = 250 ml. at N.T.P

∴ Amount of H₂O₂ that will liberate 250 ml of O₂ at N.T.P.
= \(\frac{68 \times 250}{22400}\) = 0.759 g

Question 5.
10 ml of a given solution of H₂O₂ contains 0.91 g of H₂O₂ Express its strength in volume.
Answer:
68g of H₂O₂ produce O₂ = 22400 ml at N.T.P.

∴ 0.91g of H₂O₂ will produce O₂
\(\frac{22400 \times 0.91}{68}\) = 300 ml at N.T.P.

Volume strength = \(\frac{300}{10}\) = 30

Question 6.
Calculate the strength in volumes of a solution containing 30.36 g/litre of H₂O₂.
Answer:

68g of H₂O₂ Produces 22.4 L O₂ at N.T.P.

30.36g of K₂O₂ will produce \(\frac{22.4}{68}\) × 30.46
= 10 L O₂ at N.T.P.
The given solution of H₂O₂ produces 10 L of O₂ at N.T.P.