chapter 20 lecture- electrochemistry

Electrochemistry

• Spontaneous reactions that involve electron transfer can be used to generate electricity (Ex. Battery).

• Non-spontaneous reactions that involve electron transfer can be forced to proceed by the addition of an electric current (Ex. electrolysis).

Chapter 20- Electrochemistry

• Section 20.1- Oxidation numbers

Oxidation Numbers

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

Oxidation and Reduction


• A species is oxidized when it loses electrons.


• A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.


• A species is reduced when it gains electrons.


• A species is reduced when it gains electrons.Here, each of the H+ gains an electron and they

combine to form H2.


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


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

• What is oxidized is the reducing agent.Zn reduces H+ by giving it electrons.

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation number of 0.


1. 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.


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 in which it has an oxidation number of −1.




Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.




Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

Fluorine always has an oxidation number of −1.


3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. 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.


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.

1. N of NO2- is reduced, Cr of Cr2O7

2- is oxidized2. N of NO2

- is oxidized, Cr of Cr2O72- is reduced

3. O of NO2- is oxidized, Cr of Cr2O7

2- is reduced4. Cr3+ is reduced, N of NO2

- is oxidized5. N of NO3

- is oxidized, Cr3+ is reduced

For the reaction given below, what substance is oxidized and

what is reduced? 3 NO2

- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3

- + 4 H2O

Which species is oxidized and which is reduced in the following reaction:

Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)

1. Zn, oxidized; H+, reduced2. H+, reduced; Zn, oxidized3. Zn2+, oxidized; H2, reduced4. H2, oxidized; Zn2+, reduced

Correct Answer:

The oxidation state of Zn goes from 0 to +2 while the oxidation state of H goes from +1 to 0.

1. Zn, oxidized; H+, reduced2. H+, reduced; Zn, oxidized3. Zn2+, oxidized; H2, reduced4. H2, oxidized; Zn2+, reduced

PRACTICE EXERCISEIdentify the oxidizing and reducing agents in the oxidation-reduction reaction

Answer: Al(s) is the reducing agent; MnO4–

(aq) is the oxidizing agent.

PRACTICE EXERCISEIdentify the oxidizing and reducing agents in the oxidation-reduction reaction


• Section 20.2- Balancing redox reaction by the half-reaction method.

Balancing Oxidation-Reduction Equations

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

Balancing Oxidation-Reduction Equations

This 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.

Half-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.


1. 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.


3. Balance each half-reaction.a. Balance elements other than H and O.


3. Balance each half-reaction.a. Balance elements other than H and O.b. Balance O by adding H2O.


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+.


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.


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.

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.



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

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

1. Yes2. No


Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) → Mn2+(aq) + CO2(aq)


First, we assign oxidation numbers.

MnO4− + C2O4

2- → Mn2+ + CO2

+7 +3 +4+2


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.

Oxidation Half-Reaction

C2O42− → CO2


C2O42− → CO2

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

C2O42− → 2 CO2


C2O42− → 2 CO2


C2O42− → 2 CO2

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

C2O42− → 2 CO2 + 2 e−

Reduction Half-Reaction

MnO4− → Mn2+


MnO4− → Mn2+

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

MnO4− → Mn2+ + 4 H2O


MnO4− → Mn2+ + 4 H2O


MnO4− → Mn2+ + 4 H2O

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

8 H+ + MnO4− → Mn2+ + 4 H2O


8 H+ + MnO4− → Mn2+ + 4 H2O


8 H+ + MnO4− → Mn2+ + 4 H2O

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

5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C2O42− → 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O


Now we evaluate the two half-reactions together:

C2O42− → 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

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


5 C2O42− → 10 CO2 + 10 e−

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


5 C2O42− → 10 CO2 + 10 e−

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

When we add these together, we get:

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2− →

2 Mn2+ + 8 H2O + 10 CO2 +10 e−


10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2− →

2 Mn2+ + 8 H2O + 10 CO2 +10 e−


10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2− →

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

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

16 H+ + 2 MnO4− + 5 C2O4

2− →

2 Mn2+ + 8 H2O + 10 CO2

Balance the following oxidation-reduction reaction that occurs in acidic solution:

C2O42− + MnO4

− → Mn2+ + CO2

1. 8 H+ + 5 C2O42− + MnO4

− → Mn2+ + 4 H2O + 10 CO2

2. 16 H+ + 2 C2O42− + 2 MnO4

− → 2 Mn2+ + 8 H2O + 4 CO23. 16 H+ + 5 C2O4

2− + 2 MnO4− → 2 Mn2+ + 8 H2O + 10

CO24. C2O4

2−+ MnO4− → Mn2+ + 2 CO2 + 2O2

Correct Answer:

Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

1. 8 H+ + 5 C2O42− + MnO4

− → Mn2+ + 4 H2O + 10 CO2

2. 16 H+ + 2 C2O42− + 2 MnO4

− → 2 Mn2+ + 8 H2O + 4 CO23. 16 H+ + 5 C2O4

2− + 2 MnO4− → 2 Mn2+ + 8 H2O + 10

CO24. C2O4

2−+ MnO4− → Mn2+ + 2 CO2 + 2O2

1. 1, 1, 2, 1, 12. 1, 1, 4, 1, 23. 1, 1, 2, 1, 24. 4, 1, 2, 1, 25. 4, 1, 4, 4, 2

When the following reaction is balanced, what are the coefficients

for each substance? __Ag + __O2 + __H+ __Ag+ + __H2O

PRACTICE EXERCISEComplete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution.

Balancing in Basic Solution

• If a reaction occurs in 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.



• 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.

1. CN− + MnO4− + 2 OH− → CNO− + MnO2 + H2O

2. 2 CN− + 2 MnO4− + 2 OH− →

2 CNO− + 2 MnO2 + 4 OH−

3. 2 CN− + MnO4− → 2 CNO− + MnO2 + O2

4. 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + 2 OH−

Balance the following oxidation-reduction reaction that occurs in basic solution:

CN− + MnO4− → CNO− + MnO2

Correct Answer:

Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

1. CN− + MnO4− + 2 OH− → CNO− + MnO2 + H2O

2. 2 CN− + 2 MnO4− + 2 OH− →

2 CNO− + 2 MnO2 + 4 OH−

3. 2 CN− + MnO4− → 2 CNO− + MnO2 + O2

4. 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + 2 OH−

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


• Section 20.3- Voltaic cells

Voltaic Cells

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

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.

Voltaic Cells

• A typical cell looks like this.

• The oxidation occurs at the anode.

• The reduction occurs at the cathode.

Voltaic Cells

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.

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.

Voltaic Cells• 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.

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.

because positive charge builds up in the anode and must be neutralized as oxidation takes place there.

the surface Zn atoms

SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell

is spontaneous. The voltaic cell utilizing this redox reaction 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.

The oxidation-reduction reaction

SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell

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.

The oxidation-reduction reaction

Solve: In one half-reaction, Cr2O72–(aq) is converted into Cr3+(aq). Starting with these ions and then completing

and balancing the half-reaction, we have

In the other half-reaction, I–(aq) is converted to I2(s):

Now we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first half-reaction is the reduction process (electrons shown on the reactant side of the equation), and by definition, this process occurs at the cathode. The second half-reaction is the oxidation (electrons shown on the product side of the equation), which occurs at the anode. The I– ions are the source of electrons, and the Cr2O7

2– ions accept the electrons. Hence, the electrons flow through the external circuit from the electrode immersed in the KI solution (the anode) to the electrode immersed in the K2Cr2O7 – H2SO4 solution (the cathode). The electrodes themselves do not react in any way; they merely provide a means of transferring electrons from or to the solutions. The cations move through the solutions toward the cathode, and the anions move toward the anode. The anode (from which the electrons move) is the negative electrode, and the cathode (toward which the electrons move) is the positive electrode.

(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?

PRACTICE EXERCISEThe two half-reactions in a voltaic cell are

Answer: (a) The first reaction occurs at the anode, the second reaction at the cathode. (b) The anode (Zn) is consumed in the cell reaction. (c) The cathode is positive.

(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?

PRACTICE EXERCISEThe two half-reactions in a voltaic cell are


• Section 20.4- Standard CELL EMF

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.

Electromotive Force (emf)

• 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.

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 JC

Positive cell potentials = spontaneous processes

YES, positive voltages = spontaneous

Standard Reduction Potentials

Reduction potentials for

many electrodes have been

measured and tabulated.

Given the following reaction, which is true?

1. Plating Ag onto Cu is a spontaneous process.2. Plating Cu onto Ag is a spontaneous process.3. Plating Ag onto Cu is a nonspontaneous process.4. Plating Cu onto Ag is a nonspontaneous process.5. Energy will have to be put in for the reaction to

proceed.

Cu(s)+ 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46 V

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

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) + 2 e− → H2 (g, 1 atm)

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.

Cell Potentials• For the oxidation in this cell,

• For the reduction,

Ered = −0.76 V°

Ered = +0.34 V°

Cell Potentials

Ecell° = Ered° (cathode) − Ered° (anode)= +0.34 V − (−0.76 V)= +1.10 V

Positive values = spontaneous

SAMPLE EXERCISE 20.5 Calculating Ered from Ecellº º

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:

SAMPLE EXERCISE 20.5 Calculating Ered from Ecellº º

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:

Solve:

SolutionAnalyze: We are given and for Zn2+ and asked to calculate for Cu2+.Plan: In the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given for Zn2+ is (anode). Because Cu2+ is reduced, it is in the cathode half-cell. Thus, the unknown reduction potential for Cu2+ is (cathode). Knowing and (anode), we can use Equation 20.8 to solve for (cathode).

Check: This standard reduction potential agrees with the one listed in Table 20.1.Comment: The standard reduction potential for Cu2+ can be represented as and that for Zn2+ as The subscript identifies the ion that is reduced in the reduction half-reaction.

The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate for the reduction of In3+ to In+.

PRACTICE EXERCISEA voltaic cell is based on the half-reactions

Answer: –0.40 V

The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate for the reduction of In3+ to In+.

PRACTICE EXERCISEA voltaic cell is based on the half-reactions

SAMPLE EXERCISE 20.6 Calculating Ecell from Eredº º

Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

SAMPLE EXERCISE 20.6 Calculating Ecell from Eredºº

Solution Analyze: We are given the equation for a redox reaction and asked to use data in Table 20.1 to calculate the standard emf (standard potential) for the associated voltaic cell.Plan: Our first step is to identify the half-reactions that occur at the cathode and the anode, which we did in Sample Exercise 20.4. Then we can use data from Table 20.1 and Equation 20.8 to calculate the standard emf.

Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

Solve: The half-reactions are

Although the iodide half-reaction at the anode must be multiplied by 3 in order to obtain a balanced equation for the reaction, the value of is not multiplied by 3. As we have noted, the standard reduction potential is an intensive property, so it is independent of the specific stoichiometric coefficients.

According to Table 20.1, the standard reduction potential for the reduction of Cr2O72– to Cr3+ is +1.33

V, and the standard reduction potential for the reduction of I2 to I– (the reverse of the oxidation half-reaction) is +0.54 V. We then use these values in Equation 20.8.

Check: The cell potential, 0.79 V, is a positive number. As noted earlier, a voltaic cell must have a positive emf in order to operate.

PRACTICE EXERCISEUsing data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:

Answer: +2.20 V

PRACTICE EXERCISEUsing data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:

SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF

By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential.

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

SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF

By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential.

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

Solve: (a) According to Appendix E, The standard reduction potential for Sn2+ is more positive (less negative) than that for Cd2+; hence, the reduction of Sn2+ is the reaction that occurs at the cathode.

The anode reaction therefore is the loss of electrons by Cd.

(b) The cell potential is given by Equation 20.8.

Notice that it is unimportant that the values of both half-reactions are negative; the negative values merely indicate how these reductions compare to the reference reaction, the reduction of H+(aq).Check: The cell potential is positive, as it must be for a voltaic cell.

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

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.

SAMPLE EXERCISE 20.8 Determining the Relative Strengths of Oxidizing Agents

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

–(aq), Ag+(aq), Cr2O72–(aq).

Solution Plan: The more readily an ion is reduced (the more positive its value), the stronger it is as an oxidizing agent.

Solve: From Table 20.1, we have

Because the standard reduction potential of Cr2O72– is the most positive, Cr2O7

2– is the strongest oxidizing agent of the three. The rank order is Ag+ < NO3

– < Cr2O72–.

PRACTICE EXERCISEUsing Table 20.1, rank the following species from the strongest to the weakest reducing agent: I–(aq), Fe(s), Al(s).

PRACTICE EXERCISEUsing Table 20.1, rank the following species from the strongest to the weakest reducing agent: I–(aq), Fe(s), Al(s).

Answer: Al(s) > Fe(s) > I–(aq)

Oxidizing and Reducing Agents

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

Which substance is the stronger oxidizing agent?

• Br2

• O2

• NO3-

• H+

• Cl2

Which substance is the stronger reducing agent?

• H2O2

• Mn2+

• NO• I-

• Ag

Which substance is the stronger reducing agent?

• H2O2

• Mn2+

• NO• I-• Ag

1. +0.76 V2. +1.52 V3. −0.76 V4. −1.52 V

Calculate the emf of the following cell:Zn(s)|Zn2+(aq, 1 M)|| H+(aq, 1 M)|H2(g, 1 atm)|PtE° (Zn/Zn2+)= −0.76 V.

Correct Answer:

Zn is the anode, hydrogen at the Pt wire is the cathode.

1. +0.76 V2. +1.52 V3. −0.76 V4. −1.52 V

E°cell = E°cathode − E°anode

E°cell = E°cathode − E°anode = 0.00 V − (−0.76 V)E°cell = +0.76 V

Calculate the emf produced by the following voltaic cell reaction:

Zn + 2 Fe3+ → Zn2+ + 2 Fe2+ Zn2+ + 2 e− → Zn E° = −0.76 V

Fe3+ + e− → Fe2+ E° = 0.77 V

1. +0.01 V2. +0.78 V

3. −0.78 V4. +1.53 V

Correct Answer:

Zn is being oxidized at the anode and Fe3+ is being reduced at the cathode. Thus,

E°cell = E°cathode − E°anode

E°cell = E°cathode − E°anode = 0.77 V − (−0.76 V)E°cell = +1.53 V

1. +0.01 V2. +0.78 V3. −0.78 V4. +1.53 V

As written, is the following oxidation-reduction equation spontaneous or non-spontaneous?

Zn2+ + 2 Fe2+ → Zn + 2 Fe3+

Zn2+ + 2 e− → Zn E° = −0.76 VFe3+ + e− → Fe2+ E° = 0.77 V

1. Spontaneous2. Nonspontaneous

Correct Answer:In this case, the reduction process is Zn2+ → Zn while the oxidation process is Fe2+ → Fe3+. Thus:

1. Spontaneous2. Nonspontaneous

E° = E°red (reduction) − E°red (oxidation)

E° = −0.76 V − (0.77 V) = −1.53 V

A negative E° indicates a nonspontaneous process.


• Section 20.5- Free Energy & EMF

Free Energy

Positive Ecell values indicate a spontaneous process, so they must have negative ΔG values.

SAMPLE EXERCISE 20.9 Spontaneous or Not?

Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions.

SAMPLE EXERCISE 20.9 Spontaneous or Not?

Solution Analyze: We are given two equations and must determine whether or not each is spontaneous.Plan: To determine whether a redox reaction is spontaneous under standard conditions, we first need to write its reduction and oxidation half-reactions. We can then use the standard reduction potentials and Equation 20.10 to calculate the standard emf, E°, for the reaction. If a reaction is spontaneous, its standard emf must be a positive number.

Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions.

Solve: (a) In this reaction Cu is oxidized to Cu2+ and H+ is reduced to H2. The corresponding half-reactions and associated standard reduction potentials are

Notice that for the oxidation, we use the standard reduction potential from Table 20.1 for the reduction of Cu2+ to Cu. We now calculate E° by using Equation 20.10:

SAMPLE EXERCISE 20.9 continued

Because the value of E° is positive, this reaction is spontaneous and could be used to build a voltaic cell.

Because E° is negative, the reaction is not spontaneous in the direction written. Copper metal does not react with acids in this fashion. The reverse reaction, however, is spontaneous and would have an E° of +0.34 V:

Cu2+ can be reduced by H2.

(b) We follow a procedure analogous to that in (a):

In this case

PRACTICE EXERCISEUsing the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:

Answer: Reactions (b) and (c) are spontaneous.

PRACTICE EXERCISEUsing the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:

Free Energy

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

ΔG = −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

Free Energy

Under standard conditions,

ΔG° = −nFE°

Given the following reaction, which is true?

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46V

SAMPLE EXERCISE 20.10 Determining ΔG° and K

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

(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction

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

SAMPLE EXERCISE 20.10 Determining ΔG° and K

Solution Analyze: We are asked to determine ΔG° and K for a redox reaction, using standard reduction potentials.Plan: We use the data in Table 20.1 and Equation 20.10 to determine E° for the reaction and then use E° in Equation 20.12 to calculate ΔG°. We will then use Equation 19.22, ΔG° = –RT in K, to calculate K.

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

(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction

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

Solve: (a) We first calculate E° by breaking the equation into two half-reactions, as we did in Sample Exercise 20.9, and then obtain values from Table 20.1 (or Appendix E):

Even though the second half-reaction has 4 Ag, we use the value directly from Table 20.1 because emf is an intensive property.

Using Equation 20.10, we have


K is indeed very large! This means that we expect silver metal to oxidize in acidic environments, in air, to Ag+. Notice that the voltage calculated for the reaction was 0.43 V, which is easy to measure. Directly measuring such a large equilibrium constant by measuring reactant and product concentrations at equilibrium, on the other hand, would be very difficult.

The half-reactions show the transfer of four electrons. Thus, for this reaction n = 4. We now use Equation 20.12 to calculate ΔG°:

The positive value of E° leads to a negative value of ΔG°.

Now we need to calculate the equilibrium constant, K, using ΔG° = –RT ln K. Because ΔG° is a large negative number, which means the reaction is thermodynamically very favorable, we expect K to be large.


Comment: E° is an intensive quantity, so multiplying a chemical equation by a certain factor will not affect the value of E°. Multiplying an equation will change the value of n, however, and hence the value of ΔG°. The change in free energy, in units of J/mol of reaction as written, is an extensive quantity. The equilibrium constant is also an extensive quantity.

(b) The overall equation is the same as that in part (a), multiplied by The half-reactions are

The values of are the same as they were in part (a); they are not changed by multiplying the half-reactions by Thus, E° has the same value as in part (a):

Notice, though, that the value of n has changed to n = 2, which is the value in part (a). Thus, ΔG° is half as large as in part (a).

Now we can calculate K as before:

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

PRACTICE EXERCISEFor the reaction

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

PRACTICE EXERCISEFor the reaction

Answer: (a) 6, (b) +87 kJ/mol, (c) K = 6 ×10–16


• Section 20.6- Non-standard EMF

Nernst Equation

• Remember thatΔG = ΔG° + RT ln Q

• This means−nFE = −nFE° + RT ln Q

Nernst Equation

Dividing both sides by −nF, we get the Nernst equation:

E = E° − RTnF ln Q

or, using base-10 logarithms,

E = E° − 2.303 RTnF log Q

Nernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E° − 0.0592n log Q

2.303 RTF = 0.0592 V

Calculate the emf produced by the following voltaic cell reaction.[Zn2+] = 1.0 M, [Fe2+] = 0.1 M, [Fe3+] = 1.0 M

Zn + 2 Fe3+ → Zn2+ + 2 Fe2+

Zn2+ + 2 e− → Zn E° = −0.76 VFe3+ + e− → Fe2+ E° = 0.77 V

1. +1.47 V2. +1.53 V3. +1.59 V

Correct Answer:Q

nEE log

(0.0592)° −=

[ ] [ ][ ]2 3

22 2

Fe

ZnFe log

2(0.0592)

1.53+

++

−=E

[ ] [ ][ ]2

2

1.01.00.1

log 2

(0.0592)1.53 −=E

1.59 0.0592 1.53 (0.01) log 2

(0.0592)1.53 =+=−=E

1. +1.47 V2. +1.53 V3. +1.59 V

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.

SAMPLE EXERCISE 20.13 pH of a Concentration Cell

A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate[H+] for the solution at electrode 1. What is its pH?

SAMPLE EXERCISE 20.13 pH of a Concentration Cell

A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate[H+] for the solution at electrode 1. What is its pH?

Solution Analyze: We are given the voltage of a concentration cell and the direction in which the current flows. We also have the concentrations of all reactants and products except for [H+] in half-cell 1, which is our unknown.Plan: We can use the Nernst equation to determine Q and then use Q to calculate the unknown concentration. Because this is a concentration cell, = 0 V.

Solve: Using the Nernst equation, we have

Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of H+(aq) in electrode 1 represented with the unknown x:


Comment: The concentration of H+ at electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H2 to H+(aq) increases [H+] at electrode 1.

Thus,

At electrode 1, therefore,

and the pH of the solution is

PRACTICE EXERCISEA concentration cell is constructed with two Zn(s)-Zn2+(aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 × 10–4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell?

PRACTICE EXERCISEA concentration cell is constructed with two Zn(s)-Zn2+(aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 × 10–4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell?

Answer: (a) the second half-cell, (b) 0.105 V


• Section 20.7- Applications of Oxidation-Reduction Reactions

Batteries

Alkaline Batteries

• When the anode is gone, the battery is “dead”.

• Rechargable batteries use electricity to drive the reactions in the opposite direction.

• This replaces the anode.

Hydrogen Fuel Cells

A primary battery cannot be recharged. Which of the following batteries fits this category?

1. Lead-acid battery 2. Nickel-cadmium3. Alkaline battery4. Lithium ion

Correct Answer:

In this list, only the alkaline battery is a primary battery and is thus nonrechargeable.

1. Lead-acid battery 2. Nickel-cadmium3. Alkaline battery4. Lithium ion


• Section 20.8- Corrosion

• Corrosion is the oxidation of metals to form unwanted compounds. (Ex. Rusting)

• Some metals form a protective oxide coating that prevents further corrosion. (Ex. Al forms protective Al2O3)

Corrosion and…

…Corrosion Prevention

Sacrificial anode

a.k.a. cathodic protection

Al, Zn

Based on the standard reduction potentials, which metal would not provide cathodic protection to iron?

1. Magnesium2. Nickel3. Sodium4. Aluminum

Correct Answer:

In order to provide cathodic protection, the metal that is oxidized while protecting the cathode must have a more negative standard reduction potential. Here, only Ni has a more positive reduction potential (−0.28 V) than Fe2+ (−0.44 V) and cannot be used for cathodic protection.

1. Magnesium2. Nickel3. Sodium4. Aluminum


• Section 20.9- Electrolysis

ElectrolysisVoltaic cells use spontaneous redox reactions to do work.

It is possible to use electrical energy to force non-spontaneous redox reactions to occur.

These electrolysis reactions occur in electrolytic cells.

Cathode is still reductionAnode is oxidation.

In electrolysis the electrons leave the negative terminal of the voltage source and enter the cathode where reduction occurs

Electroplating

• Two electrodes immersed in a single solution with voltage applied.

• The metal from the anode will be oxidized into ions that enter the solution.

• The cathode will be plated with a thin layer of ions from the solution.

171

Electroplating

• charge passing through circuit is measured in coulombs.

• Coulombs = Amperes x seconds• 96,500 C per mole of e-

YOU CAN DETERMINE # e- TRANSFERRED!!

• The number of electrons transferred is directly proportional to the amount of substance that is oxidized or reduced.

• Amount of solid plated is determined by the # of electrons in the reduction 1/2 reaction.

Ni2+ is electrolyzed to Ni by a current of 2.43 amperes. If current flows for 600 s, how much Ni is plated (in grams)? (AW Ni = 58.7 g/mol)

1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g

Correct Answer:

FnFWti

×

××=mass

( )

C/mol) 96,500(2g/mol) (58.7s) (600. A2.43

mass ×

××=

1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g

SAMPLE EXERCISE 20.14 Aluminum 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.

SAMPLE EXERCISE 20.14 Aluminum 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.Solution Analyze: We are told that AlCl3 is electrolyzed to form Al and asked to calculate the number of grams of Al produced in 1.00 h with 10.0 A.Plan: Figure 20.31 provides a road map of the problem. First, the product of the amperage and the time in seconds gives the number of coulombs of electrical charge being used (Equation 20.18). Second, the coulombs can be converted with the Faraday constant (F = 96,485 C/mole electrons) to tell us the number of moles of electrons being supplied. Third, reduction of 1 mol of Al3+ to Al requires three moles of electrons. Hence we can use the number of moles of electrons to calculate the number of moles of Al metal it produces. Finally, we convert moles of Al into grams.

Solve: First, we calculate the coulombs of electrical charge that are passed into the electrolytic cell:

Second, we calculate the number of moles of electrons that pass into the cell:

Third, we relate the number of moles of electrons to the number of moles of aluminum being formed, using the half-reaction for the reduction of Al3+:


Thus, three moles of electrons (3 F of electrical charge) are required to form 1 mol of Al:

Finally, we convert moles to grams:

Because each step involves a multiplication by a new factor, the steps can be combined into a single sequence of factors:

PRACTICE EXERCISE(a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is

Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 ×103s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A?

Answer: (a) 30.2 g of Mg, (b) 3.97 × 103 s

PRACTICE EXERCISE(a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is

Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 ×103s. (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 lecture- electrochemistry

Education

oxidation numbers1

positive oxidation numbers

oxidation state of h

oxidation state of zn

reduced zn

oxidized zn2

oxidized h2

negativeoxidation numbers