Study of First Element Hydrogen Exe-6D Short Answer Chemistry Class-9 ICSE Selina Publishers

Study of First Element Hydrogen Exe-6D Short Answer Chemistry Class-9 ICSE Selina Publishers Solutions Chapter-6. Step By Step ICSE Selina Concise Solutions of Chapter-6 Study of First Element Hydrogen with All Exercise including MCQs, Very Short Answer Type, Short Answer Type, Long Answer Type, Numerical and Structured/Application Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

Study of First Element Hydrogen Exe-6D Short Answer Chemistry Class-9 ICSE Concise Selina Publishers

Board ICSE
Publications Selina Publication
Subject Chemistry
Class 9th
Chapter-6 Study of First Element Hydrogen
Book Name Concise
Topics Solution of Exercise – 6D Short Answer Type
Academic Session 2023-2024

D. Exercise – 6D Short Answer Type

Study of First Element Hydrogen Class-9 Chemistry Concise Solutions  

Page-119

Question 1.

Divide the following redox reactions into oxidation and reduction half reactions.

(i) Zn + Pb2+ ⟶ Zn2+ + Pb

(ii) Zn + Cu2+ ⟶ Zn2+ + Cu

(iii) Cl2 + 2Br ⟶ Br2 + 2Cl

Answer:

(i) Zn + Pb2+ ⟶ Zn2+ + Pb

Oxidation : Zn ⟶ Zn2+ + 2e

Reduction : Pb2+ + 2e ⟶ Pb

(ii) Zn + Cu2+ ⟶ Zn2+ + Cu

Oxidation : Zn ⟶ Zn2+ + 2e

Reduction : Cu2+ + 2e ⟶ Cu

(iii) Cl2 + 2Br ⟶ Br2 + 2Cl

Oxidation : Cl2 ⟶ 2Cl + 2e

Reduction : 2Br + 2e ⟶ Br2

Question 2.

(a) Write the equation in the ionic form

CuSO4(aq)  + Fe(s)→ FeSO4(aq) + Cu(s)

(b) Divide the above equation into oxidation and reduction half reactions.

Answer:

(a) Equation in the ionic form:

Cu2+ SO42-   + Fe → Fe2+  SO42- + Cu

(b) Fe → Fe2+  2 e —- Oxidation

Cu2+ + 2 e → Cu —- Reduction

Question 3.

Select the odd one out and justify your answer.

(a) Zn, Fe, Mg and Na

(b) SO2, H2S, NH3 and CO3

(c) Fe, Zn, Cu and Mg

(d) Fe, Pb, Al and Zn

Answer:

(a) Na

The other metals react with dil. HCl liberating hydrogen gas, while sodium reacts violently with acid.

(b) NH3 is basic in nature.

(c) Cu

Metals more reactive than hydrogen can displace it from acids.

(d) Pb

Lead reacts with dilute sulphuric acid or HCl and forms an insoluble coating of lead sulphate or lead chloride.

The others react with dilute sulphuric acid or HCl to liberate hydrogen.

Question 4.

Give reasons:

(a) Hydrogen is collected by the downward displacement of water and not of air, even though it is lighter than air.

(b) A candle brought near the mouth of a jar containing hydrogen gas starts burning but is extinguished when pushed inside the jar.

(c) Apparatus for laboratory preparation of hydrogen should be air tight and away from a naked flame.

Answer:

(a) Hydrogen is collected by the downward displacement of air because
i. It is insoluble in water.
ii. It forms an explosive mixture with air and therefore cannot be collected by the downward displacement of air even though it is lighter than it.

(b) Hydrogen is combustible, but it does not support combustion. So, the candle burns in air or oxygen when brought near the mouth of a jar containing hydrogen but is extinguished when pushed inside the jar as the supply of oxygen is cut off.

(c) Apparatus for laboratory preparation of hydrogen should be airtight and away from a naked flame because a mixture of hydrogen and air explodes violently when brought near a flame.

Question 5.

Write half-reactions for the following reaction :

A+ + B ⟶ A + B+ and name the following:

(a) oxidizing agent

(b) substance oxidized

(c) reducing agent.

Answer:

Half reaction:

A+ + e ⟶ A (Reduction)

B ⟶ B + e (Oxidation)

a. A, as it is oxidizing B by accepting electrons.

b. B, as it is losing electrons.

c. B, a it is reducing A by providing electrons.

Question 6.

State whether the following conversions are oxidation or reduction:

(a) PbO2 + SO2→ PbSO4

(b) Cu2+ + 2 e→ Cu

(c) K → K+ + e

(d) 2Cl – e→ Cl2

Answer:

(a) Oxidation

(b) Reduction

(c) Oxidation

(d) Oxidation

Question 7.

State, giving reasons, whether the substances printed in bold letters have been oxidized or reduced.

(a) PbO + CO → Pb + CO2
(b) Mg + 2HCl → MgCl2 + H2
(c) H2S + Cl2→ 2HCl + S
(d) Cl2 + H2S → 2HCl + S

Answer:

(a) PbO in the given reaction is reduced to Pb by losing oxygen.

(b) Magnesium undergoes oxidation by loss of electrons (Mg – 2e→ Mg2+).

(c) H2S undergoes oxidation by loss of hydrogen to give sulphur.

(d) Chlorine undergoes reduction by the addition of hydrogen to form HCl.

—  : End of Study of First Element Hydrogen Exe-6D Short Answer Class-9 ICSE Chemistry Solutions :–

Return to  Return to Concise Selina ICSE Chemistry Class-9 

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