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Determine the oxidation number of each atom indicated in the following: a. \({H}_{2}\) f. \({HNO}_{3}\) b. \({H}_{2} {O}\) g. \({H}_{2} {SO}_{4}\) c. Al h. \({Ca}({OH})_{2}\) d. \({MgO}\) i. \({Fe}({NO}_{3})_{2}\) e. \({Al}_{2} {S}_{3}\) j. \({O}_{2}\)

Short Answer

Expert verified
a. 0, b. H: +1, O: -2, c. 0, d. Mg: +2, O: -2, e. Al: +3, S: -2, f. H: +1, N: +5, O: -2, g. H: +1, S: +6, O: -2, h. Ca: +2, O: -2, H: +1, i. Fe: +2, N: +5, O: -2, j. 0

Step by step solution

01

Determining the oxidation number for \({H}_{2}\)

In a diatomic molecule of hydrogen, each hydrogen atom has an oxidation number of 0. This is because it is in its elemental form.
02

Determining the oxidation number for \({H}_{2} {O}\)

In water \({H}_{2}O\), hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2.
03

Determining the oxidation number for Al

For elemental aluminum, the oxidation number is 0.
04

Determining the oxidation number for \({MgO}\)

In \({MgO}\), magnesium has an oxidation number of +2 and oxygen has an oxidation number of -2.
05

Determining the oxidation number for \({Al}_{2} {S}_{3}\)

In \({Al}_{2} {S}_{3}\), aluminum has an oxidation number of +3 and sulfur has an oxidation number of -2.
06

Determining the oxidation number for \({HNO}_{3}\)

In \({HNO}_{3}\), hydrogen has an oxidation number of +1, nitrogen has an oxidation number of +5, and oxygen has an oxidation number of -2.
07

Determining the oxidation number for \({H}_{2} {SO}_{4}\)

In \({H}_{2} {SO}_{4}\), hydrogen has an oxidation number of +1, sulfur has an oxidation number of +6, and oxygen has an oxidation number of -2.
08

Determining the oxidation number for \({Ca}({OH})_{2}\)

In \({Ca}({OH})_{2}\), calcium has an oxidation number of +2, oxygen has an oxidation number of -2, and hydrogen has an oxidation number of +1.
09

Determining the oxidation number for \({Fe}({NO}_{3})_{2}\)

In \({Fe}({NO}_{3})_{2}\), iron has an oxidation number of +2, nitrogen has an oxidation number of +5, and oxygen has an oxidation number of -2.
10

Determining the oxidation number for \({O}_2\)

In a diatomic molecule of oxygen, each oxygen atom has an oxidation number of 0. This is because it is in its elemental form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

chemical compounds
Chemical compounds are substances made up of two or more elements that are chemically bonded together. These compounds have unique properties distinct from the individual elements they are composed of. For instance, water (\(H_2O\)) comprises hydrogen and oxygen, but it has different properties compared to hydrogen gas (\(H_2\)) and oxygen gas (\(O_2\)). In a chemical compound, the elements are present in a fixed ratio.
Properties of chemical compounds often depend on the types of bonds between the atoms, such as ionic or covalent bonds. Understanding chemical compounds is fundamental to studying and comprehending various chemical reactions.
oxidation states
The oxidation state or oxidation number is an indicator of the degree of oxidation of an atom in a chemical compound. It denotes the number of electrons an atom gains, loses, or appears to use when it forms bonds. For instance, in water (\(H_2O\)), each hydrogen atom has an oxidation state of +1, while oxygen has an oxidation state of -2.
The sum of oxidation states in a neutral compound will always equal zero, while for ions, it will equal the ion charge. This principle helps in balancing redox reactions and making sense of different chemical processes. Learning to assign oxidation numbers is crucial in understanding how elements interact in chemical reactions.
electrochemistry
Electrochemistry is a branch of chemistry that studies the relationship between chemical reactions and electrical energy. It involves redox (oxidation-reduction) reactions where electron transfer occurs. In these reactions, oxidation involves the loss of electrons, while reduction refers to the gain of electrons.
Electrochemical cells, like batteries and fuel cells, operate on this principle. These cells convert chemical energy into electrical energy and vice versa. A fundamental understanding of oxidation numbers is necessary to grasp electrochemical reactions, as these numbers indicate how electrons are transferred during a reaction.
redox reactions
Redox reactions, short for reduction-oxidation reactions, are processes in which the oxidation state of atoms changes by the transfer of electrons. These reactions are essential in various natural and industrial processes, including metabolism, combustion, and corrosion.
Understanding redox reactions involves identifying the substances that are oxidized and reduced. Oxidation is the process of losing electrons, leading to an increase in oxidation state. Reduction, on the other hand, is the gain of electrons, resulting in a decrease in oxidation state. Mastering redox concepts is vital for analyzing and predicting the outcomes of chemical reactions.

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Most popular questions from this chapter

Performance For one day, record situations that show evidence of oxidation- reduction reactions. Identify the reactants and the products, and determine whether there is proof that a chemical reaction has taken place.

Aluminum is described in Group 13 of the Elements Handbook (Appendix A) as a self-protecting metal. This property of aluminum results from a redox reaction. a. Write the redox equation for the oxidation of aluminum. b. Write the half-reactions for this reaction, and show the number of electrons transferred. c. What problems are associated with the buildup of aluminum oxide on electrical wiring made of aluminum?

Balance the equation for the reaction in which hot, concentrated sulfuric acid reacts with zinc to form zinc sulfate, hydrogen sulfide, and water.

Which of the following are redox reactions? a. \(2 \mathrm{Na}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NaCl}\) b. \(\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) c. \(2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{H}_{2}+\mathrm{O}_{2}\) d. \(\mathrm{NaCl}+\mathrm{AgNO}_{3} \longrightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}\) e. \(\mathrm{NH}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NH}_{4}+\mathrm{Cl}^{-}\) f. \(2 \mathrm{KClO}_{3} \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2}\) g. \(\mathrm{H}_{2}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{HCl}\) h. \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) i. \(\mathrm{Zn}+\mathrm{CuSO}_{4} \longrightarrow \mathrm{ZnSO}_{4}+\mathrm{Cu}\)

Drawing Conclusions An element that disproportionates must have at least how many different oxidation states? Explain your reasoning.

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