chapter 20 electrochemistry © 2012 pearson education, inc

Chapter 20

Electrochemistry

© 2012 Pearson Education, Inc.

Electrochemistry

Electrochemical Reactions(Red-Ox)

In electrochemical reactions, electrons are transferred from one species to another.


Electrochemistry

A. (a) Bubbling is caused by water vapor escaping and (b) the exothermic reaction in water produces steam.B. (a) Bubbling is caused by water vapor escaping and (b) the endothermic reaction in water produces steam.C. (a) Bubbling is caused by hydrogen gas forming and (b) the endothermic reaction in water produces steam.D. (a) Bubbling is caused by hydrogen gas forming and (b) the exothermic reaction in water produces steam.

Electrochemistry

Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.


Electrochemistry

Oxidation and Reduction

• A species is oxidized when it loses electrons.

(the oxidation number increases)– Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

• A species is reduced when it gains electrons.

(the oxidation number decreases)

– Here, each of the H+ gains an electron, and they combine to form H2.


Electrochemistry

Oxidation and Reduction

• The species reduced is the oxidizing agent.– H+ oxidizes Zn by taking electrons from it.

• The species oxidized is the reducing agent.– Zn reduces H+ by giving it electrons.


Electrochemistry

Assigning Oxidation Numbers1. Elements in their elemental form have an oxidation number of 0.

2. The oxidation number of a monatomic ion is the same as its charge.3. Nonmetals tend to have negative oxidation numbers, although some

are positive in certain compounds or ions.– Oxygen has an oxidation number of 2, except in the peroxide

ion, which has an oxidation number of 1.– Hydrogen is 1 when bonded to a metal, and +1 when bonded to

a nonmetal.– Fluorine always has an oxidation number of 1.– The other halogens have an oxidation number of 1 when they

are negative. They can have positive oxidation numbers, however; most notably in oxyanions.

4. The sum of the oxidation numbers in a neutral compound is 0.

5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.


Electrochemistry

A. N, +1; O, –1B. N, +2; O, –2C. N, +3; O, –2D. N, +4; O, –2

Electrochemistry

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents

The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity:Cd(s) + NiO2(s) + 2 H2O(l) Cd(OH)2(s) + Ni(OH)2(s)

Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent.

Practice ExerciseIdentify the oxidizing and reducing agents in the reaction

2 H2O(l) + Al(s) + MnO4(aq) Al(OH)4

(aq) + MnO2(s)

Electrochemistry

Balancing Oxidation-Reduction Equations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

This method involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half-reactions, and then combining them to attain the balanced equation for the overall reaction.


Electrochemistry

The Half-Reaction Method1. Assign oxidation numbers to determine what is oxidized and

what is reduced.

2. Write the oxidation and reduction half-reactions.

3. Balance each half-reaction.

a. Balance elements other than H and O.

b. Balance O by adding H2O.

c. Balance H by adding H+.

d. Balance charge by adding electrons.


Electrochemistry

The Half-Reaction Method

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.7. Make sure the equation is balanced according to charge.


Electrochemistry


Consider the reaction between MnO4 and C2O4

2:

MnO4(aq) + C2O4

2(aq) Mn2+(aq) + CO2(aq)


Electrochemistry

A. MnO4- is reduced and the reducing agent is H+.

B. MnO4- is reduced and the reducing agent is C2O4

2-.C. H+ is reduced and the reducing agent is C2O4

2-.D. H+ is reduced and the reducing agent is MnO4

-.

Electrochemistry


First, we assign oxidation numbers:

MnO4 + C2O4

2 Mn2+ + CO2

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.


Electrochemistry

Oxidation Half-ReactionC2O4

2 CO2

To balance the carbon, we add a coefficient of 2:

C2O42 2CO2

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side:

C2O42 2CO2 + 2e


Electrochemistry

Reduction Half-Reaction

MnO4 Mn2+

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side:

MnO4 Mn2+ + 4H2O

To balance the hydrogen, we add 8H+ to the left side:

8H+ + MnO4 Mn2+ + 4H2O


Electrochemistry

Reduction Half-Reaction

8H+ + MnO4 Mn2+ + 4H2O

To balance the charge, we add 5e to the left side:

5e + 8H+ + MnO4 Mn2+ + 4H2O


Electrochemistry

Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C2O42 2CO2 + 2e

5e + 8H+ + MnO4 Mn2+ + 4H2O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2:


Electrochemistry


5C2O42 10CO2 + 10e

10e + 16H+ + 2MnO4 2Mn2+ + 8H2O

When we add these together, we get:

10e + 16H+ + 2MnO4 + 5C2O4

2

2Mn2+ + 8H2O + 10CO2 +10e


Electrochemistry


10e + 16H+ + 2MnO4 + 5C2O4

2 2Mn2+ + 8H2O + 10CO2 +10e

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16H+ + 2MnO4 + 5C2O4

2 2Mn2+ + 8H2O + 10CO2


Electrochemistry

A. Yes, in each half-reactionB. Yes, on both sides of the balanced equation for a

redox reactionC. No, the electrons do not show in half-reactions and

therefore they do not appear in a balanced equation for a redox reaction.

D. No, the electrons cancel in adding half-reactions to form a balanced equation for a redox reaction.

Electrochemistry

Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents

Complete and balance this equation by the method of half-reactions:Cr2O7

2(aq) + Cl(aq) Cr3+(aq) + Cl2(g) (acidic solution)

Practice ExerciseComplete and balance the following equations using the method of half-reactions. Both reactions occur in

acidic solution.

(a) Cu(s) + NO3(aq) Cu2+(aq) + NO2(g)

(b) Mn2+(aq) + NaBiO3(s) Bi3+(aq) + MnO4(aq)

Electrochemistry

Balancing in Basic Solution

• If a reaction occurs in a basic solution, one can balance it as if it occurred in acid.

• Once the equation is balanced, add OH to each side to “neutralize” the H+ in the equation and create water in its place.

• If this produces water on both sides, you might have to subtract water from each side.


Electrochemistry

Sample Exercise 20.3 Balancing Redox Equations in Basic Solution

Complete and balance this equation for a redox reaction that takes place in basic solution:CN(aq) + MnO4

(aq) CNO(aq) + MnO2(s) (basic solution)

Practice ExerciseComplete and balance the following equations for oxidation-reduction reactions that occur in basic solution:

(a) NO2(aq) + Al(s) NH3(aq) + Al(OH)4

(aq)

(b)Cr(OH)3(s) + ClO(aq) CrO42(aq) + Cl2(g)

Electrochemistry

Voltaic Cells

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.


Electrochemistry

A. The blue color of Cu2+(aq) lessens as it is reduced to Cu(s).B. The blue color of Cu(s) lessens as it is reduced to Cu2+(aq).C. The blue color of Zn2+(aq) lessens as it is reduced to Zn(s).D. The blue color of Zn(s) lessens as it is reduced to Zn(s).

Electrochemistry

Voltaic Cells

• We can use that energy to do work if we make the electrons flow through an external device.

• We call such a setup a voltaic cell.


Electrochemistry

A. Cu, because it loses electrons in the chemical reaction.B. Cu, because it gains electrons in the chemical reaction.C. Zn, because it loses electrons in the chemical reaction.D. Zn, because it gains electrons in the chemical reaction.

Electrochemistry

Voltaic Cells• A typical cell looks like this.• The oxidation occurs at the

anode.• The reduction occurs at the

cathode.• Once even one electron flows

from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.


Electrochemistry

Voltaic Cells• Therefore, we use a salt

bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.– Cations move toward the

cathode.– Anions move toward the

anode.• In the cell, then, electrons

leave the anode and flow through the wire to the cathode.

• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.


Electrochemistry

Voltaic Cells

• As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.

• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.


Electrochemistry

A. Zn2+(aq) ions formed in the reaction completely migrate through the salt bridge to the right compartment to maintain electrical charge balance.

B. Anions from the right compartment migrate through the salt bridge to the left compartment to maintain electrical charge balance.

C. NO3-(aq) ions in the salt bridge migrate into the left compartment and Zn2+(aq) ions migrate into the salt

bridge to maintain electrical charge balance.D. Na+(aq) ions in the salt bridge migrate into the left compartment and Zn2+(aq) ions migrate into the salt

bridge to maintain electrical charge balance.

Electrochemistry

Sample Exercise 20.4 Describing a Voltaic Cell

The oxidation-reduction reactionCr2O7

2(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.

Practice ExerciseThe two half-reactions in a voltaic cell are

Zn(s) Zn2+(aq) + 2 e

ClO3(aq) + 6 H+(aq) + 6 e Cl(aq) + 3 H2O(l)

(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive?

Electrochemistry

Electromotive Force (emf)• Water only spontaneously

flows one way in a waterfall.• Likewise, electrons only

spontaneously flow one way in a redox reaction—from higher to lower potential energy.

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential and is designated Ecell.


Electrochemistry

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 JC


Electrochemistry

Standard Cell Potential

The cell potential under standard conditions is denoted

Zn(s) + Cu2+(aq, 1M) Zn2+(aq,1M) + Cu(s)

= +1.10 v


cellE

cellE

Electrochemistry

Standard Reduction Potentials

Reduction potentials for many electrodes have been measured and tabulated.


Electrochemistry

A. YesB. No

Electrochemistry

Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) Ered (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.


Electrochemistry

Standard Hydrogen Electrode

• Their values are referenced to a standard hydrogen electrode (SHE).

• By definition, the reduction potential for hydrogen is 0 V:

2 H+(aq, 1M) + 2e H2(g, 1 atm)


Electrochemistry

Cell Potentials• For the oxidation in this cell,

• For the reduction,

Ered = 0.76 V

Ered = +0.34 V


Electrochemistry

A. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the increase in negative charge of nitrate ions formed in the reaction.

B. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the decreasing positive charge of hydrogen ions consumed in the reaction.

C. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the decreasing positive charge of zinc ions consumed in the reaction.

D. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the increasing positive charge of hydrogen ions formed in the reaction.

Electrochemistry

Cell Potentials

Ecell = Ered

(cathode) Ered (anode)

= +0.34 V (0.76 V)

= +1.10 V


Electrochemistry

A. 1 atm pressure for Cl2(g) and 1 M solution for Cl–

(aq).

B. 1 M solution for Cl2(g) and for Cl–(aq).

C. 1 atm pressure for Cl2(g) and for Cl–(aq).

D. 1 torr pressure for Cl2(g) and 1 M solution for Cl–

(aq).

Electrochemistry

For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have

Given that the standard reduction potential of Zn2+ to Zn(s) is 0.76 V, calculate the for the reduction of Cu2+ to Cu:

Cu2+(aq, 1 M) + 2 e Cu(s)

Sample Exercise 20.5 Calculating from

Practice ExerciseThe standard cell potential is 1.46 V for a voltaic cell based on the following half-reactions:

In+(aq) In3+(aq) + 2 e

Br2(l) + 2 e 2Br(aq)Using Table 20.1, calculate for the reduction ofIn3+ to In+.

Electrochemistry

A. The electrode with the more positive Eredo value is

associated with the cathode.B. The electrode with the less positive (more negative) Ered

o value is associated with the cathode.

Electrochemistry

Use Table 20.1 to calculate for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction Cr2O7

2(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)

Sample Exercise 20.6 Calculating from

Practice ExerciseUsing data in Table 20.1, calculate the standard emf for a cell that employs the overall cell reaction 2 Al(s) + 3 I2(s) 2 Al3+(aq) + 6 I(aq).

Electrochemistry

Sample Exercise 20.7 Determining Half-Reactions at Electrodes and Calculating Cell Potentials

A voltaic cell is based on the two standard half-reactions

Cd2+(aq) + 2 e Cd(s)

Sn2+(aq) + 2 e Sn(s)

Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential.

Practice ExerciseA voltaic cell is based on a Co2+/Co half-cell and an AgCl/Ag half-cell.(a) What half-reaction occurs at the anode? (b) What is the standard cell potential?

Electrochemistry

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.


Electrochemistry

A. A strong oxidizing agent is one that readily gains electrons and thus has a large positive value of Ered

o for its half-reaction. The larger the positive value of Ered

o the greater is the tendency for the reduction half-reaction to occur.

B. A strong oxidizing agent is one that readily gains electrons and thus has a large negative value of Ered

o for its half-reaction. The larger the negative value of Ered


C. A strong oxidizing agent is one that readily loses electrons and thus has a large positive value of Ered

o for its half-reaction. The larger the positive value Ered


D. A strong oxidizing agent is one that readily loses electrons and thus has a large positive value of Ered

o for its half-reaction. The larger the value Ered


Electrochemistry

Oxidizing and Reducing Agents

The greater the difference between the two, the greater the voltage of the cell.


Electrochemistry

Sample Exercise 20.8 Determining Relative Strengths of Oxidizing Agents

Practice ExerciseUsing Table 20.1, rank the following species from the strongestto the weakest reducing agent: I (aq), Fe(s), Al(s).

Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents:NO3

(aq), Ag+ (aq), Cr2O72 (aq).

Electrochemistry

Sample Exercise 20.9 Determining Spontaneity

Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions.(a) Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g)(b) Cl2(g) + 2 I(aq) 2 Cl(aq) + I2(s)

Practice ExerciseUsing the standard reduction potentials listed in Appendix E, determine which of the following reactions arespontaneous under standard conditions:(a) I2(s) + 5 Cu2+(aq) + 6 H2O(l) 2 IO3

(aq) + 5 Cu(s) + 12 H+(aq)(b) Hg2+(aq) + 2 I(aq) Hg(l) + I2(s)(c) H2SO3(aq) + 2 Mn(s) + 4 H+(aq) S(s) + 2 Mn2+(aq) + 3 H2O(l)

Electrochemistry

A. Hg(l) is the stronger reducing agent because Hg(l) has the more negative Eredo.

B. Pb(s) is the stronger reducing agent because Pb(s) has the more negative Eredo.

C. Hg(l) is the stronger reducing agent because Hg(l) is a liquid metal and this changes Eredo.

D. Pb(s) is the stronger reducing agent because Pb(s) is further right in the periodic table than Hg(l).

Electrochemistry

Free Energy

G for a redox reaction can be found by using the equation

G = nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday:

1 F = 96,485 C/mol = 96,485 J/V-mol


Electrochemistry

Free Energy

Under standard conditions,

G = nFE


Electrochemistry

A. The number of moles of the oxidizing agent in the balanced chemical equation.B. The number of moles of the reducing agent in the balanced chemical equation.C. The net number of moles of gas reacting in the balanced chemical equation.D. The number of moles of electrons transferred in the balanced chemical equation.

Electrochemistry

Sample Exercise 20.10 Using Standard Reduction Potentials to Calculate G and K

(a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change, G,and the equilibrium constant, K, at 298 K for the reaction

(b) Suppose the reaction in part (a) is written

What are the values of E, G, and K when the reaction is written in this way?

Practice ExerciseFor the reaction

3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH(aq) 3 Ni(s) + 2 CrO42(aq) + 8 H2O(l)

(a) What is the value of n? (b) Use the data in Appendix E to calculate G. (c) Calculate K at T = 298 K.

Electrochemistry

Nernst Equation• Remember that

G = G + RT ln Q• This means

nFE = nFE + RT ln Q• Dividing both sides by nF, we get the Nernst

equation:

• or, using base-10 logarithms,


E = E RTnF ln Q

E = E 2.303RTnF log Q

Electrochemistry

Nernst Equation

At room temperature (298 K),

Thus, the equation becomes

E = E 0.0592n

log Q

2.303RTF

= 0.0592 V


Electrochemistry

Sample Exercise 20.11 Cell Potential under Nonstandard Conditions

Calculate the emf at 298 K generated by a voltaic cell in which the reaction isCr2O7

2(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)when [Cr2O7

2] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0 105 M.

Practice ExerciseCalculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al3+] = 4.0 103 M and [I] = 0.0100 M.

Electrochemistry

Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell

If the potential of a Zn-H+ cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn2+] = 1.0 M andPH2

= 1.0 atm, what is the H+ concentration?

Practice ExerciseWhat is the pH of the solution in the cathode half-cell in Figure

20.9 when PH2 = 1.0 atm, [Zn2+] in the anode half-cell is 0.10 M,

and the cell emf is 0.542 V?

Electrochemistry

Concentration Cells

• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0.


Electrochemistry

A. The Ni2+(aq) ions and the anions in the salt bridge migrate toward the anode. The NO3-(aq)

ions and the cations in the salt bridge migrate toward the cathode. B. The Ni2+(aq) ions and the cations in the salt bridge migrate toward the anode. The NO3

-(aq) ions and the anions in the salt bridge migrate toward the cathode.

C. The Ni2+(aq) ions and the cations in the salt bridge migrate toward the cathode. The NO3-

(aq) ions and the anions in the salt bridge migrate toward the anode. D. The Ni2+(aq) ions and the anions in the salt bridge migrate toward the cathode. The NO3

-(aq) ions and the cations in the salt bridge migrate toward the anode.

Electrochemistry

Sample Exercise 20.13 Determining pH Using a Concentration CellA voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2

= 1.00 atm and an

unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 atm, [H+] =

1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate for the solution at electrode 1.What is the pH of the solution?

Practice ExerciseA concentration cell is constructed with two Zn(s)Zn2+(aq) half-cells. In one half-cell [Zn2+] = 1.35 M, and in the other [Zn2+] = 3.75 104 M. (a) Which half-cell is the anode? (b) What is the emf of the cell?

Electrochemistry

Applications of Oxidation-Reduction

Reactions


Electrochemistry

Batteries


Electrochemistry

A. +1B. +2C. +4D. +6

Electrochemistry

Alkaline Batteries


Electrochemistry

Hydrogen Fuel Cells


Electrochemistry

Corrosion and…


Electrochemistry

…Corrosion Prevention


Electrochemistry

A. O2

B. FeC. Fe2+

D. H+

Electrochemistry

A. Al and CuB. Cu and NiC. Al and ZnD. Ni and Zn

Electrochemistry

Electrolysis

• Voltaic cells are spontaneous. Electrolytic cells are non-spontaneous, and work by using an applied electric current.

• A battery or other source acts as an electron pump.


Electrochemistry

Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis

Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

Practice Exercise(a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 103 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A?

chapter 20 electrochemistry © 2012 pearson education, inc

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