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Chapter 20: Problem 16

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l)-\cdots \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (b) \(\operatorname{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots+\mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)-\cdots\) \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
All the given reactions involve oxidation-reduction. In reaction (a), Phosphorus is reduced (from +5 to +3) and Bromine is oxidized (from -1 to +1). In reaction (b), Iodine is oxidized (from -1 to +5) and Chlorine is reduced (from +1 to -1). In reaction (c), Sulfur is oxidized (from +4 to +6) and Nitrogen is reduced (from +5 to +2). In reaction (d), Bromine is oxidized (from -1 to 0) and Sulfur is reduced (from +6 to +4).

Step by step solution

01

Assign Oxidation Numbers

We assign oxidation numbers to the reactants and products using the rules of oxidation numbers. The oxidation numbers are given below in parentheses: \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l)-\cdots \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (+5)_P(-1)_Br(+1)_H(-2)_O --> (+3)_P(+1)_H(-2)_O(+1)_H(-1)_Br
02

Compare Oxidation Numbers

Comparing the assigned oxidation numbers of elements in the reactants and products, we observe changes in the oxidation numbers for P (from +5 to +3) and Br (from -1 to +1).
03

Conclusion for (a)

Since there are changes in the oxidation numbers, the reaction (a) involves oxidation-reduction. Phosphorus is reduced (from +5 to +3) and Bromine is oxidized (from -1 to +1). (b) \(\operatorname{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots+\mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\)
04

Assign Oxidation Numbers

We assign oxidation numbers to the reactants and products using the rules of oxidation numbers: \(\operatorname{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots+\mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (+1)_Na(-1)_I(+1)_H(-2)_O(+1)_Cl --> (+1)_Na(-2)_O(+5)_I(+1)_H(-1)_Cl
05

Compare Oxidation Numbers

Comparing the assigned oxidation numbers of elements in the reactants and products, we observe changes in the oxidation numbers for I (from -1 to +5) and Cl (from +1 to -1).
06

Conclusion for (b)

Since there are changes in the oxidation numbers, the reaction (b) involves oxidation-reduction. Iodine is oxidized (from -1 to +5) and Chlorine is reduced (from +1 to -1). (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)-\cdots\) \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g)\)
07

Assign Oxidation Numbers

We assign oxidation numbers to the reactants and products using the rules of oxidation numbers: \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)-\cdots\) \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g)\) (+4)_S(-2)_O(-2)_O(+5)_N(-2)_O(+1)_H(-2)_O --> (+6)_S(-2)_O(+1)_H(-2)_O(+2)_N(-2)_O
08

Compare Oxidation Numbers

Comparing the assigned oxidation numbers of elements in the reactants and products, we observe changes in the oxidation numbers for S (from +4 to +6) and N (from +5 to +2).
09

Conclusion for (c)

Since there are changes in the oxidation numbers, the reaction (c) involves oxidation-reduction. Sulfur is oxidized (from +4 to +6) and Nitrogen is reduced (from +5 to +2). (d) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)
10

Assign Oxidation Numbers

We assign oxidation numbers to the reactants and products using the rules of oxidation numbers: \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (+1)_H(-2)_O(+6)_S(-2)_O(+1)_Na(-1)_Br --> (0)_Br(+4)_S(-2)_O(+1)_Na(-2)_O(+6)_S(-2)_O(+1)_H(-2)_O
11

Compare Oxidation Numbers

Comparing the assigned oxidation numbers of elements in the reactants and products, we observe changes in the oxidation numbers for Br (from -1 to 0) and S (from +6 to +4).
12

Conclusion for (d)

Since there are changes in the oxidation numbers, the reaction (d) involves oxidation-reduction. Bromine is oxidized (from -1 to 0) and Sulfur is reduced (from +6 to +4).

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Key Concepts

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

Assigning Oxidation Numbers
Understanding how to assign oxidation numbers is crucial when studying chemical reactions, especially in the context of redox chemistry. Oxidation numbers are a way of keeping track of the electrons during a reaction and can tell us whether a redox reaction has occurred.

To assign oxidation numbers, there are several rules that we follow. For example, the oxidation number of a pure element is always zero, and for a monoatomic ion, it is equal to the charge of the ion. In compounds, the sum of the oxidation numbers must equal the overall charge. Hydrogen is usually +1 (except when it forms hydrides with metals, where it is -1), and oxygen is typically -2, except in peroxides or when bonded with fluorine.

By comparing the oxidation numbers of atoms in the reactants and products, we can identify the elements that undergo oxidation or reduction. For instance, when phosphorus goes from +5 in PBr3 to +3 in H3PO3, it has gained electrons and thus is reduced. Similarly, when bromine goes from -1 to 0, it loses electrons and is oxidized. These changes are clear indicators of a redox process.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. There are different types of chemical reactions, such as synthesis, decomposition, single replacement, double replacement, and combustion. Oxidation-reduction reactions, or redox reactions, are a special category where electrons are transferred between species, changing their oxidation states.

Identifying a redox reaction involves looking for changes in oxidation numbers, as seen in our exercise. For example, when sodium iodide reacts with hypochlorous acid, the iodine in iodide goes from -1 to +5 in NaIO3, and thus, is oxidized. Concurrently, chlorine is reduced. Similarly, in the reaction between sulfur dioxide and nitric acid, sulfur is oxidized while nitrogen is reduced. These changes signify that electron transfer is taking place.
Redox Chemistry
Redox chemistry centers around oxidation-reduction reactions, where 'oxidation' means losing electrons, and 'reduction' means gaining electrons. In any redox reaction, there is an exchange of electrons from one species to another, leading to changes in oxidation numbers.

In redox chemistry, we often speak of oxidizing and reducing agents. An oxidizing agent is a substance that accepts electrons and is reduced, while a reducing agent donates electrons and is oxidized. Understanding the role of each reactant in a redox reaction provides deeper insight into the reaction mechanisms. In the exercise, for example, PBr3 acts as a reducing agent as it gives up electrons from phosphorus, while water behaves as an oxidizing agent.
Balancing Equations
Balancing chemical equations is essential to ensure that the law of conservation of mass is satisfied, which states that matter cannot be created or destroyed in a chemical reaction. To balance equations, we must have the same number of each type of atom on both sides of the equation.

For redox reactions, balancing also involves ensuring that the electrons lost in oxidation match the electrons gained in reduction. This sometimes requires the use of coefficients to multiply the number of molecules participating in the reaction. In our example exercises, this can be observed in reactions such as the one between SO2 and HNO3, where coefficients are used to balance the sulfur and nitrogen atoms, as well as their respective oxidation states.

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Most popular questions from this chapter

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., Vol. \(50,1973,61\) ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

For the generic reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow\) \(\mathrm{A}^{-}(a q)+\mathrm{B}^{+}(a q)\) for which \(E^{\circ}\) is a positive number, answer the following questions: (a) What is being oxidized, and what is being reduced? (b) If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (c) Which half-reaction from (b) is higher in potential energy? (d) What is the sign of the free energy change for the reaction? [Sections \(20.4\) and \(20.5]\)

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for \(2.00 \mathrm{~min}, 12.3 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(25.5^{\circ} \mathrm{C}\), and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l)-\ldots\) \(3 \mathrm{Ce}^{3+}(a q)+\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}(a q)-\cdots \rightarrow\) \(\mathrm{N}_{2}(g)+5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4^{-}}(a q)\)

A common shorthand way to represent a voltaic cell is to list its components as follows: anode|anode solution || cathode solution|cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}\); sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+& 6 \mathrm{H}^{+}(a q)--\rightarrow \\ & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.
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