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Chemistry - Study of the First Element - Hydrogen (Preparation, Properties, Oxidation/Reduction)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hydrogen (H2H_2) is the first element in the periodic table, possessing a unique position because it shows properties of both alkali metals (Group 1) and halogens (Group 17).

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In the laboratory, Hydrogen is prepared by the action of dilute acids on reactive metals like zinc: Zn+2HCl→ZnCl2+H2↑Zn + 2HCl \rightarrow ZnCl_2 + H_2 \uparrow. Granulated zinc is preferred because it contains impurities like copper which act as a catalyst.

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Industrial preparation includes the Bosch Process, where water gas ((CO+H2)(CO + H_2)) is reacted with steam in the presence of Fe2O3Fe_2O_3 and Cr2O3Cr_2O_3 at 450∘C450^{\circ}C.

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Hydrogen is a powerful reducing agent. It removes oxygen from metallic oxides like CuOCuO or PbOPbO to yield the respective metals.

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Oxidation is defined as the addition of oxygen, removal of hydrogen, or the loss of electrons (De-electronation).

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Reduction is defined as the addition of hydrogen, removal of oxygen, or the gain of electrons (Electronation).

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A Redox reaction is a simultaneous process where one reactant is oxidized and the other is reduced. For example, in the reaction between H2H_2 and CuOCuO, H2H_2 is the reducing agent and CuOCuO is the oxidizing agent.

📐Formulae

Zn+H2SO4(dil.)→ZnSO4+H2↑Zn + H_2SO_4 (dil.) \rightarrow ZnSO_4 + H_2 \uparrow

C+H2O→1000∘C(CO+H2) [Water Gas]C + H_2O \xrightarrow{1000^{\circ}C} (CO + H_2) \text{ [Water Gas]}

(CO+H2)+H2O→Fe2O3,450∘CCO2+2H2+Δ(CO + H_2) + H_2O \xrightarrow{Fe_2O_3, 450^{\circ}C} CO_2 + 2H_2 + \Delta

CuO+H2→Cu+H2OCuO + H_2 \rightarrow Cu + H_2O

2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O

Fe→Fe2++2e− (Oxidation)Fe \rightarrow Fe^{2+} + 2e^- \text{ (Oxidation)}

Cu2++2e−→Cu (Reduction)Cu^{2+} + 2e^- \rightarrow Cu \text{ (Reduction)}

💡Examples

Problem 1:

Identify the oxidizing agent and the reducing agent in the following reaction: PbO+H2→Pb+H2OPbO + H_2 \rightarrow Pb + H_2O

Solution:

Oxidizing Agent: PbOPbO; Reducing Agent: H2H_2.

Explanation:

In this reaction, PbOPbO loses oxygen to become PbPb, so it undergoes reduction and acts as the oxidizing agent. H2H_2 gains oxygen to become H2OH_2O, so it undergoes oxidation and acts as the reducing agent.

Problem 2:

Explain the electronic concept of oxidation using the reaction: Mg+Cl2→MgCl2Mg + Cl_2 \rightarrow MgCl_2

Solution:

Oxidation: Mg→Mg2++2e−Mg \rightarrow Mg^{2+} + 2e^-; Reduction: Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^-.

Explanation:

According to the electronic concept, oxidation is the loss of electrons. Here, the Magnesium atom loses two electrons to form a Magnesium ion (Mg2+Mg^{2+}), thus it is oxidized. Chlorine gains those electrons, thus it is reduced.

Problem 3:

Why is concentrated sulphuric acid (H2SO4H_2SO_4) not used in the laboratory preparation of Hydrogen from Zinc?

Solution:

Concentrated H2SO4H_2SO_4 is a strong oxidizing agent.

Explanation:

If concentrated H2SO4H_2SO_4 is used, it reacts with the produced Hydrogen or the metal to produce Sulphur dioxide (SO2SO_2) gas instead of Hydrogen gas: Zn+2H2SO4(conc.)→ZnSO4+SO2+2H2OZn + 2H_2SO_4 (conc.) \rightarrow ZnSO_4 + SO_2 + 2H_2O.