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Chapter 4: Problem 51

Which element is oxidized, and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) $$ 3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q) $$ (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(I)\)

Short Answer

Expert verified
In the given reactions: (a) N is reduced and H is oxidized. (b) Fe is reduced and Al is oxidized. (c) Cl is reduced and I is oxidized. (d) S is oxidized and O is reduced.

Step by step solution

01

Assign Oxidation Numbers

First, assign oxidation numbers to each element in the reactants and products: In \(\mathrm{N}_{2}\), the oxidation number of N is 0. In \(\mathrm{H}_{2}\), the oxidation number of H is 0. In \(\mathrm{NH}_{3}\), the oxidation number of N is -3 and that of H is +1.
02

Identify Oxidation and Reduction

Compare the oxidation numbers of the elements in the reactants and products: N (oxidation number) changes from 0 to -3. H (oxidation number) changes from 0 to +1. In this reaction, N is reduced (its oxidation number decreases), and H is oxidized (its oxidation number increases). (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)
03

Assign Oxidation Numbers

First, assign oxidation numbers to each element in the reactants and products: In \(\mathrm{Fe(NO}_3)_2\), the oxidation number of Fe is +2, N is +5, and O is -2. In \(\mathrm{Al}\), the oxidation number of Al is 0. In \(\mathrm{Fe}\), the oxidation number of Fe is 0. In \(\mathrm{Al(NO}_3)_3\), the oxidation number of Al is +3, N is +5, and O is -2.
04

Identify Oxidation and Reduction

Compare the oxidation numbers of the elements in the reactants and products: Fe (oxidation number) changes from +2 to 0. Al (oxidation number) changes from 0 to +3. In this reaction, Fe is reduced (its oxidation number decreases), and Al is oxidized (its oxidation number increases). (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\)
05

Assign Oxidation Numbers

First, assign oxidation numbers to each element in the reactants and products: In \(\mathrm{Cl}_2\), the oxidation number of Cl is 0. In \(\mathrm{NaI}\), the oxidation number of Na is +1 and I is -1. In \(\mathrm{I}_2\), the oxidation number of I is 0. In \(\mathrm{NaCl}\), the oxidation number of Na is +1 and Cl is -1.
06

Identify Oxidation and Reduction

Compare the oxidation numbers of the elements in the reactants and products: Cl (oxidation number) changes from 0 to -1. I (oxidation number) changes from -1 to 0. In this reaction, Cl is reduced (its oxidation number decreases), and I is oxidized (its oxidation number increases). (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)
07

Assign Oxidation Numbers

First, assign oxidation numbers to each element in the reactants and products: In \(\mathrm{PbS}\), the oxidation number of Pb is +2 and S is -2. In \(\mathrm{H}_2\mathrm{O}_2\), the oxidation number of H is +1 and O is -1. In \(\mathrm{PbSO}_4\), the oxidation number of Pb is +2, S is +6, and O is -2. In \(\mathrm{H}_2\mathrm{O}\), the oxidation number of H is +1 and O is -2.
08

Identify Oxidation and Reduction

Compare the oxidation numbers of the elements in the reactants and products: S (oxidation number) changes from -2 to +6. O (oxidation number) changes from -1 to -2. In this reaction, S is oxidized (its oxidation number increases), and O is reduced (its oxidation number decreases).

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Key Concepts

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

Oxidation Numbers
Oxidation numbers are a way to keep track of electrons in chemical reactions. They help us figure out which elements are gaining or losing electrons. In general, the oxidation number refers to the charge an atom would have if the compound was composed of ions. Here鈥檚 how to assign them:

  • Pure elements have an oxidation number of 0. For example, in \(_2\) or \( e\), each element has an oxidation number of 0.
  • For simple ions, the oxidation number is the charge of the ion. For example, \(a^+\) has an oxidation number of +1.
  • In compounds, the more electronegative element is given negative oxidation numbers. For instance, in \(h_3\), nitrogen (more electronegative than hydrogen) has an oxidation number of -3.
  • The sum of oxidation numbers in a neutral compound is 0, and in ionic compounds, it equals the ion鈥檚 charge.
Applying these rules helps in organizing and understanding redox reactions.
Oxidizing and Reducing Agents
In a redox reaction, one element loses electrons (oxidation), while another gains electrons (reduction). The oxidizing agent is the element or compound that causes oxidation, meaning it gains electrons and is reduced in the process. Conversely, the reducing agent causes reduction by losing electrons and being oxidized. Let鈥檚 break it down further:

  • Oxidizing Agents: These elements or compounds accept electrons. For example, \(_2\) in the first reaction is an oxidizing agent as it gains electrons.
  • Reducing Agents: These donate electrons. In the decomposition reaction of \(h_3\), \(h_3\) acts as a reducing agent, losing electrons.
By identifying these agents, we can better understand the flow of electrons during a reaction. Always remember, an increase in oxidation number suggests oxidation, while a decrease indicates reduction.
Chemical Equations
Chemical equations represent chemical reactions, showing the reactants and products with their respective quantities. Each equation must be balanced to adhere to the law of conservation of mass, meaning the number of atoms for each element is equal on both sides of the equation. Let鈥檚 look at some tips to understand them:

  • Reactants are the starting substances and are usually placed on the left side of the equation.
  • Products are the new substances formed and are found on the right side.
  • State symbols like (s), (l), (g), and (aq) denote the physical state of substances: solid, liquid, gas, or aqueous.
  • Coefficients in front of compounds balance the equation, ensuring that the same number of each type of atom appears on both sides.
Understanding chemical equations assists in visualizing the occurrences within a reaction, thereby providing a clearer picture of the chemical changes involved.

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Most popular questions from this chapter

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NH}_{4} \mathrm{I}\) and \(\mathrm{CuCl}_{2},\) (b) LiOH and \(\mathrm{MnCl}_{2}\), (c) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{CoSO}_{4}\)

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4},\) (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3},(\mathbf{c})\) strontium sulfide, \(\mathrm{Sr} S,(\mathbf{d})\) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\). (e) Locate sulfur in the periodic table in Exercise 4.47 what region is it in? (f) Which region(s) of the periodic table contains elements that can adopt both positive and negative oxidation numbers?

Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the aqueous \(\mathrm{HCl}\) in the stomach and each of the following substances used in various antacids: (a) \(\mathrm{Al}(\mathrm{OH})_{3}(s),(\mathbf{b}) \mathrm{Mg}(\mathrm{OH})_{2}(s),(\mathbf{c}) \mathrm{MgCO}_{3}(s),\) (d) \(\mathrm{NaAl}\left(\mathrm{CO}_{3}\right)(\mathrm{OH})_{2}(s)\) (e) \(\mathrm{CaCO}_{3}(s)\)

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+}\). Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?
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