1 Chapter Eighteen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Electrochemistry Chapter Eighteen

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Chapter EighteenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Prentice Hall © 2005Hall © 2005

Electrochemistry

Chapter EighteenChapter Eighteen

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Oxidation–Reduction:The Transfer of Electrons

Electrons from copper metal are

transferred to silver ions.

Silver metal is formed, and the solution turns blue from copper(II)

ions formed.

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• Review Chapter Four for details of oxidation–reduction reactions.

• In any oxidation–reduction reaction, there are two half-reactions:

– Oxidation: a species loses electrons to another species.

– Reduction: a species gains electrons from another species.

Half-Reactions

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1. Separate a redox equation into two half-equations, one for oxidation and one for reduction.

2. Balance the number of atoms of each element in each half-equation. Usually we balance O and H atoms last.

3. Balance each half-reaction for charge by adding electrons to the left in the reduction half-equation and to the right in the oxidation half-equation.

4. Adjust the coefficients in the half-equations so that the same number of electrons appears in each half-equation.

5. Add together the two adjusted half-equations to obtain an overall redox equation.

6. Simplify the overall redox equation as necessary.

The Half-Reaction Method ofBalancing Redox Equations

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• Redox reactions in acidic solution and in basic solution may be very different from one another.

• If acidic solution is specified, we must add H2O and/or H+ as needed when we balance the number of atoms.

• If basic solution is specified, the final equation may have OH– and/or water molecules in it.

• A simple way to balance an equation in basic solution:– Balance the equation as though it were in acidic solution.

– Add as many OH– ions to each side as there are H+ ions in the equation.

– Combine the H+ and OH– ions to give water molecules on one side, and simplify the equation as necessary.

Redox Reactions in Acidic and in Basic Solution

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Example 18.1Permanganate ion, MnO4

–, is used in the laboratory as an oxidizing agent; thiosulfate ion, S2O3

2–, is used as a reducing agent. Write a balanced equation for the reaction of these ions in an acidic aqueous solution to produce manganese(II) ion and sulfate ion.

Example 18.2In basic solution, Br2 disproportionates to bromide ions and bromate ions. Use the half-reaction method to balance the equation for this reaction:

Br2(l) Br–(aq) + BrO3–(aq)

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• A voltaic cell uses a spontaneous redox reaction to produce electricity.

• A half-cell consists of an electrode (strip of metal or other conductor) immersed in a solution of ions.

A Qualitative Description of Voltaic Cells

Both oxidation and reduction occur at the electrode surface, and equilibrium is reached.

This Zn2+ becomes a Zn atom.

This Zn atom leaves the surface to become a Zn2+ ion.

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• An electrochemical cell consists of two half-cells with the appropriate connections between electrodes and solutions.

• Two half-cells may be joined by a salt bridge that permits migration of ions, without completely mixing the solutions.

• The anode is the electrode at which oxidation occurs.

• The cathode is the electrode at which reduction occurs.

• In a voltaic cell, a spontaneous redox reaction occurs and current is generated.

• Cell potential (Ecell) is the potential difference in volts between anode and cathode.

• Ecell is the driving force that moves electrons and ions.

Important Electrochemical Terms

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A Zinc–Copper Voltaic Cell

… the electrons produced move

through the wire …

… to the Cu(s) electrode, where they are accepted

by Cu2+ ions to form more Cu(s).

Positive and negative ions move through the salt bridge to equalize

the charge.

Reaction: Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)Zn(s) is oxidized to Zn2+ ions, and …

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• A cell diagram is “shorthand” for an electrochemical cell.

• The anode is placed on the left side of the diagram.

• The cathode is placed on the right side.

• A single vertical line ( | ) represents a boundary between phases, such as between an electrode and a solution.

• A double vertical line ( || ) represents a salt bridge or porous barrier separating two half-cells.

Cell Diagrams

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Example 18.3Describe the half-reactions and the overall reaction that occur in the voltaic cell represented by the cell diagram:

Pt(s) | Fe2+(aq), Fe3+(aq) || Cl–(aq) | Cl2(g) | Pt(s)

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Standard Electrode Potentials

• Since an electrode represents only a half-reaction, it is not possible to measure the absolute potential of an electrode.

• The standard hydrogen electrode (SHE) provides a reference for measurement of other electrode potentials.

• The SHE is arbitrarily assigned a potential of 0.000 V.

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• The standard electrode potential, E°, is based on the tendency for reduction to occur at an electrode.

• E° for the standard hydrogen electrode is arbitrarily assigned a value of 0.000 V.

• All other values of E° are determined relative to the standard hydrogen electrode.

• The standard cell potential (E°cell) is the difference between E° of the cathode and E° of the anode.

• E°cell = E°(cathode) – E°(anode)

Standard Electrode Potentials (cont’d)

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Standard hydrogen electrode

The voltmeter reading and the

direction of electron flow tell us that …

… Cu2+ is more easily reduced than H+, by

0.340 volts.

Measuring the Standard Potentialof the Cu2+/Cu Electrode

Cu2+ + 2e Cu

E° = +0.340 V

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Measuring the Standard Potentialof the Zn2+/Zn Electrode

Standard hydrogen electrode

The voltmeter reading and the

direction of electron flow tell us that …

… Zn2+ is harder to reduce than H+, by

0.763 volts.

Zn2+ + 2e Zn E° = – 0.763

V

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F2 is the strongest oxidizing agent

Li is the strongest reducing

agent

F– is the weakest reducing agent

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• Standard electrode potentials and cell voltages are intensive properties; they do not depend on the total amounts of the species present.

• A “flashlight battery” (D-cell) and a “penlight battery” (AA cell) produce the same potential—1.5 volts.

• E° does depend on the particular species in the reaction (or half-reaction).

• As we shall learn later, cell and electrode potentials can depend on concentration of the species present.

Important Points about Electrode and Cell Potentials

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Example 18.4Determine E° for the reduction half-reaction

Ce4+(aq) + e– Ce3+(aq), given that the cell voltage for the voltaic cell

Co(s) | Co2+(1 M) || Ce4+(1 M), Ce3+(1 M) | Pt(s)

is E°cell = 1.887 V.

Example 18.5Balance the following oxidation–reduction equation, and determine E°cell for the reaction.

O2(g) + H+(aq) + I–(aq) H2O(l) + I2(s)

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• An electrochemical cell does work.

welec = nFEcell

n = number of electrons in the balanced equation

F = 96,485 coulombs per mole.

Electrode Potentials, Spontaneous Change, and Equilibrium

• The amount of electrical work is also equal to –G:

G = –nFEcell

• Under standard conditions:

G° = –nFE°cell

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• If Ecell is positive, the forward reaction is spontaneous.

• If Ecell is negative, the forward reaction is nonspontaneous (the reverse reaction is _____).

• If Ecell = 0, the system is at equilibrium.

• When a cell reaction is reversed, Ecell and G change signs.

Criteria for SpontaneousChange in Redox Reactions

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Example 18.6Will copper metal displace silver ion from aqueous solution? That is, does the reaction

Cu(s) + 2 Ag+(1 M) Cu2+(1 M) + 2 Ag(s)occur spontaneously from left to right?

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The Activity Series Revisited

• In the activity series of metals (Section 4.4), any metal in the series will displace a metal below it from a solution of that metal’s ions.

• Theoretical basis: The activity series lists metals in order of their standard potentials.

• Displacement of a metal from a solution of its ions by a metal higher in the series corresponds to a positive value of Ecell and a spontaneous reaction.

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Example 18.7

A Conceptual ExampleThe photograph in Figure 18.11 shows strips of copper and zinc joined together and then dipped in HCl(aq). Explain what happens. That is, what are the gas bubbles on the zinc and on the copper, and how did they get there?

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• Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.

G° = –nFE°cell

andG° = –RT ln Keq

therefore –RT ln Keq = –nFEocell

Equilibrium Constants in Redox Reactions

RT ln Keq RTE°cell = ––––––––– = –––– ln Keq nF nF

0.025693 VE°cell = –––––––– ln Keq n

R and F are constant, therefore at 298 K:

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Example 18.8Calculate the values of ΔG° and Keq at 25 °C for the reaction

Cu(s) + 2 Ag+(1 M) Cu2+(1 M) + 2 Ag(s)

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Thermodynamics, Equilibrium, and Electrochemistry: A Summary

From any one of the three quantities Keq, ΔG°, E°cell, we

can determine the others.

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Effect of Concentrations on Cell Voltage

• A nonstandard cell differs in potential from a standard cell (1 M concentrations, 1 atm partial pressures).

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From the previous relationships we can show that:

• This Nernst equation relates a cell voltage for nonstandard conditions, Ecell, to a standard cell voltage, E°cell, and to the concentrations of reactants and products expressed through the reaction quotient, Q.

• We can use the Nernst equation to find cell potential from concentrations, or we can measure Ecell and determine the concentration of a species in the cell.

At 25 °C, and converting to common logarithms:

Effect of Concentrations on Cell Voltage

RTEcell = E°cell – –––– ln Q nF

0.0592 VEcell = E°cell – ––––––– log Q n

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Example 18.9Calculate the expected voltmeter reading for the voltaic cell pictured in Figure 18.13.

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• A concentration cell has a potential determined solely by the difference in concentrations of solutes in equilibrium with identical electrodes.

Concentration Cells

• Since the two electrodes are identical, the standard cell potential is zero.

• To calculate the cell voltage for the nonstandard conditions, the Nernst equation is used.

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• Consider a concentration cell consisting of a standard hydrogen electrode, and a hydrogen electrode with unknown [H+].

• The Nernst equation for such a cell is:

Ecell(in volts) = 0.0592 pH

pH = Ecell /0.0592

• Thus, we can measure the pH of an unknown solution by making it part of such a concentration cell.

pH Measurement

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In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!

The pH Meter

A stable reference electrode and a glass-membrane

electrode are contained within a combination pH electrode.

The electrode is merely dipped into a solution, and

the potential difference between the electrodes is

displayed as pH.

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• We often call any device that stores chemical energy for later release as electricity a battery.

• Technically, a D, C, or AA “battery” is actually a single electrochemical cell.

• A battery consists of several cells joined together to produce higher current or higher voltage.

• A 9-volt “transistor” battery, an automobile battery, and a rechargable “battery pack” are all true batteries.

Batteries: Using Chemical Reactions to Make Electricity

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• A primary cell employs an irreversible chemical reaction.

• When the reactants inside the cell are largely used up, the cell is “dead.”

• The LeClanché cell or dry cell (right) is the “ordinary” type of flashlight “battery.”

The Dry Cell

• Alkaline cells cost more than the LeClanché cell but they have a longer shelf life and longer service life.

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• A lead–acid storage battery used in an automobile uses secondary cells; they are rechargeable.

• By connecting the cell to an external electric energy source, the discharge reaction is reversed.

The Lead–Acid Storage Battery

Cell reaction: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq) 2 PbSO4(s) + 2 H2O(l)

Charging reaction: 2 PbSO4(s) + 2 H2O(l) Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq)

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• The nickel–cadmium (NiCd) cell uses a cadmium anode and a cathode containing Ni(OH)2.

• A NiCd cell can be recharged hundreds of times. It produces 1.2 V (a Leclanché cell produces 1.5 V).

• Nickel–metal hydride cells (NiMH) use a metal alloy anode that contains hydrogen.

• In use, the anode releases the hydrogen, forming water. Like the NiCd cell, a NiMH cell produces 1.2 V.

• Lithium-ion cells use a lithium–cobalt oxide or lithium–manganese oxide material as the anode. The electrolyte is an organic solvent containing a dissolved lithium salt.

• Many modern laptop computers and cellular phones use lithium-ion cells.

Other Secondary Cells

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• In a fuel cell, the cell reaction is equivalent to a combustion reaction.

Fuel Cells

• The reactants are supplied externally; the cell does not “go dead” as long as the oxidizing and reducing agents are provided.

• Fuel cells are generally operated under nonstandard conditions and at temperatures considerably higher than 25 °C.

• H2–O2 fuel cells are seeing use in some automobiles.

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• In moist air, iron is oxidized to Fe2+, particularly at scratches, nicks, or dents. These areas are referred to anodic areas.

• Other regions of the iron serve as cathodic areas, where the electrons from the anodic areas reduce O2 to OH–.

• Iron(II) ions migrate from the anodic areas to the cathodic areas where they combine with the hydroxide ions.

• The iron(II) is then further oxidized to iron(III) by atmospheric oxygen.

• Common rust is Fe2O3 · x H2O.

Corrosion: Metal LossThrough Voltaic Cells

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Corrosion of an Iron Piling

One way to minimize rusting is to provide a

different anode reaction.

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• The simplest defense against corrosion of iron is to coat it (with paint or metal) to exclude oxygen from the surface.

• An entirely different approach is to protect iron with a more active metal.

• Galvanized iron has been coated with zinc.

• The zinc provides an alternative anode reaction; the zinc corrodes, protecting the iron.

Protecting Iron from Corrosion

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• In cathodic protection, an iron object to be protected is connected to a chunk of an active metal.

Cathodic Protection

• The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.

• Water heaters often employ a magnesium anode for cathodic protection.

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Example 18.10 A Conceptual ExampleIn Figure 18.20, iron nails are placed in a warm colloidal dispersion of agar in water. Phenolphthalein indicator and potassium ferricyanide, K3[Fe(CN)6], are also present in the dispersion. When the faintly yellow dispersion cools and stands for a few hours, it solidifies into a gel, and soon the colored regions begin to develop—pink along the middle part of the nail and blue at both ends. Explain what is happening in Figure 18.20a.

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• A voltaic cell corresponds to a spontaneous cell reaction.

• An electrolytic cell corresponds to a nonspontaneous cell reaction. The reaction is called electrolysis.

• The external source of electricity acts like an “electron pump.” It pulls electrons away from the anode, where oxidation takes place, and pushes them toward the cathode, where reduction takes place.

• The polarities of the electrodes are reversed from those in the voltaic cell, because now the external source controls the flow of electrons.

Electrolytic Cells

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Molten NaCl (around 1000 °C)

The nonspontaneous reaction is driven by external potential.

2 NaCl(l) 2 Na(l) + Cl2(g)

Electrolysis of MoltenSodium Chloride

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• In an electrolytic cell, all combinations of cathode and anode half-reactions give negative values of E°cell.

• The reaction most likely to occur is the one with the least negative value of E°cell (requires the lowest applied voltage from the external electricity source). HOWEVER …

• In many half-reactions, particularly those involving gases, various interactions at electrode surfaces make the required voltage for electrolysis higher than the voltage calculated from E° data.

• Overvoltage is the excess voltage above the voltage calculated from E° values that is required in electrolysis.

Predicting Electrolysis Reactions

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• The electrodes used in the electrolysis of NaCl(l) and NaCl(aq) are examples of inert electrodes.

• An inert electrode only provides a surface on which the exchange of electrons can occur.

• In contrast, some electrodes are active electrodes; they participate directly in a half-reaction.

• Many of the electrodes we saw in voltaic cells were active electrodes.

Inert Electrodesand Active Electrodes

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Example 18.11Predict the electrolysis reaction when AgNO3(aq) is electrolyzed (a) using platinum electrodes and (b) using a silver anode and a platinum cathode.

Example 18.12 A Conceptual ExampleTwo electrochemical cells are connected as shown. Specifically, the zinc electrodes of the two cells are joined, as are the copper electrodes. Will there be (a) no current, (b) a flow of electrons in the direction of the red arrows, or (c) a flow of electrons in the direction of the blue arrows?

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• The unit of electric charge is the coulomb (C), and the charge carried by one electron is –1.6022 x 10–19 C.

• Electric current, expressed in amperes (A), is the rate of flow of electric charge (C/s).

• To calculate the quantitative outcome of an electrolysis reaction:

1. Determine the amount of charge (C)—the product of current and time.

2. Convert the amount of charge to moles of electrons.

3. Use a half-equation to relate moles of electrons to moles of a reactant or a product.

4. Convert from moles of reactant or product to the final quantity desired.

Quantitative Electrolysis

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Example 18.13We can use electrolysis to determine the gold content of a sample. The sample is dissolved, and all the gold is converted to Au3+(aq), which is then reduced back to Au(s) on an electrode of known mass. The reduction half-reaction is Au3+(aq) + 3e– Au(s). What mass of gold will be deposited at the cathode in 1.00 hour by a current of 1.50 A?

Example 18.14 An Estimation ExampleWithout doing detailed calculations, determine which of the following solutions will yield the greatest mass of metal at a platinum cathode during electrolysis by a 1.50-A electric current for 30.2 min: CuSO4(aq), AgNO3(aq), or AuCl3(aq).

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Electrolysis plays an important role in the manufacture and purification of many substances, including chlorine, copper, silver, magnesium, aluminum, lead, zinc, sodium, fluorine, titanium, sodium hydroxide, hydrogen …

Producing Chemicalsby Electrolysis

Electrolysis of NaCl(aq) is used to produce H2, Cl2, and

NaOH, all of which have important industrial uses.

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• Electrolysis can be used to coat one metal onto another, a process called electroplating.

Electroplating

• Usually, the object to be electroplated, such as a spoon, is cast of an inexpensive metal. It is then coated with a thin layer of a more attractive, corrosion-resistant, and expensive metal, such as silver or gold.

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Cumulative ExampleAn important source of silver is the process in which lead is purified from lead-containing ores. The percent Ag in a 1.050-g lead sample is determined as follows: The sample is dissolved in nitric acid to produce 500.0 mL of a solution of Pb2+(aq) containing a small quantity of Ag+(aq). A strip of pure Ag(s) is immersed in the solution, and the potential difference between this half-cell as the cathode and a standard silver–silver chloride anode is found to be 0.281 V. What is the mass percent silver in the lead sample?

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1 Chapter Eighteen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Electrochemistry Chapter Eighteen