Chapter 20 Electrochemistry Dr. Subhash Goel South GA State College Douglas, GA Lecture Presentation © 2012 Pearson Education, Inc

Chapter 20

Electrochemistry

Dr. Subhash GoelSouth GA State College

Douglas, GA

Lecture Presentation

© 2012 Pearson Education, Inc.

Electrochemistry

Oxidation Numbers• An oxidation occurs when an atom or ion loses

electrons.• A reduction occurs when an atom or ion gains

electrons.• One cannot occur without the other.• In electrochemical reactions, electrons are

transferred from one species to another.• 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.– Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.

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

combine to form H2.


Electrochemistry

Oxidation and Reduction

• What is reduced is the oxidizing agent.

– H+ oxidizes Zn by taking electrons from it, and hence it is oxidizing agent.

• What is oxidized is the reducing agent.

– Zn reduces H+ by giving it electrons, and hence it is reducing agent.


Electrochemistry

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents

SolutionSolve

The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO2 is converted into Ni(OH)2. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent.

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.

Electrochemistry

Oxidation NumbersTo determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

Electrochemistry

Assigning Oxidation Numbers

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.


Electrochemistry


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.


Electrochemistry


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

– 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.


Electrochemistry


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

Balancing Oxidation-Reduction Equations

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


Electrochemistry

Balancing Oxidation-Reduction Equations

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 Method

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

2. Write the oxidation and reduction half-reactions.


Electrochemistry


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


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


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-Reaction

C2O42 CO2

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

C2O42 2CO2


Electrochemistry

Oxidation Half-Reaction

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


Electrochemistry


MnO4 Mn2+ + 4H2O

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

8H+ + MnO4 Mn2+ + 4H2O


Electrochemistry


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

Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents

SolutionAnalyze We are given an incomplete, unbalanced (skeleton) equation for a redox reaction occurring in acidic solution and asked to complete and balance it.

Plan We use the half-reaction procedure we just learned.

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

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

Solve Step 1: We divide the equation into two half-reactions:

Step 2: We balance each half-reaction. In the first half-reaction the presence of one Cr2O7

2 among the reactants requires two Cr3+ among the products. The seven oxygen atoms in Cr2O7

2 are balanced by adding seven H2O to the products. The 14 hydrogen atoms in 7 H2O are then balanced by adding 14 H+ to the reactants:

We then balance the charge by adding electrons to the left side of the equation so that the total charge is the same on the two sides:

Cr2O72(aq) Cr3+(aq)

Cl(aq) Cl2(g)

14 H+(aq) + Cr2O72(aq) 2 Cr3+(aq) + 7 H2O(l)

6 e + 14 H+(aq) + Cr2O72(aq) 2 Cr3+(aq) + 7

H2O(l)

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

SolutionAnalyze We are given an incomplete equation for a basic redox reaction and asked to balance it.

Plan We go through the first steps of our procedure as if the reaction were occurring in acidic solution. We then add the appropriate number of OH ions to each side of the equation, combining H+ and OH to form H2O.We complete the process by simplifying the equation.

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)

SolveStep 1: We write the incomplete, unbalanced half-reactions:

Step 2: We balance each half-reaction as if it took place in acidic solution:

CN(aq) CNO(aq)MnO4

(aq) MnO2(s)

CN(aq) + H2O(l) CNO(aq) + 2 H+(aq) + 2 e

3 e + 4 H+(aq) + MnO4(aq) MnO2(s) + 2 H2O(l)

Electrochemistry

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

• 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

Voltaic Cells• A voltaic cell consist of two half cells that are

electrically connected. Each half cell is the portion of an electrochemical cell in which half-reaction takes place.

• A simple half cell can be made from a metal strip that dips into a solution of its metal ions.

• Example zinc in zinc salt solution (zinc electrode) or copper in copper salt solution (copper electrode).

• The oxidation occurs at the anode.• The reduction occurs at the cathode.


Electrochemistry

Electrochemistry

Voltaic CellsOnce 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.

• 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.


Electrochemistry

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.


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

Sample Exercise 20.4 Describing a Voltaic CellThe oxidation-reduction reaction

Cr2O72(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.

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

Cr2O72(aq) + 14 H+(aq) + 6 e 2 Cr3+(aq) + 7 H2O(l)

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

6 I(aq) 3 I2(s) + 6 e

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 on the reactant side of the equation). By definition, the reduction process occurs at the cathode. The second half-reaction is the oxidation process (electrons on the product side of the equation), which occurs at the anode. The I ions are the source of electrons, and the Cr2O72 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 Cr2O72H2SO4 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.

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.


Electrochemistry

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.


Electrochemistry

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 JC


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

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 (do not depend on mass).


Electrochemistry

Standard Reduction Potentials

Reduction potentials for many electrodes have been measured and tabulated.


Electrochemistry

Cell Potentials• For the oxidation in this cell,

• For the reduction,

Ered = 0.76 V

Ered = +0.34 V


Electrochemistry

Cell Potentials

Ecell = Ered

(cathode) Ered (anode)

= +0.34 V (0.76 V)

= +1.10 V


Electrochemistry

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).

Solve

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

Electrochemistry

SolutionAnalyze We are given the equation for a redox reaction and asked to use data in Table 20.1 to calculate the standard cell potential for the associated voltaic cell.

Plan Our first step is to identify the half-reactions that occur at the cathode and anode, which we did in Sample Exercise 20.4. Then we use Table 20.1 and Equation 20.8 to calculate the standard cell potential.

Solve The half-reactions areCathode: Cr2O7

2(aq) + 14 H+(aq) + 6 e 2 Cr3+(aq) + 7 H2O(l)

Anode: 6 I(aq) 3 I2(s) + 6 e

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 use these values in Equation 20.8:

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

Electrochemistry

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

SolutionAnalyze We have to look up for two half-reactions. We then use these values first to determine the cathode and the anode and then to calculate the standard cell potential, .

Plan The cathode will have the reduction with the more positive value, and the anode will have the less positive . To write the half-reaction at the anode, we reverse the half-reaction written for the reduction, so that the half-reaction is written as an oxidation.

Solve (a) According to Appendix E, (Cd2+/Cd) = 0.403 V and (Sn2+/Sn) = 0.136 V. 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:

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.

Electrochemistry

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

(b) The cell potential is given by the difference in the standard reduction potentials at the cathode and anode (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).

Continued

Electrochemistry

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.


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

SolutionPlan 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.

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

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

Electrochemistry

Free Energy

• Eo = Eo(reduction process) - Eo

(oxidation process)

• This equation can be used to determine spontaneity of the reaction.

+ ve Eo indicates spontaneous reaction

- ve Eo indicates a non-spontaneous rx.

Electrochemistry

Sample Exercise 20.9 Determining Spontaneity

Solution

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)

Electrochemistry

Continued

Solve(a) For oxidation of Cu to Cu2+ and reduction of H+ to H2, the half-reactions and standard reduction potentials areNotice 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: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 has a positive E value:Thus, Cu2+ can be reduced by H2.


Cu2+(aq) + H2(g) Cu(s) + 2 H+(aq) E = +0.34 V

Electrochemistry

Continued

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

In this case


E = (1.36 V) (0.54 V) = +0.82 V

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

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, E is cell energy and F is a constant, the Faraday:

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


Electrochemistry

Free Energy

Under standard conditions,


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?

Electrochemistry


Solve (a) We first calculate E by breaking the equation into two half-reactions and obtaining 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

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

Continued

E = (1.23 V) (0.80 V) = 0.43 V

The positive value of E leads to a negative value of G. The per mol part of the unit relates to the balanced equation, 4 Ag(s) + O2(g) + 4 H+(aq) 4 Ag+(aq) + 2 H2O(l). Thus, 170 kJ is associated with 4 mol Ag, 1 mol O2 and 4 mol H+, and so forth, corresponding to the coefficients in the balanced equation.

Electrochemistry


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

Continued

K is indeed very large! This means that we expect silver metal to oxidize in acidic aqueous environments,

in air, to Ag+. Notice that the emf calculated for the reaction was E = 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.(b) The overall equation is the same asthat in part (a), multiplied by . Thehalf-reactions are

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

E = +0.43 V

Electrochemistry


Notice, though, that the value of n haschanged to n = 2, which is one-half thevalue in part (a). Thus, G is half aslarge as in part (a):

The value of G is half that in part (a) because the coefficients in the chemical equation are half those in (a).

Now we can calculate K as before:

Continued

G = (2)(96,485 J/V-mol)(+0.43 V) = 83 kJ/mol

Electrochemistry

Nernst Equation

• Remember that

G = G + RT ln Q

• This means

nFE = nFE + RT ln Q


Electrochemistry

Nernst Equation

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

E = E RTnF

ln Q

or, using base-10 logarithms,

E = E 2.303RTnF

log Q


Electrochemistry

Nernst EquationAt room temperature (298 K),

Thus, the equation becomes

E = E 0.0592n

log Q

2.303RTF

= 0.0592 V


We can use this equation to find emf E produced by a cell under nonstandard conditions or to determine concentration of reactant or product by measuring E for the cell.

Electrochemistry

Nernst Equation• Consider following example:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

In this case n = 2 and standard emf is +1.10 V. Hence at 298 K

E= 1.10 V – 0.0592V/2.log [Zn2+/C2+]

For [Cu2+] = 5.0 M and [Zn2+] = 0.05M

E = 1.10V – 0.0592V/2.log[0.050/5.0]

E = 1.16 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

Sample Exercise 20.11 Cell Potential under Nonstandard Conditions

SolutionSolve We calculate E for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E = 0.79 V. As that exercise shows, six electrons are transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is

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.

Electrochemistry

Sample Exercise 20.11 Cell Potential under Nonstandard Conditions

Using Equation 20.18, we have

Continued

Electrochemistry

Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell

SolutionPlan We write the equation for the cell reaction and use standard reduction potentials to calculate E for the reaction. After determining the value of n from our reaction equation, we solve the Nernst equation, Equation 20.18, for Q. Finally, we use the equation for the cell reaction to write an expression for Q that contains [H+] to determine [H+].

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?

Solve The cell reaction isThe standard emf is

Because each Zn atom loses two electrons,

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

n = 2

Electrochemistry

Sample Exercise 20.12 Calculating Concentrations in a Voltaic CellContinued

Using Equation 20.18, we can solve for Q:

Q has the form of the equilibrium constant for the reaction:

Solving for [H+], we have

Comment A voltaic cell whose cell reaction involves can be used to measure or pH. A pH meter is a specially designed voltaic cell with a voltmeter calibrated to read pH directly. (Section 16.4)

Electrochemistry

Sample Exercise 20.13 Determining pH Using a Concentration Cell

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, .

A 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?

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:

Electrochemistry

Sample Exercise 20.13 Determining pH Using a Concentration CellContinued

Thus,

At electrode 1, therefore, the pH of the solution is pH = log[H+] = log(2.7 104) = 3.57

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.

Electrochemistry

Applications of Oxidation-Reduction

Reactions


Electrochemistry

Batteries


Electrochemistry

74

Lead Storage Battery

• six cells in series

• electrolyte = 30% H2SO4

• anode = PbPb(s) + SO4

2−(aq) PbSO4(s) + 2 e−

• cathode = Pb coated with PbO2

• PbO2 is reducedPbO2(s) + 4 H+(aq) + SO4

2−(aq) + 2 e−

PbSO4(s) + 2 H2O(l)

• cell voltage = 2.09 V• rechargeable, heavy

Electrochemistry

75

Alkaline Dry Cell• same basic cell as acidic dry cell,

except electrolyte is alkaline KOH paste

• anode = Zn (or Mg)Zn(s) Zn2+(aq) + 2 e−

• cathode = brass rod

• MnO2 is reduced2 MnO2(s) + 2 NH4

+(aq) + 2 H2O(l) + 2 e−

2 NH4OH(aq) + 2 Mn(O)OH(s)

• cell voltage = 1.54 V• longer shelf life than acidic dry cells

and rechargeable; little corrosion of zinc

Electrochemistry

76

NiCad Battery

• electrolyte is concentrated KOH solution• anode = CdCd(s) + 2 OH−(aq) Cd(OH)2(s) + 2 e− E0 = 0.81 V

• cathode = Ni coated with NiO2

• NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e− Ni(OH)2(s) + 2OH− E0 = 0.49 V

• cell voltage = 1.30 V• rechargeable, long life, light—however,

recharging incorrectly can lead to battery breakdown

Electrochemistry

77

NiMH Battery• electrolyte is concentrated KOH solution• anode = metal alloy with dissolved hydrogen

– oxidation of H from H0 to H+

M∙H(s) + OH−(aq) M(s) + H2O(l) + e− E° = 0.89 V

• cathode = Ni coated with NiO2

• NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e− Ni(OH)2(s) + 2 OH− E0 = 0.49 V

• cell voltage = 1.30 V• rechargeable, long life, light, more environmentally

friendly than NiCad, greater energy density than NiCad

Electrochemistry

78

Lithium Ion Battery• electrolyte is concentrated KOH

solution• anode = graphite impregnated with

Li ions• cathode = Li ─ transition metal

oxide– reduction of transition metal

• work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode

• rechargeable, long life, very light, more environmentally friendly, greater energy density

Electrochemistry

Hydrogen Fuel Cells


Electrochemistry

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Electrolytic Cell• uses electrical energy to overcome the energy

barrier and cause a nonspontaneous reaction– must be DC source

• the + terminal of the battery = anode• the − terminal of the battery = cathode• cations attracted to the cathode, anions to the

anode• Cations pick up electrons from the cathode and are

reduced; anions release electrons to the anode and are oxidized.

• Some electrolysis reactions require more voltage than Etot, called the overvoltage.

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Electroplating

In electroplating, the work piece is the cathode.Cations are reduced at the cathode and plate to the surface of the work piece.The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution.

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Faraday’s Law

• The amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs.– charge that flows through the cell = current x time

Electrochemistry

Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis

SolutionAnalyze 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 20.27 provides a road map for this problem. First, the product of the amperage and the time in seconds gives the number of coulombs of electrical charge being used (Equation 20.21).

Second, the coulombs can be converted with Faraday’s constant (F = 96,485 C/mol electrons) to tell us the number of moles of electrons being supplied. Third, reduction of 1 mol of to Al requires 3 mol 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.

Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl 3 if theelectrical current is 10.0 A.

Solve First, we calculate the coulombs of electrical charge passed into the electrolytic cell:Second, we calculate the number of moles of electrons that pass into the cell:

Electrochemistry

Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis

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

Thus, 3 mol of electrons are required to form 1 mol of Al:

Finally, we convert moles to grams:

Because each step involves multiplication by a new factor, we can combine all the steps:

Continued

Al3+ + 3 e Al

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Electrolysis

• Electrolysis is the process of using electricity to break apart a compound.

• Electrolysis is done in an electrolytic cell.

• Electrolytic cells can be used to separate elements from their compounds.– generate H2 from water for fuel

cells– recover metals from their ores

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Electrolysis of Pure Compounds

• must be in molten (liquid) state

• electrodes normally graphite

• cations reduced at cathode to metal element

• anions oxidized at anode to nonmetal element

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• Corrosion is the spontaneous oxidation of a metal by chemicals in the environment.

• Since many materials we use are active metals, corrosion can be a very big problem.

Corrosion

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Rusting• Rust is hydrated iron(III) oxide.• Moisture must be present.

– Water is a reactant.– required for flow between cathode and anode

• Electrolytes promote rusting.– enhance current flow

• Acids promote rusting.– lower pH = lower E°red

Electrochemistry

Corrosion and…


Documents

Chapter 20 Electrochemistry Dr. Subhash Goel South GA State College Douglas, GA Lecture Presentation © 2012 Pearson Education, Inc