Chapter 20 Electrochemistry

Chapter 20 Electrochemistry
Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 20 Electrochemistry © 2009, Prentice-Hall, Inc.

Overview of Electrochemistry
review of redox reactions and their usefulness how to balance redox reactions, especially in acidic or basic conditions voltaic cells emf (electromotive force) of voltaic cells under standard conditions free energy and redox reactions (emf and DG) emf of voltaic cells under nonstandard conditions batteries and fuel cells (on your own) electrolysis © 2009, Prentice-Hall, Inc.

Electrochemical Reactions
In electrochemical reactions, electrons are transferred from one species to another. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Oxidizing and Reducing Agents
Examples of good oxidizing agents: H2O2, MnO4-, Cr2O72-, Ce4+, O3, halogens Examples of good reducing agents: alkali and alkaline earth metals © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Cd(s) + NiO2(s) + 2 H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)
Sample Exercise 20.1 (p. 859) The nickel-cadmium (nicad) battery, a rechargeable “dry cell” used in battery-operated devices, 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. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.1) What is the reducing agent in the following reaction? i.e. What is oxidized? 2 Br-(aq) + H2O2(aq) + 2 H+(aq)  Br2(aq) + 2 H2O(l) Br-(aq) H2O2(aq) H+(aq) Br2(aq) Na+(aq) © 2009, Prentice-Hall, Inc.

2 H2O(l) + Al(s) + MnO4-(aq)  Al(OH)4-(aq) + MnO2(s)
Practice Exercise 2 (20.1) Identify the oxidizing (that which is reduced) and reducing (that which is oxidized) agents in the following oxidation-reduction equation: 2 H2O(l) + Al(s) + MnO4-(aq)  Al(OH)4-(aq) + MnO2(s) © 2009, Prentice-Hall, Inc.

Balancing Oxidation-Reduction Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Practice – Simple half-reactions
Write half-reactions for a reaction between Cu2+ and Zn metal. Overall: Zn + Cu2+  Zn2+ + Cu Oxidation: Zn  Zn e-’s Reduction: Cu e-’s  Cu © 2009, Prentice-Hall, Inc.

The Half-Reaction Method (acidic solution)
Write the overall unbalanced reaction. (i) Identify oxidized and reduced substances. (ii) Write the two incomplete half- reactions. © 2009, Prentice-Hall, Inc.

The Half-Reaction Method
Balance each half-reaction. Balance elements other than H and O. Balance O by adding H2O to side deficient in O atoms. Balance H by adding H+. Balance charge by adding electrons. Multiply the half-reactions by integers so that the electrons gained and lost are the same. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Secondly, 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. © 2009, Prentice-Hall, Inc.

Oxidation Half-Reaction
2. (ii) Write the half-reactions. In this case, begin with oxidation half-reaction: C2O42−  CO2 To balance the carbon, we add a coefficient of 2: C2O42−  2 CO2 © 2009, Prentice-Hall, Inc.

Oxidation Half-Reaction
C2O42−  2 CO2 3. 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− © 2009, Prentice-Hall, Inc.

Reduction Half-Reaction
2. (ii) and 3. Reduction half-reaction MnO4−  Mn2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO4−  Mn H2O © 2009, Prentice-Hall, Inc.

3. Combining the Half-Reactions
Now we evaluate the two half-reactions together: C2O42−  2 CO2 + 2 e− 5 e−+ 8 H+ + MnO4−  Mn H2O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2. © 2009, Prentice-Hall, Inc.

Combining the Half-Reactions
5 C2O42−  10 CO e− 10 e− + 16 H+ + 2 MnO4−  2 Mn H2O When we add these together, we get: 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn H2O + 10 CO2 +10 e− © 2009, Prentice-Hall, Inc.

Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn 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 C2O42−  2 Mn H2O + 10 CO2 © 2009, Prentice-Hall, Inc.

Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(g)
Sample Exercise 20.2 (p. 862) Complete and balance the following equation by the method of half-reactions: Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) (acidic solution) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.2 (p. 862) Complete and balance the following equation by the method of half-reactions: Cr2O72-(aq) + 6 Cl-(aq) + 14 H+(aq)  2 Cr3+(aq) + 3 Cl2(g) + 7 H2O(l) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.2) If you complete and balance the following equation in acidic solution Mn2+(aq) + NaBiO3(s)  Bi3+(aq) + MnO4-(aq) + Na+(aq) how many water molecules are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)? Four on the reactants side Three on the product side One on the reactant side Seven on the product side Two on the product side © 2009, Prentice-Hall, Inc.

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. If this produces water on both sides, you might have to subtract water from each side. © 2009, Prentice-Hall, Inc.

CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s)
Sample Exercise 20.3 (p. 850) 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) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.3 (p. 850) Complete and balance this equation for a redox reaction that takes place in basic solution: 3 CN-(aq) + 2 MnO4-(aq) + H2O(l)  3 CNO-(aq) + 2 MnO2(s) + 2 OH-(aq) (basic solution) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.3) If you complete and balance the following oxidation-reduction reaction in basic solution: NO2-(aq) + Al(s)  NH3(aq) + Al(OH)4-(aq) how many hydroxide ions are there in the balanced equation? One on the reactant side One on the product side Four on the reactant side Seven on the product side none © 2009, Prentice-Hall, Inc.

Practice Exercise 20.3 Complete and balance the following oxidation-reduction reaction in basic solution: 2 Cr(OH)3(s)+ 6 ClO-(aq)  2 CrO42-(aq) + 3 Cl2(g) + 2 H2O(l) + 2 OH-(aq) © 2009, Prentice-Hall, Inc.

Voltaic Cells A typical cell looks like this.
The oxidation occurs at the anode. The reduction occurs at the cathode. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, 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. © 2009, Prentice-Hall, 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. © 2009, Prentice-Hall, Inc.

Sample Exercise 20.4 (p. 867) The following oxidation-reduction reaction is spontaneous: Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) I2(s) H2O(l) 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 and ion migrations. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.4) The following two half-reactions occur in a voltaic cell: Ni(s)  Ni2+(aq) e- (electrode = Ni) Cu2+(aq)+ 2 e-  Cu(s) (electrode = Cu) Which one of the following descriptions most accurately describes what is occurring in the half-cell containing the Cu electrode and Cu2+(aq) solution? The electrode is losing mass and cations from the salt bridge are flowing into the half-cell. The electrode is is gaining mass and cations from the salt bridge are flowing into the half-cell. The electrode is losing mass and anions from the salt bridge are flowing into the half-cell. The electrode is is gaining mass and anions from the salt bridge are flowing into the half-cell. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.4) The two half-reactions in a voltaic cell are Zn(s)  Zn2+(aq) e- ClO3-(aq) + 6 H+(aq)+ 6 e-  Cl-(aq) + 3 H2O(l) a) Indicate which reaction occurs at the anode and which at the cathode. b) Which electrode is consumed in the cell reaction? c) Which electrode is positive? Don’t do © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Cell Potential Cell potential = difference in electrical potential. Cell potential is measured in volts (V). 1 V = 1 J C One volt = the potential difference required to impart one joule of energy to a charge of one coulomb. One Coulomb = 6.25 x 1018 units of charge 1 unit of charge = 1 e- or 1 p+ (derived from the ampere) © 2009, Prentice-Hall, Inc.

Standard Reduction Potentials
Reduction potentials for many electrodes have been measured and tabulated. See p.2 of AP Chem reference packet. © 2009, Prentice-Hall, Inc.

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) © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Cell Potentials  Ered = −0.76 V  Ered = +0.34 V
For the oxidation in this cell, For the reduction, Ered = −0.76 V  Ered = V  © 2009, Prentice-Hall, Inc.

Cell Potentials Ecell  = Ered (cathode) − (anode)
= V − (−0.76 V) = V © 2009, Prentice-Hall, Inc.

Sample Exercise 20.5 (p. 872) For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have Zn(s) + Cu2+(aq, 1M)  Zn2+(aq, 1M) + Cu(s) Eocell = 1.10 V Given that the standard reduction potential of Zn2+ to Zn is V, calculate the Eored for the reduction of Cu2+ to Cu. Cu2+(aq, 1M) e-  Cu(s) (0.34 V) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.5) A voltaic cell based on the reaction 2 Eu2+(aq) + Ni2+(aq)  2 Eu3+(aq) + Ni(s) generates Eo = 0.07 V. Given the standard reduction potential of Ni2+ given in Table 20.1 (-0.28 V) what is he standard reduction potential for the reaction Eu3+(aq) + e-  Eu2+(aq)? -0.35 V 0.35 V -0.21 V 0.21 V 0.07 V © 2009, Prentice-Hall, Inc.

Practice Exercise 20.5 A voltaic cell is based on the following half-reactions: In+(aq)  In3+(aq) e- Br2(l) e-  2 Br-(aq) The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate Eored for the reduction of In3+ to In+. (-0.40 V) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.6 (p. 872) 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 following reaction: Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) I2(s) H2O(l) (0.79 V) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.6) Using the data in Table 20.1 what value would you calculate for the standard emf (Eocell) for a voltaic cell that employs the overall cell reaction 2 Ag+(aq) + Ni(s)  2 Ag(s) + Ni2+(aq)? +0.52 V -0.52 V +1.08 V -1.08 V +0.80 V © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.6) Using the data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction: 2 Al(s) I2(s)  2 Al3+(aq) I-(aq) (+2.20 V) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.7 (p. 873) A voltaic cell is based on the following two standard half-reactions: Cd2+(aq) e-  Cd(s) Sn2+(aq) e-  Sn(s) By using the data in Appendix E or your AP Chem packet, determine a) the half-reactions that occur at the cathode and the anode, and b) the standard cell potential (0.267 V) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.7) Consider three voltaic cells, each similar to the one shown in Figure In each voltaic cell, one half-cell contains a 1.0 M Fe(NO3)2(aq) solution with an Fe electrode. The contents of the other half-cells are as follows: Cell 1: a 1.0 M CuCl2(aq) solution with a Cu electrode Cell 2: a 1.0 M NiCl2(aq) solution with a Ni electrode Cell 3: a 1.0 M ZnCl2(aq) solution with a Zn electrode In which voltaic cell(s) does iron act as the anode? Cell 1 Cell 2 Cell 3 Cells 1 and 2 All three cells. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.7) A 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? (0.499 V) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.8 (p. 875) Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3-(aq), Ag+(aq), Cr2O72-(aq). (i.e. which ions are most strongly going to be reduced?) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.8 (p. 875) Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: Ag+(aq), NO3-(aq), Cr2O72-(aq). From lowest reduction potential to highest (i.e. which ions are most strongly going to be reduced?) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.8) Based on the data in Table 20.1, which of the following species would you expect to be the strongest oxidizing agent (i.e. most likely to be reduced)? Cl-(aq) Cl2(g) O2(g) H+(aq) Na+(aq) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.8) Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I-(aq), Fe(s), Al(s). (i.e. most likely to be reduced to the least likely to be reduced) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.8) Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I-(aq), Fe(s), Al(s). Opposite direction (i.e. most likely to be reduced to the least likely to be reduced) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.9 (p. 876) Using standard reduction potentials (Table 20.1), 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) © 2009, Prentice-Hall, Inc.

Sample Exercise 20.9 (p. 876) Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions: a) Cu(s) + 2 H+(aq)  Cu2+(aq) + H2(g) NO b) Cl2(g) + 2 I-(aq)  2 Cl-(aq) + I2(s) YES © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.9) Which of the following elements is capable of oxidizing Fe2+(aq) ions to Fe3+(aq) ions: chlorine, bromine, iodine? I2 Cl2 Cl2 and I2 Cl2 and Br2 all three elements © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.9) Using standard reduction potentials (Appendix E), determine whether the following reactions are spontaneous under standard conditions: I2(s) + 5 Cu2+(aq) + 6 H2O(l)  2 IO3-(aq) Cu(s) + H+(aq) Hg2+(aq) + 2 I-(aq)  Hg(l) + I2(s) H2SO3(aq) + 2 Mn(s) + 4 H+(aq)  S(s) Mn2+(aq) H2O(l) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.9) Using standard reduction potentials (Appendix E), determine whether the following reactions are spontaneous under standard conditions: I2(s) + 5 Cu2+(aq) + 6 H2O(l) NO  2 IO3-(aq) Cu(s) + H+(aq) Hg2+(aq) + 2 I-(aq)  Hg(l) + I2(s) YES H2SO3(aq) + 2 Mn(s) + 4 H+(aq) YES  S(s) Mn2+(aq) H2O(l) © 2009, Prentice-Hall, Inc.

Where we’ve been Where we’re going Electrochemistry –
cell potentials in voltaic cells under standard conditions Thermodynamics – Gibbs Free Energy (Sec. 19.5) Equilibrium combined with Gibbs Free Energy (Sec. 19.7) Apply Gibbs Free Energy to voltaic cells Determine cell potential of voltaic cells under non-standard conditions Electrolytic cells Batteries and Corrosion (on your own) Electrical work © 2009, Prentice-Hall, Inc.

Gibbs Free Energy (DG) - remember
Predicts whether a particular reaction with occur, involves both enthalpy and entropy If DG is negative, the reaction is spontaneous in the forward direction. If DG = 0, the reaction is at equilibrium. If DG is positive, the reaction in the forward direction is nonspontaneous. Appendix C - Tables of Thermodynamics Quantities under standard conditions – 1 M, 1 atm, 25oC. © 2009, Prentice-Hall, Inc.

Free Energy, Equilibrium and Electrochemistry
Combine DGo and K or Q to find DG or K under nonstandard conditions. Determine DGo of a voltaic cell using E, n and F. (electrochem) 3. Determine E under non-standard conditions using the Nernst equation, which includes K or Q. (electrochem) © 2009, Prentice-Hall, Inc.

Free Energy and Redox Reactions
Remember: Eo = Eored(reduction process) – Eored(oxidation process) A positive Eo = a spontaneous process (galvanic or voltaic cell) A negative Eo = a nonspontaneous process © 2009, Prentice-Hall, Inc.

Electrochemistry and Free Energy
G for a redox reaction can be found by using the equation G = −n F E where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,485 C = 96,485 J mol V-mol © 2009, Prentice-Hall, Inc.

Electrochemistry and Free Energy
G = −n F E Since n and F are positive, the sign of DG relies on E. If E > 0, DG will be <0 and the reaction will be spontaneous. (fits with positive Eo  direction of voltaic cell) © 2009, Prentice-Hall, Inc.

Free Energy Under standard conditions, G = −n F E
© 2009, Prentice-Hall, Inc.

4 Ag(s) + O2(g) + 4 H+(aq)  4 Ag+(aq) + 2 H2O(l)
Sample Exercise (p. 878) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, DGo, and the equilibrium constant, K, at 298 K for the following reaction: 4 Ag(s) + O2(g) + 4 H+(aq)  4 Ag+(aq) + 2 H2O(l) (DGo = -170 kJ/mol, K = 9 x 1029) © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 878) b) Suppose the reaction in part (a) were written as 2 Ag(s) + ½ O2(g) + 2 H+(aq)  2 Ag+(aq) + H2O(l) What are the values of Eo, DGo and K when the reaction is written in this way? (Eo = V, DGo = - 83 kJ/mol, K = 4 x 1014) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.10) For the reaction 3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH-(aq)  3 Ni(s) + 2 CrO42-(aq) + 8 H2O(l) DGo = +87 kJ/mol. Given the standard reduction potential of Ni2+(aq) in Table 20.1, what value do you calculate for the standard reduction potential of the half-reaction CrO42-(aq) + 4 H2O(l) + 3 e-  Cr(OH)3(s) + 5 OH-(aq)? -0.43 V -0.28 V 0.02 V -0.13 V -0.15 V © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.10) Consider the reaction 2 Ag+(aq) + H2(g)  2 Ag(s) + 2 H+(aq) Calculate DGof for the Ag+(aq) ion from the standard reduction potentials in Table 20.1 and the fact that DGof for H2(g), Ag(s) and H+(aq) are all zero. Compare your answer with the value given in Appendix C. © 2009, Prentice-Hall, Inc.

Cell EMF under nonstandard conditions
Many voltaic cells don’t operate under standard conditions (Eo, Q = 1) The cell potential is the driving force towards equilibrium – the further away the reaction is from equilibrium, the greater the magnitude of the cell potential. © 2009, Prentice-Hall, Inc.

Deviations from standard conditions that take the cell further away from equilibrium than Q = 1 will increase the cell potential relative to Eo. Deviations from the standard conditions that take the cell closer to equilibrium than Q = 1 will decrease the cell potential relative to Eo. © 2009, Prentice-Hall, Inc.

The quantitative way to describe this situation is with the Nernst equation. We can also do so more broadly by examining Q, which uses relative concentrations of ions in the two cell compartments. © 2009, Prentice-Hall, Inc.

Let’s begin with the more qualitative description
Some principles: Low concentrations of the ion of the oxidized species in the anode compartment is favorable High concentrations of the ion of the reduced species in the cathode compartment is favorable. © 2009, Prentice-Hall, Inc.

Let’s begin with the qualitative description
As the electrochemical cell continues, E gradually decreases as the potential difference between the relevant parts of the two cell compartments decreases. E.g. [Cu2+] decreases over time as Cu2+ ions are reduced and plated onto the Cu electrode. © 2009, Prentice-Hall, Inc.

Let’s begin with the qualitative description
Take note of the initial concentrations of relevant ions – this will give you some indication about how long the electrochemical cell will run. e.g. Higher concentrations in the cathode compartment  more ions available to be reduced, longer run time. © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 881) Calculate the emf at 298 K generated by a voltaic cell in which the reactions is Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) I2(s) H2O(l) when [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, and [Cr3+] = 1.0 x 10-5 M. (E = V) © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 866) Let’s address this exercise qualitatively: Analyze for anode and cathode ion concentrations and cathode ion concentrations. Anode: [Cr3+] = 1.0 x 10-5 M (very low) Cathode: [Cr2O72-] = 2.0 M (high), [H+] = 1.0 M, [I-] = 1.0 M Do we expect E to be higher than, the same as, or lower than Eo for this cell? Why? © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 866) Do we expect E to be higher than, the same as, or lower than Eo for this cell? Why? Low anode ion concentrations – favorable High cathode ion concentrations - favorable © 2009, Prentice-Hall, Inc.

Quantitative = Nernst Equation (named after Walther Hermann Nernst – Nobel Prize in Chemistry 1920)
Remember that G = G + RT ln Q This means −n F E = −n F E + RT ln Q © 2009, Prentice-Hall, Inc.

Nernst Equation To put this equation into terms of E, we can divide both sides by −n F, resulting in the Nernst equation: E = E − RT n F ln Q or, using base-10 logarithms, E = E − 2.303 RT n F log Q © 2009, Prentice-Hall, Inc.

Nernst Equation At room temperature (298 K), 2.303 RT F = 0.0592 V
Thus the equation becomes E = E − 0.0592 n log Q © 2009, Prentice-Hall, Inc.

Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
Nernst Equation Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s) If [Cu2+] = 5.0 M and [Zn2+] = M, Ecell = 1.10 V – log 0.050 = 1.16 V Note: products = ions in anode compartment © 2009, Prentice-Hall, Inc.

Let’s return to Sample Exercise 20.11 (p. 881)
Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, and [Cr3+] = 1.0 x 10-5 M. Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) I2(s) H2O(l) (E = V) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.11) (opt)
Consider a voltaic cell whose overall reaction is Pb2+(aq) + Zn(s)  Pb(s) + Zn2+(aq). What is the emf generated by this voltaic cell when the ion concentrations are [Pb2+] = 1.5 x 10-3 M and [Zn2+] = 0.55 M? 0.71 V 0.56 V 0.49 V 0.79 V 0.64 V © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.11) Consider a voltaic cell whose overall reaction is Pb2+(aq) + Zn(s)  Pb(s) + Zn2+(aq). What is the emf generated by this voltaic cell when the ion concentrations are [Pb2+] = 1.5 x 10-3 M and [Zn2+] = 0.55 M? 0.71 V 0.56 V 0.49 V 0.79 V 0.64 V © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.11) Calculate the emf generated by the cell described below when [Al3+] = 4.0 x 10-3 M and [I-] = M. 2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq) (E = V) © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 866) If the voltage of a Zn-H+ cell (like that in Figure 20.11) is 0.45 V at 25oC when [Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the concentration of H+? ([H+] = 5.8 x 10-6 M) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.12) Consider a voltaic cell where the anode half-reaction is Sn2+(aq) + 2 e-  Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 x 10-3 M and the cell emf is V? Use the reduction potentials in Appendix E that are reported to three significant figures. a) 3.3 x 10-2 M b) 1.9 x 10-4 M c) 9.0 x 10-3 M d) 6.9 x 10-4 M e) 7.6 x 10-3 M © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.12) What is the pH of the solution in the cathode compartment of the cell pictured in Figure 20.11 when PH2 = 1.0 atm, [Zn2+] in the anode compartment is 0.10 M, and cell emf is V? (pH = 4.23) © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Concentration Cells The driving force is the difference in [Ni2+]
Anode (dilute Ni2+): Ni(s)  Ni2+(aq) + 2 e- There is very little Ni2+, so this reaction proceeds in the direction that increases Ni2+, by oxidizing Ni. Cathode (concentrated Ni2+): Ni2+(aq) + 2 e-  Ni(s) There is much Ni2+, so this reaction proceeds in the direction of decreasing Ni2+, by reducing it to Ni(s). © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 870) 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 ([H+] = 1.00 M, PH2 = 1.00 atm). At 298 K the measured cell voltage is 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? (pH = 3.57) © 2009, Prentice-Hall, Inc.

Practice Exercise 20.13 A 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 x 10-4 M. a) Which half-cell is the anode of the cell? b) What is the emf of the cell? (0.105 V) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (20.13) A concentration cell constructed from two hydrogen electrodes, both with PH2 = One electrode is immersed in pure H2O and the other in 6.0 M hydrochloric acid. What is the emf generated by the cell and what is the identify of the electrode that is immersed in hydrochloric acid? -0.23 V, cathode 0.46 V, anode 0.023 V, anode 0.23 V, cathode 0.23 V, anode © 2009, Prentice-Hall, Inc.

Cell EMF and Chemical Equilibrium
From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = Keq) 0 = Eo – log Keq n log Keq = nEo 0.0592 Thus, if we know the cell emf, we can calculate the equilibrium constant (Keq). © 2009, Prentice-Hall, Inc.

Applications of Oxidation-Reduction Reactions

Lead Battery Cathode: PO2 on metal grid in H2SO4
PbO2(s) + HSO4-(aq) + 3 H+(aq) + 2 e-  PbSO4(s)+ 2 H2O(l) Anode: Pb Pb(s)+ HSO4-(aq) PbSO4(s)+ H+(aq)+ 2 e- Overall: PbO2(s) + Pb(s) + 2 HSO4-(aq) + 2 H+(aq)  2 PbSO4(s) + 2 H2O(l) Cell Potential: each cell – about 2 V © 2009, Prentice-Hall, Inc.

Alkaline Batteries Cathode: MnO2(s) + 2 H2O(l) + 2 e- 
2 MnO(OH)(s) + 2 OH-(aq) source of “alkaline” Anode: Zn(s)  Zn2+(aq) + 2 e- Cell potential: about 1.55 V © 2009, Prentice-Hall, Inc.

Nickel–Cadmium, Nickel–Metal–Hydride, and Lithium–Ion Batteries
Cathode: 2 NiO(OH)(s) + 2 H2O(l) + 2 e-  2Ni(OH)2(s) +2 OH-(aq) Anode: Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e- Cell potential - about 1.30 V Cadmium – toxic – difficult disposal © 2009, Prentice-Hall, Inc.

Nickel–Cadmium, Nickel–Metal–Hydride, and Lithium–Ion Batteries
Other rechargeable batteries have been developed: NiMH batteries (nickel–metal–hydride). Li–ion batteries (lithium–ion batteries). © 2009, Prentice-Hall, Inc.

Hydrogen Fuel Cells Cathode: reduction of O2
2 H2O(l) + O2(g) + 4 e-  4 OH-(aq) Anode: oxidation of H2 2 H2(g) + 4 OH-(aq)  4 H2O(l) + 4 e- Overall: O2(g) + 2 H2(g)  2 H2O(l) © 2009, Prentice-Hall, Inc.

…Corrosion Prevention

Protection of Underground Pipes

Electrolysis In both voltaic and electrolytic cells,
reduction occurs at the cathode, and oxidation occurs at the anode. © 2009, Prentice-Hall, Inc.

Electrolysis However, in electrolytic cells, electrons are forced to flow from the anode to the cathode. In electrolytic cells the anode is positive and the cathode is negative. In voltaic cells the anode is negative and the cathode is positive. © 2009, Prentice-Hall, Inc.

Electrolysis Example: decomposition of molten NaCl.
Cathode: 2Na+(l) + 2e-  2Na(l) (reduction) Anode: 2Cl-(l)  Cl2(g) + 2e- (oxidation) Industrially, electrolysis is used to produce metals like Al. © 2009, Prentice-Hall, Inc.

Electrolysis Electrolysis of high-melting ionic substances requires very high temperatures. Do we get the same products if we electrolyze an aqueous solution of the salt? Water complicates the issue! © 2009, Prentice-Hall, Inc.

Electrolysis Example: Consider the electrolysis of NaF(aq):
Na+(aq) + e-  Na(s) Eored = V 2H2O(l) + 2e-  H2(g) + 2OH-(aq) Eored = V Thus water is more easily reduced the sodium ion. 2F-(aq)  F2(g) + 2e- Eored = V 2H2O(l)  O2(g) + 4H+(aq) + 4e- Eored = V Thus it is easier to oxidize water than the fluoride ion. © 2009, Prentice-Hall, Inc.

Electrolysis with Active Electrodes
Active electrodes: electrodes that take part in electrolysis. Example: electroplating. Anode (nickel strip): Ni(s)  Ni2+(aq) + 2e- (active) Cathode (steel strip): Ni2+(aq) + 2e-  Ni(s) (inert) Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion. © 2009, Prentice-Hall, Inc.

Electrolytic cell with an active metal electrode.
Nickel dissolves from the anode to form Ni2+(aq). At the cathode Ni2+(aq) is reduced and forms a nickel “plate” on the cathode. © 2009, Prentice-Hall, Inc.

Quantitative Aspects of Electrolysis
How much material can we obtain with electrolysis? e.g. the reduction of Cu2+ to Cu: Cu2+(aq) + 2e-  Cu(s). Two mol of electrons will plate 1 mol of Cu. The charge of one mol of electrons is 96,500 C (1 F). 1 coulomb = amount of charge passing a point in 1 s when the current is one A. The amount of Cu can be calculated from the current (I) and time (t) required to plate: Q = I t © 2009, Prentice-Hall, Inc.

Quantitative Aspects of Electrolysis

Practice Exercise 1 (20.14) How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of CrCl3 using a current of 1.5 A? 3.8 x 10-2 s 21 min 62 min 139 min 3.2 x 103 min © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (20.14) a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e-  Mg Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 x 103 s. (30.2 g Mg) b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is A? (3.97 x 103 s) © 2009, Prentice-Hall, Inc.

Electrical Work Free energy = a measure of the maximum amount of useful work that can be obtained from a system. DG = wmax and DG = -nFE Thus wmax = -nFE If Ecell is positive, wmax will be negative. Work is done by the system on the surroundings. © 2009, Prentice-Hall, Inc.

Electrical Work The emf = a measure of the driving force for a redox process. In an electrolytic cell an external source of energy is required to force the reaction to proceed. © 2009, Prentice-Hall, Inc.

Electrical Work w = nFEexternal
In order to drive the nonspontaneous reaction the external emf must be greater than Ecell. From physics: Work is measured in units of watts: 1 W = 1 J/s Electric utilities use units of kilowatt-hours: © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 880) Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 103 kg of aluminum by electrolysis of Al3+ if the applied emf is 4.50 V. (1.34 x 104 kWh) © 2009, Prentice-Hall, Inc.

Practice Exercise 20.15 Calculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg by electrolysis of molten MgCl2 if the applied emf is 5.00 V. Assume that the process is 100% efficient. (11.0 kWh) © 2009, Prentice-Hall, Inc.

Sample Integrative Exercise 20 (p. 880)
The Ksp at 298 K for iron (II) fluoride is 2.4 x 10-6. Write a half-reaction that gives the likely products of the two-electron reduction of FeF2(s) in water. b) Use the Ksp value and the standard reduction potential of Fe2+(aq) to calculate the standard reduction potential for the half-reaction in part (a). ( V) c) Rationalize the difference in the reduction potential for the half-reaction in part (a) with that for Fe2+(aq). © 2009, Prentice-Hall, Inc.

Chapter 20 Electrochemistry

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