Chapter 20 Homework Transparency Pack Figure From Transparency Pack

Chapter 20. Electrochemistry

Media Resources

Figures and Tables in Transparency Pack: Section:

Figure 20.5 A Voltaic Cell that Uses a Salt Bridge 20.3 Voltaic Cells

to Complete the Electrical Circuit

Figure 20.9 A Voltaic Cell Using a Standard 20.4 Cell Potentials Under Standard Conditions

Hydrogen Electrode (SHE)

Activities: Section:

Balancing Redox Equations in Acid 20.2 Balancing Redox Equations

Balancing Redox Equations in Base 20.2 Balancing Redox Equations

Nernst Equation 20.6 Cell Potentials Under Nonstandard

Conditions

Batteries 20.7 Batteries and Fuel Cells

Electrolysis 20.9 Electrolysis

Electrolysis Calculation Example 20.9 Electrolysis

Prevention of Corrosion 20.8 Corrosion

Animations: Section:

Oxidation-Reduction Reactions—Part I 20.1 Oxidation States and Oxidation Reduction

Movies: Section:

Redox Chemistry of Iron and Copper 20.3 Voltaic Cells

Formation of Silver Crystals 20.3 Voltaic Cells

Electroplating 20.9 Electrolysis

Chapter 20

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3-D Models: Section:

Hydronium Ion 20.2 Balancing Redox Equations

Hydrogen 20.7 Batteries and Fuel Cells

Other Resources

Further Readings: Section:

Redox Balancing without Puzzling 20.2 Balancing Redox Equations

Ask the Historian: Balancing Redox Equations 20.2 Balancing Redox Equations

Common Student Misconceptions in 20.3 Voltaic Cells

Electrochemistry: Galvanic, Electrolytic

and Concentration Cells”

Live Demonstrations: Section:

Visible Oxidation-Reduction in Electrochemical 20.3 Voltaic Cells

Cells

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Chapter 20. Electrochemistry

Common Student Misconceptions

• Students often think that oxidation must necessarily mean adding oxygen.

• Students often have trouble balancing redox equations.

• Students often think that pure polar solvents, such as water, conduct electricity.

• Students commonly think that electrons flow through the salt-bridge (or the porous barrier) and

through solutions.

Teaching Tips

• Students should be encouraged to review section 4.4.

Lecture Outline

20.1 Oxidation States and Oxidation-Reduction Reactions

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• Chemical reactions in which the oxidation state of one or more substances change are called

oxidation-reduction reactions (redox reactions).

• Recall:

• Oxidation involves loss of electrons (OIL).

• Electrochemistry is the branch of chemistry that deals with relationships between electricity and

chemical reactions.

• Consider the spontaneous reaction that occurs when Zn is added to HCl.

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

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• The oxidation number of H has decreased from +1 to 0.

• Therefore, Zn is oxidized to Zn2+, while H+ is reduced to H2.

• Note that the reducing agent is oxidized and the oxidizing agent is reduced.

FORWARD REFERENCES

• Oxidation numbers of transition metals will be covered again in Chapters 23 (section 23.1)

for metal complexes.

20.2 Balancing Redox Equations

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• Recall the law of conservation of mass: The amount of each element present at the beginning of the

Half-Reactions

• Half-reactions are a convenient way of separating oxidation and reduction reactions.

• Consider the reaction: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)

Balancing Equations by the Method of Half-Reactions

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• Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep

purple).

• MnO4– is reduced to Mn2+ (pale pink), while the C2O42– is oxidized to CO2.

• The equivalence point is indicated by the presence of a pale pink color.

• If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

• What is the balanced chemical equation for this reaction?

“Hydronium Ion” 3–D Model from Instructor’s Resource CD/DVD

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291

• For the permanganate half-reaction, note that there is a charge of 7+ on the left and

2+ on the right.

• Therefore, 5 electrons need to be added to the left:

5e– + 8H+(aq) + MnO4– (aq) → Mn2+(aq) + 4H2O(l)

• In the oxalate half-reaction, there is a 2– charge on the left and a 0 charge on the

right, so we need to add two electrons to the products:

C2O42– (aq) → 2CO2(g) + 2e–

• Multiply each half-reaction to make the number of electrons equal.

Balancing Equations for Reactions Occurring in Basic Solution

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• The same method as above is used, but OH– is added to “neutralize” the H+ used.

• The equation must again be simplified by canceling like terms on both sides of the equation.

FORWARD REFERENCES

• Redox reactions involving transition metals will be covered in Chapter 23 (section 23.1).

20.3 Voltaic Cells

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“Balancing Redox Equations in Base” Activity from Instructor’s Resource CD/DVD

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“Common Student Misconceptions in Electrochemistry: Galvanic, Electrolytic and Concentration Cells”

from Further Readings

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“Alleviating the Common Confusion Caused by Polarity in Electrochemistry” from Further Readings

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“Voltaic Cells I: The Copper–Zinc Cell” Animation from Instructor’s Resource CD/DVD

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“Visible Oxidation-Reduction in Electrochemical Cells” from Live Demonstrations

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“Redox Chemistry of Iron and Copper” Movie from Instructor’s Resource CD/DVD

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Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

• Zn is spontaneously oxidized to Zn2+ by Cu2+.

• The Cu2+ is spontaneously reduced to Cu0 by Zn.

• The entire process is spontaneous.

• Reduction takes place at the cathode.

• As oxidation occurs, Zn is converted to Zn2+ and 2e–.

• The electrons flow toward the cathode where they are used in the reduction reaction.

• We expect the Zn electrode to lose mass and the Cu electrode to gain mass.

• Electrons flow from the anode to the cathode.

• Therefore, the anode is negative and the cathode is positive.

• Electrons cannot flow through the solution; they have to be transported through an external wire.

20.4 Cell Potentials Under Standard Conditions

• The flow of electrons from anode to cathode is spontaneous.

• What is the “driving force”?

• Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than

the anode.

• Potential difference is the difference in electrical potential.

• The potential difference is measured in volts.

• One volt (V) is the potential difference required to impart one joule (J) of energy to a charge of

one coulomb (C):

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Standard Reduction Potentials

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• We can conveniently tabulate electrochemical data.

• Standard reduction potentials, E°red are measured relative to a standard.

• The emf of a cell can be calculated from standard reduction potentials:

E°cell = E°red(cathode) – E°red(anode)

0.76 V = 0 V – E°red(anode).

• Therefore, E°red(anode) = – 0.76 V.

• Standard reduction potentials must be written as reduction reactions:

Zn2+(aq, 1M) + 2e– → Zn(s) E°red = –0.76 V.

• Since E°red = –0.76 V, we conclude that the reduction of Zn2+ in the presence of the SHE is not

spontaneous.

• However, the oxidation of Zn with the SHE is spontaneous.

• The standard reduction potential is an intensive property.

Strengths of Oxidizing and Reducing Agents

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• Consider a table of standard reduction potentials.

• We can use this table to determine the relative strengths of reducing (and oxidizing) agents.

• The more positive the E°red, the stronger the oxidizing agent (written in the table as a reactant).

• The more negative the E°red, the stronger the reducing agent (written as a product in the table).

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“Voltaic Cells II: The Zinc–Hydrogen Cell” Animation from Instructor’s Resource CD/DVD

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Table 20.1 from Transparency Pack

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Figure 20.9 from Transparency Pack

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“Standard Reduction Potential” Animation from Instructor’s Resource CD/DVD

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Figure 20.12 from Transparency Pack

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20.5 Free Energy and Redox Reactions

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• For any electrochemical process

E° = E°red(reduction process) – E°red(oxidation process).

• A positive E° indicates a spontaneous process (galvanic cell).

EMF, Free Energy and the Equilibrium Constant

• We can show that:

G = −nFE

• where G is the change in free energy, n is the number of moles of electrons transferred, F is

Faraday’s constant, and E is the emf of the cell.

Electrical Work

• Free energy is a measure of the maximum amount of useful work that can be obtained from a system.

• We know:

G = wmax

• and: G = −nFE

• thus: wmax = −nFE

• If Ecell is positive, wmax will be negative.

• Work is done by the system on the surroundings.

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20.6 Cell Potentials Under Nonstandard Conditions

• A voltaic cell is functional until E = 0 at which point equilibrium has been reached.

• The cell is then “dead.”

• The point at which E = 0 is determined by the concentrations of the species involved in the redox

reaction.

The Nernst Equation

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• We can calculate the cell potential under nonstandard conditions.

• The Nernst equation can be simplified by collecting all the constants together and using a temperature

of 298 K:

Concentration Cells

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• A concentration cell is one whose emf is generated solely because of a concentration difference.

• Example: Consider a cell with two compartments, each with a Ni(s) electrode but with different

concentrations of Ni2+(aq).

• One cell has [Ni2+] = 1.0 M and the other has [Ni2+] = 0.001 M.

• Using the Nernst equation we can calculate a cell potential of +0.0888 V for this concentration cell.

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“Nernst Equation” Activity from Instructor’s Resource CD/DVD

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“Using the Biological Cell in Teaching Electrochemistry” from Further Readings

Q

nF

RT

EE log

303.2

−=

Q

n

EE log

0592.0

−=

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20.7 Batteries and Fuel Cells

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• A battery is a portable, self-contained electrochemical power source consisting of one or more

voltaic cells.

• Primary cells: cannot be recharged.

• Secondary cells: can be recharged from an external power source after its voltage has dropped.

Lead-Acid Battery

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• A 12 V car battery consists of six cathode/anode pairs each producing 2 V.

• Cathode: PbO2 on a metal grid in sulfuric acid:

• Wood or glass-fiber spacers are used to prevent the electrodes from touching.

• An advantage of these cells is that they can be recharged.

• An external source of energy is used to reverse the process.

Alkaline Battery

• The most common nonrechargeable battery is the alkaline battery.

• Powdered zinc metal is immobilized in a gel in contact with a concentrated solution of KOH.

Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries

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• A common rechargeable battery is the nickel-cadmium (NiCad) battery.

• The reaction at the cathode is:

2NiO(OH)(s) + 2H2O(l) + 2e– → 2Ni(OH)2(s) +2OH−(aq)

• The reaction at the anode is:

Cd(s) + 2OH– (aq) → Cd(OH)2(s) + 2e–

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• NiMH batteries (nickel-metal-hydride).

• Li-ion batteries (lithium-ion batteries).

Hydrogen Fuel Cells

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• Direct production of electricity from fuels occurs in a fuel cell.

• An example is a hydrogen fuel cell.

20.8 Corrosion

• An example of an undesirable redox reaction is the corrosion of metals.

• Metal is attacked by a substance in the environment and converted to an unwanted compound.

Corrosion of Iron (Rusting)

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• Consider the rusting of iron:

• Since E°red(Fe2+) < E°red(O2), iron can be oxidized by oxygen.

Preventing Corrosion of Iron

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• Corrosion can be prevented by coating the iron with paint or another metal.

• This prevents oxygen and water from reacting at the surface of the iron.

• Galvanized iron is coated with a thin layer of zinc.

• Zinc protects the iron since Zn is the anode and Fe is the cathode:

Zn2+(aq) +2e– → Zn(s) E°red = −0.76 V

Fe2+(aq) + 2e– → Fe(s) E°red = −0.44 V

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“Vehicle of Change” from Further Readings

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“Hydrogen” 3–D Model from Instructor’s Resource CD/DVD

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“Evaluation of Corrosion Susceptibility of a Metal: Student Corrosion Experiment II” from Further

Readings

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20.9 Electrolysis

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• Electrolysis reactions are nonspontaneous reactions that require an external current in order to force

the reaction to proceed.

• They take place in electrolytic cells.

• In voltaic and electrolytic cells, reduction occurs at the cathode and oxidation occurs at the anode.

• Do we get the same products if we electrolyze an aqueous solution of the salt?

• Water complicates the issue!

• Example: Consider the electrolysis of NaF(aq):

Na+ (aq) + e– → Na(s) E°red = −2.71 V

2H2O(l) + 2e– → H2(g) + 2OH−(aq) E°red = −0.83 V

• Thus, water is more easily reduced than the sodium ion.

2F– (aq) → F2(g) + 2e– E°red = +2.87 V

2H2O(l) → O2(g) + 4H+(aq) + 4e– E°red = +1.23 V

Quantitative Aspects of Electrolysis

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• We want to know how much material we obtain with electrolysis.

• Consider the reduction of Cu2+ to Cu.

Cu2+(aq) + 2e– →Cu(s).

• 2 moles of electrons will plate 1 mol of Cu.

• The charge of one mol of electrons is 96,500 C (1 F).

• A coulomb is the amount of charge passing a point in one second when the current is one ampere.

• The amount of Cu can be calculated from the current (amperes) and time required to plate.

Coulombs = amperes



seconds

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