1. Chapter 20 Electrochemistry
Lecture Presentation
Chapter 20 Electrochemistry
Dr. Subhash Goel
South GA State College
Douglas, GA
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2. © 2012 Pearson Education, Inc.
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.
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3. 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.
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4. 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.
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5. 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.
Solution
Solve
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.

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

7. Assigning Oxidation Numbers
Elements in their elemental form have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.
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8. Assigning Oxidation Numbers
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.
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9. Assigning Oxidation Numbers
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.
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10. Assigning Oxidation Numbers
The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
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11. Balancing Oxidation-Reduction Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.
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12. 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.
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13. The Half-Reaction Method
Assign oxidation numbers to determine what is oxidized and what is reduced.
Write the oxidation and reduction half-reactions.
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14. The Half-Reaction Method
Balance each half-reaction.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
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15. The Half-Reaction Method
Multiply the half-reactions by integers so that the electrons gained and lost are the same.
Add the half-reactions, subtracting things that appear on both sides.
Make sure the equation is balanced according to mass.
Make sure the equation is balanced according to charge.
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16. The Half-Reaction Method
Consider the reaction between MnO4 and C2O42:
MnO4(aq) + C2O42(aq)  Mn2+(aq) + CO2(aq)
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17. The Half-Reaction Method
First, we assign oxidation numbers:
MnO4 + C2O42  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.
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18. Oxidation Half-Reaction
C2O42  CO2
To balance the carbon, we add a coefficient of 2:
C2O42  2CO2
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19. 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
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20. 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
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21. Reduction Half-Reaction
MnO4  Mn2+ + 4H2O
To balance the hydrogen, we add 8H+ to the left side:
8H+ + MnO4  Mn2+ + 4H2O
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22. Reduction Half-Reaction
8H+ + MnO4  Mn2+ + 4H2O
To balance the charge, we add 5e to the left side:
5e + 8H+ + MnO4  Mn2+ + 4H2O
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23. 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:
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24. Combining the Half-Reactions
5C2O42  10CO2 + 10e
10e + 16H+ + 2MnO4  2Mn2+ + 8H2O
When we add these together, we get:
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
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25. Combining the Half-Reactions
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2
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26. Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents
Complete and balance this equation by the method of half-reactions:
Cr2O72(aq) + Cl(aq)  Cr3+(aq) + Cl2(g) (acidic solution)
Solution
Analyze 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.
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 Cr2O72 among the reactants requires two Cr3+ among the products. The seven oxygen atoms in Cr2O72 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) 26

27. 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.
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28. 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)
Solution
Analyze 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.
Solve
Step 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) 28

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Voltaic Cells
In 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.
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30. © 2012 Pearson Education, Inc.
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.
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32. © 2012 Pearson Education, Inc.
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.
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.
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33. © 2012 Pearson Education, Inc.
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.
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34. © 2012 Pearson Education, Inc.
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.
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35. Sample Exercise 20.4 Describing a Voltaic Cell
The 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 Cr2O72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. 35

36. 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.
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37. 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.
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38. © 2012 Pearson Education, Inc.
Cell Potential
Cell potential is measured in volts (V).
1 V = 1 J C
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39. 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)
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40. 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).
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41. Standard Reduction Potentials
Reduction potentials for many electrodes have been measured and tabulated.
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Cell Potentials
For the oxidation in this cell,
For the reduction,
Ered = 0.76 V 
Ered = V 
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43. © 2012 Pearson Education, Inc.
Cell Potentials
Ecell  =
Ered
(cathode) 
(anode)
= V  (0.76 V)
= V
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44. Sample Exercise 20.5 Calculating from
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)
Solution
Analyze 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 44

45. Sample Exercise 20.6 Calculating from
Use Table 20.1 to calculate for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
Solution
Analyze 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 Then we use Table 20.1 and Equation 20.8 to calculate the standard cell potential.
Solve The half-reactions are
Cathode: Cr2O72(aq) + 14 H+(aq) + 6 e  2 Cr3+(aq) + 7 H2O(l)
Anode: I(aq)  3 I2(s) + 6 e
According to Table 20.1, the standard reduction potential for the reduction of Cr2O72 to Cr3+ is V and the standard reduction potential for the reduction of I2 to I (the reverse of the oxidation half-reaction) is V.We use these values in Equation 20.8: 45

46. 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.
Solution
Analyze 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: 46

47. Sample Exercise 20.7 Determining Half-Reactions at Electrodes and Calculating Cell Potentials
Continued
(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). 47

48. Oxidizing and Reducing Agents
The strongest oxidizers have the most positive reduction potentials.
The strongest reducers have the most negative reduction potentials.
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49. Oxidizing and Reducing Agents
The greater the difference between the two, the greater the voltage of the cell.
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50. Sample Exercise 20.8 Determining 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, Cr2O72 is the strongest oxidizing agent of the three. The rank order is Ag+ < NO3 < Cr2O72. 50

51. 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.

52. Solution 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)
Solution 52

53. Sample Exercise 20.9 Determining Spontaneity
Continued
Solve
(a) For oxidation of Cu to Cu2+ and reduction of H+ to H2, the half-reactions and 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:
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 = V 53

54. Sample Exercise 20.9 Determining Spontaneity
Continued
(b) We follow a procedure analogous to that in (a):
In this case
E = (1.36 V)  (0.54 V) = V
Because the value of E is positive, this reaction is spontaneous and could be used to build a voltaic cell. 54

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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
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56. © 2012 Pearson Education, Inc.
Free Energy
Under standard conditions,
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57. 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? 57

58. Sample Exercise 20.10 Using Standard Reduction Potentials to Calculate G and K
Continued
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 to calculate G:
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. 58

59. Sample Exercise 20.10 Using Standard Reduction Potentials to Calculate G and K
Continued
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.
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 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):
E = V 59

60. Sample Exercise 20.10 Using Standard Reduction Potentials to Calculate G and K
Continued
Notice, though, that the value of n has
changed to n = 2, which is one-half the
value in part (a). Thus, G is half as
large 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:
G = (2)(96,485 J/V-mol)(+0.43 V) = 83 kJ/mol 60

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Nernst Equation
Remember that
G = G + RT ln Q
This means
nFE = nFE + RT ln Q
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62. © 2012 Pearson Education, Inc.
Nernst Equation
Dividing both sides by nF, we get the Nernst equation:
E = E  RT nF
ln Q
or, using base-10 logarithms,
E = E 
2.303RT nF
log Q
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63. © 2012 Pearson Education, Inc.
Nernst Equation
At room temperature (298 K),
2.303RT F
= V
Thus, the equation becomes
E = E 
0.0592 n
log Q
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.
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64. Nernst Equation Consider following example:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this case n = 2 and standard emf is V. Hence at 298 K
E= 1.10 V – V/2.log [Zn2+/C2+]
For [Cu2+] = 5.0 M and [Zn2+] = 0.05M
E = 1.10V – V/2.log[0.050/5.0]
E = 1.16 V

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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.
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66. Sample Exercise 20.11 Cell Potential under Nonstandard Conditions
Calculate the emf at 298 K generated by a voltaic cell in which the reaction is
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
when [Cr2O72] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0  105 M.
Solution
Solve 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 66

67. Sample Exercise 20.11 Cell Potential under Nonstandard Conditions
Continued
Using Equation 20.18, we have 67

68. 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 and
PH2 = 1.0 atm, what is the H+ concentration?
Solution
Plan 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+].
Solve The cell reaction is
The standard emf is
Because each Zn atom loses two electrons,
Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g)
n = 2 68

69. Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell
Continued
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) 69

70. Sample Exercise 20.13 Determining pH Using 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 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?
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,
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: 70

71. Sample Exercise 20.13 Determining pH Using a Concentration Cell
Continued
Thus,
At electrode 1, therefore, the pH of the solution is
pH = log[H+] = log(2.7  104) = 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. 71

72. Applications of Oxidation-Reduction Reactions
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73. © 2012 Pearson Education, Inc.
Batteries
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74. PbO2(s) + 4 H+(aq) + SO42−(aq) + 2 e−
Lead Storage Battery
six cells in series
electrolyte = 30% H2SO4
anode = Pb
Pb(s) + SO42−(aq) ® PbSO4(s) + 2 e−
cathode = Pb coated with PbO2
PbO2 is reduced
PbO2(s) + 4 H+(aq) + SO42−(aq) + 2 e−
® PbSO4(s) + 2 H2O(l)
cell voltage = 2.09 V
rechargeable, heavy

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 reduced
2 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

76. NiCad Battery electrolyte is concentrated KOH solution anode = Cd
Cd(s) + 2 OH−(aq) ® Cd(OH)2(s) + 2 e− E0 = 0.81 V
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(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

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 reduced
NiO2(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

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

79. © 2012 Pearson Education, Inc.
Hydrogen Fuel Cells
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80. 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.

82. 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.

83. 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

84. 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.
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 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.
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: 84

85. Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis
Continued
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:
Al e  Al 85

86. 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

87. 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

88. Corrosion
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.

89. 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

90. © 2012 Pearson Education, Inc.
Corrosion and…
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