1749. Fuel Cells -
A fuel cell is a galvanic cell that uses traditional combustive fuels, most often hydrogen or methane, that are continuously fed into the cell along with an oxidant. (An alternative, but not very popular, name for a fuel cell is a flow battery.) Within the cell, fuel and oxidant undergo the same redox chemistry as when they are combusted, but via a catalyzed electrochemical that is significantly more efficient. For example, a typical hydrogen fuel cell uses graphite electrodes embedded with platinum-based catalysts to accelerate the two halfcell reactions:

FIGURE 16.14 In this hydrogen fuel cell, oxygen from the air reacts with hydrogen, producing water and electricity.

$$
\begin{aligned}
& \text { Anode: } \quad 2 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \\
& \text {Cathode: } \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) \\
& \hline \text { Cell: } \quad 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) \quad \\
& \hline \text { cell } \sim 1.2 \mathrm{~V}
\end{aligned}
$$

These types of fuel cells generally produce voltages of approximately 1.2 V . Compared to an internal combustion engine, the energy efficiency of a fuel cell using the same redox reaction is typically more than double ( $\sim 20 \%-25 \%$ for an engine versus $\sim 50 \%-75 \%$ for a fuel cell). Hydrogen fuel cells are commonly used on extended space missions, and prototypes for personal vehicles have been developed, though the technology remains relatively immature.


1750. LINK TO LEARNING -
Check out this link (http://openstax.org/l/16fuelcells) to learn more about fuel cells.


1750. LINK TO LEARNING - 1750.1. Corrosion
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Define corrosion
- List some of the methods used to prevent or slow corrosion

Corrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year.


1752. Statue of Liberty: Changing Colors -
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure 16.15). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper "skin." So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide ( $\mathrm{Cu}_{2} \mathrm{O}$ ), which is red, and then to copper(II) oxide, which is black

$$
\begin{aligned}
2 \mathrm{Cu}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s) \\
\mathrm{Cu}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CuO}(s)
\end{aligned}
$$

Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the CuO

$$
\begin{gathered}
2 \mathrm{CuO}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Cu}_{2} \mathrm{CO}_{3}(\mathrm{OH})_{2}(s) \\
3 \mathrm{CuO}(s)+2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Cu}_{2}\left(\mathrm{CO}_{3}\right)_{2}(\mathrm{OH})_{2}(s) \\
4 \mathrm{CuO}(s)+\mathrm{SO}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Cu}_{4} \mathrm{SO}_{4}(\mathrm{OH})_{6}(s)
\end{gathered}
$$


1752. Statue of Liberty: Changing Colors -
These three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation, a phenomenon discussed further in another chapter of this text.

FIGURE 16.15 (a) The Statue of Liberty is covered with a copper skin, and was originally brown, as shown in this painting. (b) Exposure to the elements has resulted in the formation of the blue-green patina seen today.

Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure 16.15. The relevant redox reactions are described by the following equations:

$$
\begin{array}{llr}
\text { anode: } & \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} & E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V} \\
\text { cathode: } & \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{O}_{2} / \mathrm{O}^{2}}^{\circ}=+1.23 \mathrm{~V} \\
\text { overall: } & 2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{cell}}^{\circ}=+1.67 \mathrm{~V}
\end{array}
$$

Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:


1752. Statue of Liberty: Changing Colors -
Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:

$$
4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+(4+2 x) \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}(s)+8 \mathrm{H}^{+}(a q)
$$

The stoichiometry of the hydrate varies, as indicated by the use of $x$ in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.

FIGURE 16.16 Corrosion can occur when a painted iron or steel surface is exposed to the environment by a scratch through the paint. A galvanic cell results that may be approximated by the simplified cell schematic $\mathrm{Fe}(s) \mid \mathrm{Fe}^{2+}(\mathrm{aq})$ $\| \mathrm{O}_{2}(\mathrm{aq}), \mathrm{H}_{2} \mathrm{O}(\mathrm{D}) \mid \mathrm{Fe}(\mathrm{s})$.

One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.

Other strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron.


1752. Statue of Liberty: Changing Colors -
Iron and other metals may also be protected from corrosion by galvanization, a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph.

Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure 16.17). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.

FIGURE 16.17 Cathodic protection is a useful approach to electrochemically preventing corrosion of underground storage tanks.


1753. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the process of electrolysis
- Compare the operation of electrolytic cells with that of galvanic cells
- Perform stoichiometric calculations for electrolytic processes

Electrochemical cells in which spontaneous redox reactions take place (galvanic cells) have been the topic of discussion so far in this chapter. In these cells, electrical work is done by a redox system on its surroundings as electrons produced by the redox reaction are transferred through an external circuit. This final section of the chapter will address an alternative scenario in which an external circuit does work on a redox system by imposing a voltage sufficient to drive an otherwise nonspontaneous reaction, a process known as electrolysis. A familiar example of electrolysis is recharging a battery, which involves use of an external power source to drive the spontaneous (discharge) cell reaction in the reverse direction, restoring to some extent the composition of the half-cells and the voltage of the battery. Perhaps less familiar is the use of electrolysis in the refinement of metallic ores, the manufacture of commodity chemicals, and the electroplating of metallic coatings on various products (e.g., jewelry, utensils, auto parts). To illustrate the essential concepts of electrolysis, a few specific processes will be considered.


1754. The Electrolysis of Molten Sodium Chloride -
Metallic sodium, Na , and chlorine gas, $\mathrm{Cl}_{2}$, are used in numerous applications, and their industrial production relies on the large-scale electrolysis of molten sodium chloride, $\mathrm{NaCl}(1)$. The industrial process typically uses a Downs cell similar to the simplified illustration shown in Figure 16.18. The reactions associated with this process are:

| anode: | $2 \mathrm{Cl}^{-}(l) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-}$ |
| :--- | :---: |
| cathode: | $\mathrm{Na}^{+}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Na}(l)$ |
| cell: | $2 \mathrm{Na}^{+}(l)+2 \mathrm{Cl}^{-}(l) \longrightarrow 2 \mathrm{Na}(l)+\mathrm{Cl}_{2}(g)$ |

The cell potential for the above process is negative, indicating the reaction as written (decomposition of liquid $\mathrm{NaCl})$ is not spontaneous. To force this reaction, a positive potential of magnitude greater than the negative cell potential must be applied to the cell.

FIGURE 16.18 Cells of this sort (a cell for the electrolysis of molten sodium chloride) are used in the Downs process for production of sodium and chlorine, and they typically use iron cathodes and carbon anodes.


1755. The Electrolysis of Water -
Water may be electrolytically decomposed in a cell similar to the one illustrated in Figure 16.19. To improve electrical conductivity without introducing a different redox species, the hydrogen ion concentration of the water is typically increased by addition of a strong acid. The redox processes associated with this cell are

| anode: | $2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-}$ | $E_{\text {anode }}^{\circ}$ |
| :--- | ---: | :--- |
| cathode: $2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)$ | $E_{\text {cathode }}^{\circ}$ | $=0 \mathrm{~V}$ |
| cell: | $2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 229 \mathrm{~V}$ |  |
|  |  | $E_{\text {cell }}^{\circ}$ |

Again, the cell potential as written is negative, indicating a nonspontaneous cell reaction that must be driven by imposing a cell voltage greater than +1.229 V . Keep in mind that standard electrode potentials are used to inform thermodynamic predictions here, though the cell is not operating under standard state conditions. Therefore, at best, calculated cell potentials should be considered ballpark estimates.

FIGURE 16.19 The electrolysis of water produces stoichiometric amounts of oxygen gas at the anode and hydrogen at the anode.


1756. The Electrolysis of Aqueous Sodium Chloride -
When aqueous solutions of ionic compounds are electrolyzed, the anode and cathode half-reactions may involve the electrolysis of either water species $\left(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}^{+}, \mathrm{OH}^{-}\right)$or solute species (the cations and anions of the compound). As an example, the electrolysis of aqueous sodium chloride could involve either of these two anode reactions:

$$
\begin{array}{ll}
\text { (i) } 2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} & E_{\text {anode }}^{\circ}=+1.35827 \mathrm{~V} \\
\text { (ii) } 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & E_{\text {anode }}^{\circ}=+1.229 \mathrm{~V}
\end{array}
$$

The standard electrode (reduction) potentials of these two half-reactions indicate water may be oxidized at a less negative/more positive potential $(-1.229 \mathrm{~V})$ than chloride ion $(-1.358 \mathrm{~V})$. Thermodynamics thus predicts that water would be more readily oxidized, though in practice it is observed that both water and chloride ion are oxidized under typical conditions, producing a mixture of oxygen and chlorine gas.

Turning attention to the cathode, the possibilities for reduction are:


1756. The Electrolysis of Aqueous Sodium Chloride -
Turning attention to the cathode, the possibilities for reduction are:

$$
\begin{array}{ll}
\text { (iii) } 2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & E_{\text {cathode }}^{\circ}=0 \mathrm{~V} \\
\text { (iv) } 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q) & E_{\text {cathode }}^{\circ}=-0.8277 \mathrm{~V} \\
\text { (v) } \mathrm{Na}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Na}(s) & E_{\text {cathode }}^{\circ}=-2.71 \mathrm{~V}
\end{array}
$$

Comparison of these standard half-reaction potentials suggests the reduction of hydrogen ion is
thermodynamically favored. However, in a neutral aqueous sodium chloride solution, the concentration of hydrogen ion is far below the standard state value of 1 M (approximately $10^{-7} \mathrm{M}$ ), and so the observed cathode reaction is actually reduction of water. The net cell reaction in this case is then

$$
\text { cell: } 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{OH}^{-}(a q) \quad E_{\mathrm{cell}}^{\circ}=-2.186 \mathrm{~V}
$$

This electrolysis reaction is part of the chlor-alkali process used by industry to produce chlorine and sodium hydroxide (lye).


1757. Chemistry in Everyday Life Electroplating -
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. The silver plating of eating utensils is used here to illustrate the process. (Figure 16.20).

FIGURE 16.20 This schematic shows an electrolytic cell for silver plating eating utensils.
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. Applying a sufficient potential results in the oxidation of the silver anode

$$
\text { anode }: \mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-}
$$

and reduction of silver ion at the (spoon) cathode:

$$
\text { cathode: } \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)
$$

The net result is the transfer of silver metal from the anode to the cathode. Several experimental factors must be carefully controlled to obtain high-quality silver coatings, including the exact composition of the electrolyte solution, the cell voltage applied, and the rate of the electrolysis reaction (electrical current).


1758. Quantitative Aspects of Electrolysis -
Electrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second ( $\mathrm{A}=1 \mathrm{C} / \mathrm{s}$ ). The charge transferred, $Q$, by passage of a constant current, $I$, over a specified time interval, $t$, is then given by the simple mathematical product

$$
Q=I t
$$

When electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process

$$
\mathrm{M}^{\mathrm{n}+}(a q)+\mathrm{ne}^{-} \longrightarrow \mathrm{M}(s)
$$

involves the transfer of $n$ mole of electrons. The charge transferred is, therefore,

$$
Q=n F
$$

where $F$ is Faraday's constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction.


1760. Converting Current to Moles of Electrons -
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?


1761. Solution -
Faraday's constant can be used to convert the charge ( $Q$ ) into moles of electrons (n). The charge is the current (I) multiplied by the time

$$
n=\frac{Q}{F}=\frac{\frac{10.23 \mathrm{C}}{\mathrm{~s}} \times 1 \mathrm{hr} \times \frac{60 \mathrm{~min}}{\mathrm{hr}} \times \frac{60 \mathrm{~s}}{\mathrm{~min}}}{96,485 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-}}=\frac{36,830 \mathrm{C}}{96,485 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-}}=0.3817 \mathrm{~mol} \mathrm{e}^{-}
$$

From the problem, the solution contains $\mathrm{AgNO}_{3}$, so the reaction at the cathode involves 1 mole of electrons for each mole of silver

$$
\text { cathode: } \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)
$$

The atomic mass of silver is $107.9 \mathrm{~g} / \mathrm{mol}$, so

$$
\text { mass } \mathrm{Ag}=0.3817 \mathrm{~mol} \mathrm{e} ~-~ \times \frac{1 \mathrm{~mol} \mathrm{Ag}}{1 \mathrm{~mol} \mathrm{e}^{-}} \times \frac{107.9 \mathrm{~g} \mathrm{Ag}}{1 \mathrm{~mol} \mathrm{Ag}}=41.19 \mathrm{~g} \mathrm{Ag}
$$


1762. Check Your Learning -
Aluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?

Answer:
$\mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s) ; 0.0777 \mathrm{~mol} \mathrm{Al}=2.10 \mathrm{~g} \mathrm{Al}$.


1764. Time Required for Deposition -
In one application, a $0.010-\mathrm{mm}$ layer of chromium must be deposited on a part with a total surface area of 3.3 $\mathrm{m}^{2}$ from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is $7.19 \mathrm{~g} / \mathrm{cm}^{3}$.


1765. Solution -
First, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):

$$
\text { volume }=\left(0.010 \mathrm{~mm} \times \frac{1 \mathrm{~cm}}{10 \mathrm{~mm}}\right) \times\left(3.3 \mathrm{~m}^{2} \times\left(\frac{10,000 \mathrm{~cm}^{2}}{1 \mathrm{~m}^{2}}\right)\right)=33 \mathrm{~cm}^{3}
$$

Use the computed volume and the provided density to calculate the molar amount of chromium required:

$$
\begin{gathered}
\text { mass }=\text { volume } \times \text { density }=33 \mathrm{~cm}^{3} \times \frac{7.19 \mathrm{~g}}{\mathrm{~cm}^{3}}=237 \mathrm{~g} \mathrm{Cr} \\
\\
\mathrm{~mol} \mathrm{Cr}=237 \mathrm{~g} \mathrm{Cr} \times \frac{1 \mathrm{~mol} \mathrm{Cr}}{52.00 \mathrm{~g} \mathrm{Cr}}=4.56 \mathrm{~mol} \mathrm{Cr}
\end{gathered}
$$

The stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:

$$
Q=4.56 \mathrm{~mol} \mathrm{Cr} \times \frac{3 \mathrm{~mol} \mathrm{e}^{-}}{1 \mathrm{~mol} \mathrm{Cr}} \times \frac{96485 \mathrm{C}}{\mathrm{~mol} \mathrm{e}^{-}}=1.32 \times 10^{6} \mathrm{C}
$$

Finally, if this charge is passed at a rate of $33.46 \mathrm{C} / \mathrm{s}$, the required time is:


1765. Solution -
Finally, if this charge is passed at a rate of $33.46 \mathrm{C} / \mathrm{s}$, the required time is:

$$
t=\frac{Q}{I}=\frac{1.32 \times 10^{6} \mathrm{C}}{33.46 \mathrm{C} / \mathrm{s}}=3.95 \times 10^{4} \mathrm{~s}=11.0 \mathrm{hr}
$$


1766. Check Your Learning -
What mass of zinc is required to galvanize the top of a $3.00 \mathrm{~m} \times 5.50 \mathrm{~m}$ sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of $\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}$ and the current is 25.5 A , how long will it take to galvanize the top of the iron? The density of zinc is $7.140 \mathrm{~g} / \mathrm{cm}^{3}$.


1767. Answer: -
11.8 kg Zn requires 382 hours.


1768. Key Terms -
active electrode electrode that participates as a reactant or product in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidationreduction reaction
alkaline battery primary battery similar to a dry cell that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an improved replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell
anode electrode in an electrochemical cell at which oxidation occurs
battery single or series of galvanic cells designed for use as a source of electrical power
cathode electrode in an electrochemical cell at which reduction occurs
cathodic protection approach to preventing corrosion of a metal object by connecting it to a sacrificial anode composed of a more readily oxidized metal
cell notation (schematic) symbolic representation of the components and reactions in an electrochemical cell
cell potential ( $\boldsymbol{E}_{\text {cell }}$ ) difference in potential of the cathode and anode half-cells
concentration cell galvanic cell comprising halfcells of identical composition but for the concentration of one redox reactant or product
corrosion degradation of metal via a natural electrochemical process
dry cell primary battery, also called a zinc-carbon battery, based on the spontaneous oxidation of zinc by manganese(IV)
electrode potential ( $E_{\mathrm{X}}$ ) the potential of a cell in which the half-cell of interest acts as a cathode when connected to the standard hydrogen electrode
electrolysis process using electrical energy to cause a nonspontaneous process to occur
electrolytic cell electrochemical cell in which an external source of electrical power is used to drive an otherwise nonspontaneous process
Faraday's constant (F) charge on 1 mol of electrons; $F=96,485 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-}$
fuel cell devices similar to galvanic cells that require a continuous feed of redox reactants; also called a flow battery
galvanic (voltaic) cell electrochemical cell in
which a spontaneous redox reaction takes place; also called a voltaic cell
galvanization method of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.


1768. Key Terms -
which a spontaneous redox reaction takes place; also called a voltaic cell
galvanization method of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.
half cell component of a cell that contains the redox conjugate pair ("couple") of a single reactant
inert electrode electrode that conducts electrons to and from the reactants in a half-cell but that is not itself oxidized or reduced
lead acid battery rechargeable battery commonly used in automobiles; it typically comprises six galvanic cells based on Pb half-reactions in acidic solution
lithium ion battery widely used rechargeable battery commonly used in portable electronic devices, based on lithium ion transfer between the anode and cathode
Nernst equation relating the potential of a redox system to its composition
nickel-cadmium battery rechargeable battery based on Ni/Cd half-cells with applications similar to those of lithium ion batteries
primary cell nonrechargeable battery, suitable for single use only
sacrificial anode electrode constructed from an easily oxidized metal, often magnesium or zinc, used to prevent corrosion of metal objects via cathodic protection
salt bridge tube filled with inert electrolyte solution
secondary cell battery designed to allow recharging
standard cell potential ( $E_{\text {cell }}^{\circ}$ ) the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K
standard electrode potential $\left(\left(E_{X}^{\circ}\right)\right)$ electrode potential measured under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K
standard hydrogen electrode (SHE) half-cell based on hydrogen ion production, assigned a potential of exactly 0 V under standard state conditions, used as the universal reference for measuring electrode potential


1769. Key Equations -
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$E_{\text {cell }}^{\circ}=\frac{R T}{n F} \ln K$
$E_{\text {cell }}^{\circ}=\frac{0.0257 \mathrm{~V}}{n} \ln K=\frac{0.0592 \mathrm{~V}}{n} \log K$
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q \quad$ (Nernst equation)
$E_{\text {cell }}=E^{\circ}{ }_{\text {cell }}-\frac{0.0592 \mathrm{~V}}{n} \log Q$
$\Delta G=-n F E_{\text {cell }}$
$\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$
$w_{\text {ele }}=w_{\text {max }}=-n F E_{\text {cell }}$
$Q=I \times t=n \times F$


1770. Summary - 1770.1. Review of Redox Chemistry
Redox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.


1770. Summary - 1770.2. Galvanic Cells
Galvanic cells are devices in which a spontaneous redox reaction occurs indirectly, with the oxidant and reductant redox couples contained in separate half-cells. Electrons are transferred from the reductant (in the anode half-cell) to the oxidant (in the cathode half-cell) through an external circuit, and inert solution phase ions are transferred between half-cells, through a salt bridge, to maintain charge neutrality. The construction and composition of a galvanic cell may be succinctly represented using chemical formulas and others symbols in the form of a cell schematic (cell notation).


1770. Summary - 1770.3. Electrode and Cell Potentials
The property of potential, $E$, is the energy associated with the separation/transfer of charge. In electrochemistry, the potentials of cells and halfcells are thermodynamic quantities that reflect the driving force or the spontaneity of their redox processes. The cell potential of an electrochemical cell is the difference in between its cathode and anode. To permit easy sharing of half-cell potential data, the standard hydrogen electrode (SHE) is
assigned a potential of exactly 0 V and used to define a single electrode potential for any given half-cell. The electrode potential of a half-cell, $E_{X}$, is the cell potential of said half-cell acting as a cathode when connected to a SHE acting as an anode. When the half-cell is operating under standard state conditions, its potential is the standard electrode potential, $E^{\circ}{ }_{\mathrm{X}}$. Standard electrode potentials reflect the relative oxidizing strength of the half-reaction's reactant, with stronger oxidants exhibiting larger (more positive) $E^{o}{ }_{X}$ values. Tabulations of standard electrode potentials may be used to compute standard cell potentials, $E^{o}$ cell, for many redox reactions. The arithmetic sign of a cell potential indicates the spontaneity of the cell reaction, with positive values for spontaneous reactions and negative values for nonspontaneous reactions (spontaneous in the reverse direction).


1770. Summary - 1770.4. Potential, Free Energy, and Equilibrium
Potential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction's cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of $K$ ). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient, $Q$, as
described by the Nernst equation.


1770. Summary - 1770.5. Batteries and Fuel Cells
Galvanic cells designed specifically to function as electrical power supplies are called batteries. A variety of both single-use batteries (primary cells) and rechargeable batteries (secondary cells) are commercially available to serve a variety of applications, with important specifications including voltage, size, and lifetime. Fuel cells, sometimes called flow batteries, are devices that harness the energy of spontaneous redox reactions normally associated with combustion processes. Like batteries, fuel cells enable the reaction's electron transfer via an external circuit, but they require continuous input of the redox reactants (fuel and oxidant) from an external reservoir. Fuel cells are typically much more efficient in converting the energy released by the reaction to useful work in comparison to internal combustion engines.


1770. Summary - 1770.6. Corrosion
Spontaneous oxidation of metals by natural
electrochemical processes is called corrosion, familiar examples including the rusting of iron and the tarnishing of silver. Corrosion process involve the creation of a galvanic cell in which different sites on the metal object function as anode and cathode, with the corrosion taking place at the anodic site. Approaches to preventing corrosion of metals include use of a protective coating of zinc (galvanization) and the use of sacrificial anodes connected to the metal object (cathodic protection).


1770. Summary - 1770.7. Electrolysis
Nonspontaneous redox processes may be forced to occur in electrochemical cells by the application of an appropriate potential using an external power source-a process known as electrolysis. Electrolysis is the basis for certain ore refining processes, the industrial production of many chemical commodities, and the electroplating of metal coatings on various products. Measurement of the current flow during electrolysis permits stoichiometric calculations.


1771. Exercises - 1771.1. Review of Redox Chemistry
1. Identify each half-reaction below as either oxidation or reduction.
(a) $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}$
(b) $\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 \mathrm{e}^{-}$
(c) $\mathrm{MnO}_{4}{ }^{2-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{e}^{-}$
(d) $\mathrm{Li}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Li}$
2. Identify each half-reaction below as either oxidation or reduction.
(a) $\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}$
(b) $\mathrm{Mn}^{2+} \longrightarrow \mathrm{MnO}_{2}$
(c) $\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}$
(d) $\mathrm{NO}_{3}{ }^{-} \longrightarrow \mathrm{NO}$
3. Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.
(a) $\mathrm{Ca} \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}, \mathrm{F}_{2}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~F}^{-}$
(b) $\mathrm{Li} \longrightarrow \mathrm{Li}^{+}+\mathrm{e}^{-}, \mathrm{Cl}_{2}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}$
(c) $\mathrm{Fe} \longrightarrow \mathrm{Fe}^{3+}+3 \mathrm{e}^{-}, \mathrm{Br}_{2}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}$


1771. Exercises - 1771.1. Review of Redox Chemistry
(d) $\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}, \mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
4. Balance the equations below assuming they occur in an acidic solution.
(a) $\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Sn}^{2+} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{Sn}^{4+}$
(b) $\mathrm{PbO}_{2}+\mathrm{Hg} \longrightarrow \mathrm{Hg}_{2}{ }^{2+}+\mathrm{Pb}^{2+}$
(c) $\mathrm{Al}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Al}^{3+}+\mathrm{Cr}^{3+}$
5. Identify the oxidant and reductant of each reaction of the previous exercise.
6. Balance the equations below assuming they occur in a basic solution.
(a) $\mathrm{SO}_{3}{ }^{2-}(a q)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{Cu}(\mathrm{OH})(s)$
(b) $\mathrm{O}_{2}(g)+\mathrm{Mn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MnO}_{2}(s)$
(c) $\mathrm{NO}_{3}{ }^{-}(a q)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NO}(g)$
(d) $\mathrm{Al}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{Cr}(\mathrm{OH})_{4}{ }^{-}(a q)$
7. Identify the oxidant and reductant of each reaction of the previous exercise.


1771. Exercises - 1771.1. Review of Redox Chemistry
7. Identify the oxidant and reductant of each reaction of the previous exercise.
8. Why don't hydroxide ions appear in equations for half-reactions occurring in acidic solution?
9. Why don't hydrogen ions appear in equations for half-reactions occurring in basic solution?
10. Why must the charge balance in oxidation-reduction reactions?


1771. Exercises - 1771.2. Galvanic Cells
11. Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.
(a) $\mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Ni}(s)$
(b) $2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$
(c) $\mathrm{Mn}(s)+\mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Sn}(s)$
(d) $3 \mathrm{CuNO}_{3}(a q)+\mathrm{Au}\left(\mathrm{NO}_{3}\right)_{3}(a q) \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Au}(s)$
12. Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.
(a) $\mathrm{Mg}(s)\left|\mathrm{Mg}^{2+}{ }^{2+}(a q) \| \mathrm{Cu}^{2+}(a q)\right| \mathrm{Cu}(s)$
(b) $\mathrm{Ni}(s)\left|\mathrm{Ni}^{2+}(a q) \| \mathrm{Ag}^{+}(a q)\right| \operatorname{Ag}(s)$
13. Write a balanced equation for the cell reaction of each cell in the previous exercise.
14. Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.
(a) $\mathrm{Al}(s)+\mathrm{Zr}^{4+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Zr}(s)$


1771. Exercises - 1771.2. Galvanic Cells
(a) $\mathrm{Al}(s)+\mathrm{Zr}^{4+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Zr}(s)$
(b) $\mathrm{Ag}^{+}(a q)+\mathrm{NO}(g) \longrightarrow \mathrm{Ag}(s)+\mathrm{NO}_{3}{ }^{-}(a q) \quad$ (acidic solution)
(c) $\mathrm{SiO}_{3}{ }^{2-}(a q)+\mathrm{Mg}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{Mg}(\mathrm{OH})_{2}(s)$
(basic solution)
(d) $\mathrm{ClO}_{3}^{-}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \quad$ (basic solution)
15. Identify the oxidant and reductant in each reaction of the previous exercise.
16. From the information provided, use cell notation to describe the following systems:
(a) In one half-cell, a solution of $\mathrm{Pt}\left(\mathrm{NO}_{3}\right)_{2}$ forms Pt metal, while in the other half-cell, Cu metal goes into a
$\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ solution with all solute concentrations 1 M .
(b) The cathode consists of a gold electrode in a $0.55 \mathrm{M} \mathrm{Au}\left(\mathrm{NO}_{3}\right)_{3}$ solution and the anode is a magnesium electrode in $0.75 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}$ solution.
(c) One half-cell consists of a silver electrode in a $1 \mathrm{M} \mathrm{AgNO}_{3}$ solution, and in the other half-cell, a copper electrode in $1 \mathrm{MCu}\left(\mathrm{NO}_{3}\right)_{2}$ is oxidized.
17. Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?


1771. Exercises - 1771.2. Galvanic Cells
17. Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?
18. An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
19. An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
20. The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode.


1771. Exercises - 1771.3. Electrode and Cell Potentials
21. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
(a) $\mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Ni}(s)$
(b) $2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$
(c) $\mathrm{Mn}(s)+\mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Sn}(s)$
(d) $3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Au}\left(\mathrm{NO}_{3}\right)_{3}(a q) \longrightarrow 3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Au}(s)$
22. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
(a) $\mathrm{Mn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Ni}(s)$
(b) $3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Cu}(s)$
(c) $\mathrm{Na}(s)+\mathrm{LiNO}_{3}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{Li}(s)$
(d) $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ba}(s) \longrightarrow \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ca}(s)$


1771. Exercises - 1771.3. Electrode and Cell Potentials
23. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
$\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q) \| \mathrm{Au}^{3+}(a q)\right| \mathrm{Au}(s)$
24. Determine the cell reaction and standard cell potential at $25^{\circ} \mathrm{C}$ for a cell made from a cathode half-cell consisting of a silver electrode in $1 M$ silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?
25. Determine the cell reaction and standard cell potential at $25^{\circ} \mathrm{C}$ for a cell made from an anode half-cell containing a cadmium electrode in 1 M cadmium nitrate and a cathode half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?
26. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
$\operatorname{Pt}(s)\left|\mathrm{H}_{2}(g)\right| \mathrm{H}^{+}(a q) \| \mathrm{Br}_{2}(a q), \mathrm{Br}^{-}(a q) \mid \operatorname{Pt}(s)$


1771. Exercises - 1771.4. Potential, Free Energy, and Equilibrium
27. For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).
(a) $0.000 \mathrm{~V}, \mathrm{n}=2$
(b) $+0.434 \mathrm{~V}, \mathrm{n}=2$
(c) $-2.439 \mathrm{~V}, \mathrm{n}=1$
28. For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.
(a) $12 \mathrm{~kJ} / \mathrm{mol}, \mathrm{n}=3$
(b) $-45 \mathrm{~kJ} / \mathrm{mol}, \mathrm{n}=1$
29. Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K .
(a) $\mathrm{Hg}(l)+\mathrm{S}^{2-}(a q, 0.10 M)+2 \mathrm{Ag}^{+}(a q, 0.25 M) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{HgS}(s)$
(b) The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
(c) The cell comprised of a half-cell in which aqueous bromine $(1.0 \mathrm{M})$ is being oxidized to bromide ion $(0.11 M)$ and a half-cell in which $\mathrm{Al}^{3+}(0.023 M)$ is being reduced to aluminum metal.
30. Determine $\Delta G$ and $\Delta G^{\circ}$ for each of the reactions in the previous problem.
31. Use the data in Appendix $L$ to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.
(a) $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)$


1771. Exercises - 1771.4. Potential, Free Energy, and Equilibrium
(a) $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)$
(b) $\mathrm{CdS}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q) \quad$ at 377 K
(c) $\mathrm{Hg}^{2+}(a q)+4 \mathrm{Br}^{-}(a q) \rightleftharpoons\left[\mathrm{HgBr}_{4}\right]^{2-}(a q)$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad$ at $25^{\circ} \mathrm{C}$


1771. Exercises - 1771.5. Batteries and Fuel Cells
32. Consider a battery made from one half-cell that consists of a copper electrode in $1 \mathrm{MCuSO}_{4}$ solution and another half-cell that consists of a lead electrode in $1 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ solution.
(a) What is the standard cell potential for the battery?
(b) What are the reactions at the anode, cathode, and the overall reaction?
(c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.
(d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some $\mathrm{PbSO}_{4}(s)$ forms. Would the cell potential increase, decrease, or remain the same?
33. Consider a battery with the overall reaction: $\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$.
(a) What is the reaction at the anode and cathode?
(b) A battery is "dead" when its cell potential is zero. What is the value of $Q$ when this battery is dead?
(c) If a particular dead battery was found to have $\left[\mathrm{Cu}^{2+}\right]=0.11 \mathrm{M}$, what was the concentration of silver ion?
34. Why do batteries go dead, but fuel cells do not?
35. Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.
36. Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.


1771. Exercises - 1771.6. Corrosion
37. Which member of each pair of metals is more likely to corrode (oxidize)?
(a) Mg or Ca
(b) Au or Hg
(c) Fe or Zn
(d) Ag or Pt
38. Consider the following metals: $\mathrm{Ag}, \mathrm{Au}, \mathrm{Mg}, \mathrm{Ni}$, and Zn . Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use -0.447 V as the standard reduction potential for steel.
39. Aluminum $\left(E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\circ}=-2.07 \mathrm{~V}\right)$ is more easily oxidized than iron $\left(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\circ}=-0.477 \mathrm{~V}\right)$, and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. What might explain this observation?
40. If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.
41. Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?
42. Why would a sacrificial anode made of lithium metal be a bad choice


1771. Exercises - 1771.7. Electrolysis
43. If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?
44. For the scenario in the previous question, how many electrons moved through the circuit?
45. Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.
(a) $\mathrm{CaCl}_{2}$
(b) LiH
(c) $\mathrm{AlCl}_{3}$
(d) $\mathrm{CrBr}_{3}$
46. What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of $3.33 \times 10^{5} \mathrm{C}$ passes through each cell?
47. How long would it take to reduce 1 mole of each of the following ions using the current indicated?
(a) $\mathrm{Al}^{3+}, 1.234 \mathrm{~A}$
(b) $\mathrm{Ca}^{2+}, 22.2 \mathrm{~A}$
(c) $\mathrm{Cr}^{5+}, 37.45 \mathrm{~A}$
(d) $\mathrm{Au}^{3+}, 3.57 \mathrm{~A}$
48. A current of 2.345 A passes through the cell shown in Figure 16.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm ? (Hint: Is hydrogen the only gas present above the water?)
49. An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a $\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}$ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123 -mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is $7.140 \mathrm{~g} / \mathrm{cm}^{3}$.

79816 • Exercises


1772. CHAPTER 17 Kinetics -
Figure 17.1 An agama lizard basks in the sun. As its body warms, the chemical reactions of its metabolism speed up.


1773. CHAPTER OUTLINE - 1773.1. Chemical Reaction Rates
17.2 Factors Affecting Reaction Rates


1773. CHAPTER OUTLINE - 1773.2. Rate Laws
17.4 Integrated Rate Laws
17.5 Collision Theory
17.6 Reaction Mechanisms
17.7 Catalysis

INTRODUCTION The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun's rays is critical to the lizard's survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators.

From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: "Will the reaction produce the desired products in useful quantities?" The second question is: "How rapidly will the reaction occur?" A third question is often asked when investigating reactions in greater detail: "What specific molecular-level processes take place as the reaction occurs?" Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.

The study of chemical kinetics concerns the second and third questions-that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors
that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur.


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Define chemical reaction rate
- Derive rate expressions from the balanced equation for a given chemical reaction
- Calculate reaction rates from experimental data

A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.

The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution's conductivity.

For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, in an aqueous solution changes slowly over time as it decomposes according to the equation:

$$
2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)
$$

The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:

$$
\text { rate of decomposition of } \begin{aligned}
\mathrm{H}_{2} \mathrm{O}_{2} & =-\frac{\text { change in concentration of reactant }}{\text { time interval }} \\
& =-\frac{\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{t_{2}}-\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{t_{1}}}{t_{2}-t_{1}} \\
& =-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}
\end{aligned}
$$

This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta ( $\Delta$ ) indicates "change in." Thus, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{t_{1}}$ represents the molar concentration of hydrogen peroxide at some time $t_{1}$; likewise, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{t_{2}}$ represents the molar concentration of hydrogen peroxide at a later time $t_{2}$; and $\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ represents the change in molar concentration of hydrogen peroxide during the time interval $\Delta t$ (that is, $t_{2}-t_{1}$ ). Since the reactant concentration decreases as the reaction proceeds, $\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by -1 . Figure 17.2 provides an example of data collected during the decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$.


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
| Time (h) | $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left(\mathrm{mol} \mathrm{L}^{-1}\right)$ | $\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left(\mathrm{mol} \mathrm{L}^{-1}\right)$ | $\Delta t(\mathrm{~h})$ | $\begin{aligned} & \text { Rate of Decomposition, } \\ & \left(\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}\right) \end{aligned}$ |
| :---: | :---: | :---: | :---: | :---: |
| 0.00 | 1.000 |  |  |  |
|  |  | $\geq-0.500$ | 6.00 | 0.0833 |
| 6.00 | $0.500=$ |  |  |  |
|  |  | $\geq-0.250$ | 6.00 | 0.0417 |
| 12.00 | $0.250=$ |  |  |  |
|  | - | $\geq-0.125$ | 6.00 | 0.0208 |
| 18.00 | $0.125=$ |  |  |  |
| 24.00 | - | $\geq-0.062$ | 6.00 | 0.010 |

FIGURE 17.2 The rate of decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ in an aqueous solution decreases as the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ decreases.

To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of $40^{\circ} \mathrm{C}$. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
$$
\frac{-\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=\frac{-(0.500 \mathrm{~mol} / \mathrm{L}-1.000 \mathrm{~mol} / \mathrm{L})}{(6.00 \mathrm{~h}-0.00 \mathrm{~h})}=0.0833 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
$$

Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:

$$
\frac{-\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=\frac{-(0.0625 \mathrm{~mol} / \mathrm{L}-0.125 \mathrm{~mol} / \mathrm{L})}{(24.00 \mathrm{~h}-18.00 \mathrm{~h})}=0.010 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
$$


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at "time zero," when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle's initial rate-analogous to the beginning of a chemical reaction-would be the speedometer reading at the moment the driver begins pressing the brakes ( $t_{0}$ ). A few moments later, the instantaneous rate at a specific moment-call it $t_{1}$-would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car's average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop ( $\Delta t$ ). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.

The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ at any time $t$ is given by the slope of a straight line that is tangent to the curve at that time (Figure 17.3). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.


1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates
FIGURE 17.3 This graph shows a plot of concentration versus time for a 1.000 M solution of $\mathrm{H}_{2} \mathrm{O}_{2}$. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at $t=0 \mathrm{~h}$ ("initial rate") and at $t=12 \mathrm{~h}$ ("instantaneous rate" at 12 h ).


1775. Reaction Rates in Analysis: Test Strips for Urinalysis -
Physicians often use disposable test strips to measure the amounts of various substances in a patient's urine (Figure 17.4). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.

The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:

$$
\begin{gathered}
\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \xrightarrow{\text { catalyst }} \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{6}+\mathrm{H}_{2} \mathrm{O}_{2} \\
2 \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{I}^{-} \xrightarrow{\text { catalyst }} \mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}
\end{gathered}
$$

The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.

The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.


1775. Reaction Rates in Analysis: Test Strips for Urinalysis -
FIGURE 17.4 Test strips are commonly used to detect the presence of specific substances in a person's urine. Many test strips have several pads containing various reagents to permit the detection of multiple substances on a single strip. (credit: Iqbal Osman)


1776. Relative Rates of Reaction -
The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction

$$
\mathrm{aA} \longrightarrow \mathrm{bB}
$$

can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction:

$$
\text { rate }=-\left(\frac{1}{\mathrm{a}}\right)\left(\frac{\Delta \mathrm{A}}{\Delta t}\right)=\left(\frac{1}{\mathrm{~b}}\right)\left(\frac{\Delta \mathrm{B}}{\Delta t}\right)
$$

Consider the reaction represented by the following equation:

$$
2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)
$$

The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:

$$
-\frac{\Delta \mathrm{mol} \mathrm{NH}_{3}}{\Delta t} \times \frac{1 \mathrm{~mol} \mathrm{~N}_{2}}{2 \mathrm{~mol} \mathrm{NH}_{3}}=\frac{\Delta \mathrm{mol} \mathrm{~N}_{2}}{\Delta t}
$$

This may be represented in an abbreviated format by omitting the units of the stoichiometric factor:

$$
-\frac{1}{2} \frac{\Delta \mathrm{~mol} \mathrm{NH}}{3} \text { } \quad \Delta t \quad=\frac{\Delta \mathrm{mol} \mathrm{~N}_{2}}{\Delta t}
$$

Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:


1776. Relative Rates of Reaction -
$$
-\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_{3}\right]}{\Delta t}=\frac{\Delta\left[\mathrm{N}_{2}\right]}{\Delta t}
$$

Similarly, the rate of formation of $\mathrm{H}_{2}$ is three times the rate of formation of $\mathrm{N}_{2}$ because three moles of $\mathrm{H}_{2}$ are produced for each mole of $\mathrm{N}_{2}$ produced.

$$
\frac{1}{3} \frac{\Delta\left[\mathrm{H}_{2}\right]}{\Delta t}=\frac{\Delta\left[\mathrm{N}_{2}\right]}{\Delta t}
$$

Figure 17.5 illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen
and hydrogen at $1100^{\circ} \mathrm{C}$. Slopes of the tangent lines at $t=500 \mathrm{~s}$ show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:

$$
\frac{2.91 \times 10^{-6} \mathrm{M} / \mathrm{s}}{9.70 \times 10^{-7} \mathrm{M} / \mathrm{s}} \approx 3
$$

FIGURE 17.5 Changes in concentrations of the reactant and products for the reaction $2 \mathrm{NH}_{3} \longrightarrow \mathrm{~N}_{2}+3 \mathrm{H}_{2}$. The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at $t=500 \mathrm{~s}$.

EXAMPLE 17.1


1777. Expressions for Relative Reaction Rates -
The first step in the production of nitric acid is the combustion of ammonia:

$$
4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)
$$

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.


1778. Solution -
Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:

$$
-\frac{1}{4} \frac{\Delta\left[\mathrm{NH}_{3}\right]}{\Delta t}=-\frac{1}{5} \frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=\frac{1}{4} \frac{\Delta[\mathrm{NO}]}{\Delta t}=\frac{1}{6} \frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}\right]}{\Delta t}
$$

Check Your Learning
The rate of formation of $\mathrm{Br}_{2}$ is $6.0 \times 10^{-6} \mathrm{~mol} / \mathrm{L} / \mathrm{s}$ in a reaction described by the following net ionic equation:

$$
5 \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{Br}_{2}+3 \mathrm{H}_{2} \mathrm{O}
$$

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.


1779. Answer: -
$-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{BrO}_{3}{ }^{-}\right]}{\Delta t}=-\frac{1}{6} \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[\mathrm{Br}_{2}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}\right]}{\Delta t}$


1781. Reaction Rate Expressions for Decomposition of $\mathbf{H}_{\mathbf{2}} \mathbf{O}_{\mathbf{2}}$ -
The graph in Figure 17.3 shows the rate of the decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ over time:

$$
2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}
$$

Based on these data, the instantaneous rate of decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ at $t=11.1 \mathrm{~h}$ is determined to be $3.20 \times 10^{-2} \mathrm{~mol} / \mathrm{L} / \mathrm{h}$, that is:

$$
-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=3.20 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
$$

What is the instantaneous rate of production of $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{O}_{2}$ ?


1782. Solution -
The reaction stoichiometry shows that

$$
-\frac{1}{2} \frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}\right]}{\Delta t}=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}
$$

Therefore:

$$
\frac{1}{2} \times 3.20 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}
$$

and

$$
\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=1.60 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
$$


1783. Check Your Learning -
If the rate of decomposition of ammonia, $\mathrm{NH}_{3}$, at 1150 K is $2.10 \times 10^{-6} \mathrm{~mol} / \mathrm{L} / \mathrm{s}$, what is the rate of production of nitrogen and hydrogen?


1784. Answer: -
$1.05 \times 10^{-6} \mathrm{~mol} / \mathrm{L} / \mathrm{s}, \mathrm{N}_{2}$ and $3.15 \times 10^{-6} \mathrm{~mol} / \mathrm{L} / \mathrm{s}, \mathrm{H}_{2}$.


1785. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates

The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.


1786. The Chemical Nature of the Reacting Substances -
The rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air
overnight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.


1787. The Physical States of the Reactants -
A chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder (Figure 17.6). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.

(a)

(b)

FIGURE 17.6 (a) Iron powder reacts rapidly with dilute hydrochloric acid and produces bubbles of hydrogen gas: $2 \mathrm{Fe}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{FeCl}_{3}(a q)+3 \mathrm{H}_{2}(g)$. (b) An iron nail reacts more slowly because the surface area exposed to the acid is much less.


1788. LINK TO LEARNING -
Watch this video (http://openstax.org/l/16cesium) to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.


1789. Temperature of the Reactants -
Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by $10^{\circ} \mathrm{C}$.


1790. Concentrations of the Reactants -
The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate $\left(\mathrm{CaCO}_{3}\right)$ deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure 17.7). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:

$$
\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)
$$

Calcium carbonate reacts with sulfurous acid as follows:

$$
\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{CaSO}_{3}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)
$$

In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about $20 \%$ oxygen.

FIGURE 17.7 Statues made from carbonate compounds such as limestone and marble typically weather slowly over time due to the actions of water, and thermal expansion and contraction. However, pollutants like sulfur dioxide can accelerate weathering. As the concentration of air pollutants increases, deterioration of limestone occurs more rapidly. (credit: James P Fisher III)


1791. LINK TO LEARNING -
Phosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this video (http://openstax.org/l/16phosphor) to see an example.


1792. The Presence of a Catalyst -
Relatively dilute aqueous solutions of hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:

$$
2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g})
$$

Under typical conditions, this decomposition occurs very slowly. When dilute $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})$ is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound's exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter.


1793. (C) LINK TO LEARNING -
Chemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions \& Rates interactive (http://openstax.org/l/16PHETreaction) to explore how temperature, concentration, and the nature of the reactants affect reaction rates.


1794. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the form and function of a rate law
- Use rate laws to calculate reaction rates
- Use rate and concentration data to identify reaction orders and derive rate laws

As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants.
Rate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation

$$
a A+b B \longrightarrow \text { products }
$$

where $a$ and $b$ are stoichiometric coefficients. The rate law for this reaction is written as:

$$
\text { rate }=k[A]^{m}[B]^{n}
$$

in which $[A]$ and $[B]$ represent the molar concentrations of reactants, and $k$ is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents $m$ and $n$ are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant $k$ and the reaction orders $m$ and $n$ must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant $k$ is independent of the reactant concentrations, but it does vary with temperature.

The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is $m$ order with respect to $A$ and $n$ order with respect to $B$. For example, if $m=1$ and $n=2$, the reaction is first order in $A$ and second order in $B$. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall $(1+2=3)$. A few specific examples are shown below to further illustrate this concept.

The rate law:

$$
\text { rate }=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]
$$


1794. LEARNING OBJECTIVES -
The rate law:

$$
\text { rate }=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]
$$

describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:

$$
\text { rate }=k\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]^{2}
$$

describes a reaction that is second order in $\mathrm{C}_{4} \mathrm{H}_{6}$ and second order overall. The rate law:

$$
\text { rate }=k\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]
$$

describes a reaction that is first order in $\mathrm{H}^{+}$, first order in $\mathrm{OH}^{-}$, and second order overall.


1796. Writing Rate Laws from Reaction Orders -
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:

$$
\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)
$$

is second order in $\mathrm{NO}_{2}$ and zero order in CO at $100^{\circ} \mathrm{C}$. What is the rate law for the reaction?


1797. Solution -
The reaction will have the form:

$$
\text { rate }=k\left[\mathrm{NO}_{2}\right]^{m}[\mathrm{CO}]^{n}
$$

The reaction is second order in $\mathrm{NO}_{2}$; thus $m=2$. The reaction is zero order in CO; thus $n=0$. The rate law is:

$$
\text { rate }=k\left[\mathrm{NO}_{2}\right]^{2}[\mathrm{CO}]^{0}=k\left[\mathrm{NO}_{2}\right]^{2}
$$

Remember that a number raised to the zero power is equal to 1 , thus $[\mathrm{CO}]^{0}=1$, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of $\mathrm{NO}_{2}$. A later chapter section on reaction mechanisms will explain how a reactant's concentration can have no effect on a reaction rate despite being involved in the reaction.


1798. Check Your Learning -
The rate law for the reaction:

$$
\mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)
$$

has been determined to be rate $=k[N O]^{2}\left[\mathrm{H}_{2}\right]$. What are the orders with respect to each reactant, and what is the
overall order of the reaction?


1799. Answer: -
order in $\mathrm{NO}=2$; order in $\mathrm{H}_{2}=1$; overall order $=3$


1800. Check Your Learning -
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)$ and ethyl acetate $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCOCH}_{3}\right)$ as a sample reaction before studying the chemical reactions that produce biodiesel:

$$
\mathrm{CH}_{3} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCOCH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{OCOCH}_{3}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}
$$

The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:

$$
\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]
$$

What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?


1801. Answer: -
order in $\mathrm{CH}_{3} \mathrm{OH}=1$; order in $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCOCH}_{3}=0$; overall order $=1$

A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.


1803. Determining a Rate Law from Initial Rates -
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure 17.8). One such reaction is the combination of nitric oxide, NO, with ozone, $\mathrm{O}_{3}$ :

FIGURE 17.8 A contour map showing stratospheric ozone concentration and the "ozone hole" that occurs over Antarctica during its spring months. (credit: modification of work by NASA)

$$
\mathrm{NO}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})
$$

This reaction has been studied in the laboratory, and the following rate data were determined at $25^{\circ} \mathrm{C}$.

| Trial | $\mathrm{NO}$ | $\left\mathrm{O}_{3}\right$ |
| :--- | :--- | :--- |
| 1 | $\frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t}\left(\mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\right)$ |  |
| $1.00 \times 10^{-6}$ | $3.00 \times 10^{-6}$ | $6.60 \times 10^{-5}$ |
| 2 | $1.00 \times 10^{-6}$ | $6.00 \times 10^{-6}$ |
| 3 | $1.00 \times 10^{-6}$ | $9.00 \times 10^{-6}$ |
| 4 | $2.00 \times 10^{-6}$ | $9.00 \times 10^{-6}$ |
| 5 | $3.00 \times 10^{-6}$ | $9.00 \times 10^{-6}$ |

Determine the rate law and the rate constant for the reaction at $25^{\circ} \mathrm{C}$.


1804. Solution -
The rate law will have the form:

$$
\text { rate }=k[\mathrm{NO}]^{m}\left[\mathrm{O}_{3}\right]^{n}
$$

Determine the values of $m, n$, and $k$ from the experimental data using the following three-part process:
Step 1.
Determine the value of m from the data in which [ NO ] varies and $\left[\mathrm{O}_{3}\right.$ ] is constant. In the last three experiments, [ NO ] varies while $\left[\mathrm{O}_{3}\right]$ remains constant. When [NO] doubles from trial 3 to 4 , the rate doubles, and when [ NO ] triples from trial 3 to 5 , the rate also triples. Thus, the rate is also directly proportional to [NO], and $m$ in the rate law is equal to 1 .

Step 2.
Determine the value of n from data in which $\left[\mathrm{O}_{3}\right]$ varies and [NO] is constant. In the first three experiments, [ NO ] is constant and $\left[\mathrm{O}_{3}\right]$ varies. The reaction rate changes in direct proportion to the change in [ $\mathrm{O}_{3}$ ]. When $\left[\mathrm{O}_{3}\right]$ doubles from trial 1 to 2, the rate doubles; when $\left[\mathrm{O}_{3}\right]$ triples from trial 1 to 3 , the rate increases also triples. Thus, the rate is directly proportional to $\left[\mathrm{O}_{3}\right]$, and $n$ is equal to 1.The rate law is thus:

$$
\text { rate }=k[\mathrm{NO}]^{1}\left[\mathrm{O}_{3}\right]^{1}=k[\mathrm{NO}]\left[\mathrm{O}_{3}\right]
$$

Step 3.
Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:


1804. Solution -
Step 3.
Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:

$$
\begin{aligned}
& k=\frac{\text { rate }}{[\mathrm{NO}]\left[\mathrm{O}_{3}\right]} \\
&=\frac{6.60 \times 10^{-5} \overline{\mathrm{molL}}^{-1}}{\mathrm{~s}}-1 \\
&\left(1.00 \times 10^{-6} \mathrm{molL}^{-1}\right)\left(3.00 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\right)
\end{aligned}
$$


1805. Check Your Learning -
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:

$$
\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{CO}(\mathrm{~g})
$$

Determine the rate law and the rate constant for the reaction from the following experimental data:

| Trial | $\left\mathrm{CH}_{3} \mathrm{CHO}\right$ | $-\frac{\Delta\left[\mathrm{CH}_{3} \mathrm{CHO}\right]}{\Delta t}\left(\mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\right)$ |
| :--- | :--- | :--- |
| 1 | $1.75 \times 10^{-3}$ | $2.06 \times 10^{-11}$ |
| 2 | $3.50 \times 10^{-3}$ | $8.24 \times 10^{-11}$ |
| 3 | $7.00 \times 10^{-3}$ | $3.30 \times 10^{-10}$ |


1806. Answer: -
rate $=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{2}$ with $k=6.73 \times 10^{-6} \mathrm{~L} / \mathrm{mol} / \mathrm{s}$


1808. Determining Rate Laws from Initial Rates -
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:

$$
2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)
$$

| Trial | $\mathrm{NO}$ | $\left\mathrm{Cl}_{2}\right$ | $-\frac{\Delta[\mathrm{NO}]}{\Delta t}\left(\mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\right)$ |
| :--- | :--- | :--- | :--- |
| 1 | 0.10 | 0.10 | 0.00300 |
| 2 | 0.10 | 0.15 | 0.00450 |
| 3 | 0.15 | 0.10 | 0.00675 |


1809. Solution -
The rate law for this reaction will have the form:

$$
\text { rate }=k[\mathrm{NO}]^{m}\left[\mathrm{Cl}_{2}\right]^{n}
$$

As in Example 17.4, approach this problem in a stepwise fashion, determining the values of $m$ and $n$ from the experimental data and then using these values to determine the value of $k$. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of $m$ and $n$ :


1810. Step 1. -
Determine the value of m from the data in which $[\mathrm{NO}]$ varies and $\left[\mathrm{Cl}_{2}\right]$ is constant. Write the ratios with the subscripts $x$ and $y$ to indicate data from two different trials:

$$
\frac{\operatorname{rate}_{x}}{\operatorname{rate}_{y}}=\frac{k[\mathrm{NO}]_{x}^{m}\left[\mathrm{Cl}_{2}\right]_{x}^{n}}{k[\mathrm{NO}]_{y}^{m}\left[\mathrm{Cl}_{2}\right]_{y}^{n}}
$$

Using the third trial and the first trial, in which $\left[\mathrm{Cl}_{2}\right]$ does not vary, gives:

$$
\frac{\text { rate } 3}{\text { rate } 1}=\frac{0.00675}{0.00300}=\frac{k(0.15)^{m}(0.10)^{n}}{k(0.10)^{m}(0.10)^{n}}
$$

Canceling equivalent terms in the numerator and denominator leaves:

$$
\frac{0.00675}{0.00300}=\frac{(0.15)^{m}}{(0.10)^{m}}
$$

which simplifies to:

$$
2.25=(1.5)^{m}
$$

Use logarithms to determine the value of the exponent $m$ :

$$
\begin{aligned}
\ln (2.25) & =m \ln (1.5) \\
\frac{\ln (2.25)}{\ln (1.5)} & =m \\
2 & =m
\end{aligned}
$$

Confirm the result

$$
1.5^{2}=2.25
$$

Step 2.
Determine the value of n from data in which [ $\mathrm{Cl}_{2}$ ] varies and [ NO ] is constant.

$$
\frac{\text { rate } 2}{\text { rate } 1}=\frac{0.00450}{0.00300}=\frac{k(0.10)^{m}(0.15)^{n}}{k(0.10)^{m}(0.10)^{n}}
$$


1810. Step 1. -
Cancelation gives:

$$
\frac{0.0045}{0.0030}=\frac{(0.15)^{n}}{(0.10)^{n}}
$$

which simplifies to:

$$
1.5=(1.5)^{n}
$$

Thus $n$ must be 1 , and the form of the rate law is:

$$
\text { rate }=k[\mathrm{NO}]^{m}\left[\mathrm{Cl}_{2}\right]^{n}=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]
$$

Step 3.
Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are $\mathrm{mol} / \mathrm{L} / \mathrm{s}$. The units for $k$ are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are $\mathrm{mol}^{3} / \mathrm{L}^{3}$. The units for $k$ should be $\mathrm{mol}^{-2} \mathrm{~L}^{2} / \mathrm{s}$ so that the rate is in terms of $\mathrm{mol} / \mathrm{L} / \mathrm{s}$.

To determine the value of $k$ once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for $k$ :

$$
\begin{aligned}
0.00300 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} & =k\left(0.10 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}\left(0.10 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{1} \\
k & =3.0 \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}
\end{aligned}
$$


1811. Check Your Learning -
Use the provided initial rate data to derive the rate law for the reaction whose equation is:

| $\mathrm{OCl}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{OI}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ |  |  |  |
| :--- | :--- | :--- | :--- |
| Trial | $\left\mathrm{OCl}^{-}\right$ | $\left\mathrm{I}^{-}\right$ | Initial Rate (mol/L/s) |
| 1 | 0.0040 | 0.0020 | 0.00184 |
| 2 | 0.0020 | 0.0040 | 0.00092 |
| 3 | 0.0020 | 0.0020 | 0.00046 |

Determine the rate law expression and the value of the rate constant $k$ with appropriate units for this reaction.


1812. Answer: -
$$
\begin{aligned}
& \frac{\text { rate } 2}{\text { rate } 3}=\frac{0.00092}{0.00046}=\frac{k(0.0020)^{x}(0.0040)^{y}}{k(0.0020)^{x}(0.0020)^{y}} \\
& 2.00=2.00^{y} \\
& y=1 \\
& \frac{\text { rate } 1}{\text { rate } 2}=\frac{0.00184}{0.00092}=\frac{k(0.0040)^{x}(0.0020)^{y}}{k(0.0020)^{x}(0.0040)^{y}} \\
& 2.00=\frac{2^{x}}{2^{y}} \\
& 2.00=\frac{2^{x}}{2^{1}} \\
& 4.00=2^{x} \\
& x=2
\end{aligned}
$$

Substituting the concentration data from trial 1 and solving for $k$ yields:

$$
\begin{aligned}
\text { rate } & =k\left[\mathrm{OCl}^{-}\right]^{2}\left[\mathrm{I}^{-}\right]^{1} \\
0.00184 & =k(0.0040)^{2}(0.0020)^{1} \\
k & =5.75 \times 10^{4} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}
\end{aligned}
$$


1813. Reaction Order and Rate Constant Units -
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the
chemical equation for the reaction. This is merely a coincidence and very often not the case.
Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:

$$
\begin{array}{lc}
\mathrm{NO}_{2}+\mathrm{CO} \longrightarrow \mathrm{NO}+\mathrm{CO}_{2} & \text { rate }=k\left[\mathrm{NO}_{2}\right]^{2} \\
\mathrm{CH}_{3} \mathrm{CHO} \longrightarrow \mathrm{CH}_{4}+\mathrm{CO} & \text { rate }=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{2} \\
2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2} & \text { rate }=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\
2 \mathrm{NO}_{2}+\mathrm{F}_{2} \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F} & \text { rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right] \\
2 \mathrm{NO}_{2} \mathrm{Cl} \longrightarrow 2 \mathrm{NO}_{2}+\mathrm{Cl}_{2} & \text { rate }=k\left[\mathrm{NO}_{2} \mathrm{Cl}\right]
\end{array}
$$

It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.


1813. Reaction Order and Rate Constant Units -
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.

The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example 17.4 was determined to be $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$. For the third-order reaction described in Example 17.5, the unit for $k$ was derived to be $\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$. Dimensional analysis requires the rate constant unit for a reaction whose overall order is $x$ to be $\mathrm{L}^{x-1} \mathrm{~mol}^{1-x} \mathrm{~s}^{-1}$. Table 17.1 summarizes the rate constant units for common reaction orders.

Rate Constant Units for Common Reaction Orders

| Overall Reaction Order $(\boldsymbol{x})$ | Rate Constant Unit ( $\mathrm{L}^{\boldsymbol{x - 1}} \mathrm{mol}^{1-x} \mathrm{~s}^{-1}$ ) |
| :--- | :--- |
| 0 (zero) | $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ |
| 1 (first) | $\mathrm{s}^{-1}$ |
| 2 (second) | $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$ |
| 3 (third) | $\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$ |

TABLE 17.1

Note that the units in this table were derived using specific units for concentration ( $\mathrm{mol} / \mathrm{L}$ ) and time ( s ), though any valid units for these two properties may be used.


1813. Reaction Order and Rate Constant Units - 1813.1. Integrated Rate Laws
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Explain the form and function of an integrated rate law
- Perform integrated rate law calculations for zero-, first-, and second-order reactions
- Define half-life and carry out related calculations
- Identify the order of a reaction from concentration/time data

The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.

Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first, second-, and zero-order reactions.


1814. First-Order Reactions -
Integration of the rate law for a simple first-order reaction (rate $=k[A]$ ) results in an equation describing how the reactant concentration varies with time:

$$
[A]_{t}=[A]_{0} e^{-k t}
$$

where $[A] t$ is the concentration of $A$ at any time $t,[A]_{0}$ is the initial concentration of $A$, and $k$ is the first-order rate constant.

For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:

$$
\ln \left(\frac{[A]_{t}}{[A]_{0}}\right)=-k t \quad \text { or } \quad \ln \left(\frac{[A]_{0}}{[A]_{t}}\right)=k t
$$

and a format showing a linear dependence of concentration in time:

$$
\ln [A]_{t}=\ln [A]_{0}-k t
$$


1816. The Integrated Rate Law for a First-Order Reaction -
The rate constant for the first-order decomposition of cyclobutane, $\mathrm{C}_{4} \mathrm{H}_{8}$ at $500^{\circ} \mathrm{C}$ is $9.2 \times 10^{-3} \mathrm{~s}^{-1}$ :

$$
\mathrm{C}_{4} \mathrm{H}_{8} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}
$$

How long will it take for $80.0 \%$ of a sample of $\mathrm{C}_{4} \mathrm{H}_{8}$ to decompose?


1817. Solution -
Since the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:

$$
\ln \left(\frac{[A]_{0}}{[A]_{t}}\right)=k t
$$

The initial concentration of $\mathrm{C}_{4} \mathrm{H}_{8},[A]_{0}$, is not provided, but the provision that $80.0 \%$ of the sample has decomposed is enough information to solve this problem. Let $x$ be the initial concentration, in which case the concentration after $80.0 \%$ decomposition is $20.0 \%$ of $x$ or $0.200 x$. Rearranging the rate law to isolate $t$ and substituting the provided quantities yields:

$$
\begin{aligned}
t & =\ln \frac{[x]}{[0.200 x]} \times \frac{1}{k} \\
& =\ln 5 \times \frac{1}{9.2 \times 10^{-3} \mathrm{~s}^{-1}} \\
& =1.609 \times \frac{1}{9.2 \times 10^{-3} \mathrm{~s}^{-1}} \\
& =1.7 \times 10^{2} \mathrm{~s}
\end{aligned}
$$


1818. Check Your Learning -
Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:

$$
\mathrm{I}-131 \longrightarrow \mathrm{Xe}-131+\text { electron }
$$

The decay is first-order with a rate constant of $0.138 \mathrm{~d}^{-1}$. How many days will it take for $90 \%$ of the iodine -131 in a 0.500 M solution of this substance to decay to Xe-131?


1819. Answer: -
16.7 days

In the next example exercise, a linear format for the integrated rate law will be convenient:

$$
\begin{aligned}
\ln [A]_{t} & =(-k)(t)+\ln [A]_{0} \\
y & =m x+b
\end{aligned}
$$

A plot of $\ln [A]_{t}$ versus $t$ for a first-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $\ln [A]_{0}$. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.


1821. Graphical Determination of Reaction Order and Rate Constant -
Show that the data in Figure 17.2 can be represented by a first-order rate law by graphing $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ versus time. Determine the rate constant for the decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ from these data.


1822. Solution -
The data from Figure 17.2 are tabulated below, and a plot of $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ is shown in Figure 17.9.

| Trial |  | Time (h) | $\left\mathrm{H}_{2} \mathrm{O}_{2}\right$ |
| :--- | :--- | :--- | :--- |
| 1 | 0.00 | 1.000 | 0.000 |
| 2 | 6.00 | 0.500 | -0.693 |
| 3 | 12.00 | 0.250 | -1.386 |
| 4 | 18.00 | 0.125 | -2.079 |
| 5 | 24.00 | 0.0625 | -2.772 |

FIGURE 17.9 A linear relationship between $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ and time suggests the decomposition of hydrogen peroxide is a first-order reaction.

The plot of $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ versus time is linear, indicating that the reaction may be described by a first-order rate law.

According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot's slope.

$$
\text { slope }=\frac{\text { change in } y}{\text { change in } x}=\frac{\Delta y}{\Delta x}=\frac{\Delta \ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}
$$

The slope of this line may be derived from two values of $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ at different values of $t$ (one near each end of the line is preferable). For example, the value of $\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ when $t$ is 0.00 h is 0.000 ; the value when $t=24.00 \mathrm{~h}$ is $-2.772$


1822. Solution -
$$
\begin{aligned}
\text { slope } & =\frac{-2.772-0.000}{24.00-0.00 \mathrm{~h}} \\
& =\frac{-2.772}{24.00 \mathrm{~h}} \\
& =-0.116 \mathrm{~h}^{-1} \\
k & =- \text { slope }=-\left(-0.116 \mathrm{~h}^{-1}\right)=0.116 \mathrm{~h}^{-1}
\end{aligned}
$$


1823. Check Your Learning -
Graph the following data to determine whether the reaction $A \longrightarrow B+C$ is first order.

| Trial |  | Time (s) |
| :--- | :--- | :--- |
| $A]$ |  |  |
| 1 | 4.0 | 0.220 |
| 2 | 8.0 | 0.144 |
| 3 | 12.0 | 0.110 |
| 4 | 16.0 | 0.088 |
| 5 | 20.0 | 0.074 |


1824. Answer: -
The plot of $\ln [A]_{t}$ vs. $t$ is not linear, indicating the reaction is not first order:


1825. Second-Order Reactions -
The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:

$$
\text { rate }=k[A]^{2}
$$

For these second-order reactions, the integrated rate law is:

$$
\frac{1}{[A]_{t}}=k t+\frac{1}{[A]_{0}}
$$

where the terms in the equation have their usual meanings as defined earlier.


1826. EXAMPLE 17.8 -
The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas $\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)$ to yield $\mathrm{C}_{8} \mathrm{H}_{12}$ gas is described by the equation:

$$
2 \mathrm{C}_{4} \mathrm{H}_{6}(g) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(g)
$$

This "dimerization" reaction is second order with a rate constant equal to $5.76 \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~min}^{-1}$ under certain conditions. If the initial concentration of butadiene is 0.200 M , what is the concentration after 10.0 min?