Electrochemistry Pulling the Plug on the Power Grid

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1 Electrochemistry 18.1 Pulling the Plug on the Power Grid 18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions 18.4 Standard Electrode Potentials 18.7 Batteries: Using Chemistry to Generate Electricity 18.8 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity 18.9 Corrosion: Undesirable Redox Reactions Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-1

2 18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions FIGURE 18.1 A Spontaneous Oxidation-Reduction Reaction Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-2

5 Rate of flow of electrons 1 A = 1 C/s Driving force 1 V = 1 J/C Cell Potential (E cell ) or cell emf The difference in potential of the two half cells Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-5

6 Standard Cell Potential (E cell ) 1 M concentration for reactants and 1 bar pressure for gases. 25 C unless otherwise noted All relative to some standard reference (the Hydrogen Electrode, discussed later) Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) E cell = V Ni (s) + Cu 2+ (aq) Ni 2+ (aq) + Cu (s) E cell = V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-6

7 Electrochemical Cell Notation phase separations Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) oxidation half reaction on the left reduction half reaction on the right salt bridge Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-7

Fe(s) Fe 2+ (aq) MnO 4 (aq), H + (aq), Mn 2+ (aq) Pt(s) FIGURE 18.

18.4 Standard Electrode Potentials E cell depends on the specific half-reactions FIGURE 18.

2 H (aq) + 2 e H 2 (g) E cell = 0.00 V FIGURE 18.

13 Summarizing Standard Electrode Potentials: The electrode potential of the standard hydrogen electrode (SHE) is exactly zero. The electrode in any half-cell with a greater tendency to undergo reduction is positively charged relative to the SHE and therefore has a positive E. The electrode in any half-cell with a lesser tendency to undergo reduction (or greater tendency to undergo oxidation) is negatively charged relative to the SHE and therefore has a negative E. The cell potential of any electrochemical cell (E cell ) is the difference between the electrode potentials of the cathode and the anode (E cell = E cat E an ). E cell is positive for spontaneous reactions and negative for nonspontaneous reactions. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-13

16 EXAMPLE 18.3 Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials 1 Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at (The equation is balanced.) Al (s) + NO 3 (aq) + 4 H + (aq) Al 3+ (aq) + NO (g) + 2 H 2 O (l) Solution E cell =? oxidation Al (s) Al 3+ (aq) + 3 e reduction NO 3 (aq) + 4 H + (aq) + 3 e NO (g) + 2 H 2 O (l) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-16

17 EXAMPLE 18.3 continued oxidation (anode) Al 3+ (aq) + 3 e Al (s) E = 1.66 V reduction (cathode) Al (s) Al 3+ (aq) + 3 e E = V NO 3 (aq) + 4 H + (aq) + 3 e NO (g) + 2 H 2 O (l) E = V Al (s) + NO 3 (aq) + 4 H + (aq) Al 3+ (aq) + NO (g) + 2 H 2 O (l) E cell = E cathode E anode = +0.96V ( 1.66 V) = V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-17

18 Predicting the Spontaneous Direction of an Oxidation- Reduction Reaction : cathode with greater E Ni 2+ (aq) or Mn 2+ (aq)? Ni cathode Ni 2+ (aq) + 2 e Ni (s) E = 0.23 V Mn anode Mn 2+ (aq) + 2 e Mn (s) E = 1.18 V oxid n at anode Mn (s) Mn 2+ (aq) + 2 e E = V Ni (s) + Mn 2+ (aq) Ni 2+ (aq) + Mn (s) E = V E cell = E cathode E anode = 0.23 V ( 1.18 V) = V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-18

20 Summarizing the Prediction of Spontaneous Direction for Redox Reactions: The half-reaction with the more positive electrode potential attracts electrons more strongly and will undergo reduction (so substances listed at the top of Table 18.1 tend to undergo reduction; they are good oxidizing agents). The half-reaction with the more negative electrode potential repels electrons more strongly and will undergo oxidation (so substances listed near the bottom of Table 18.1 tend to undergo oxidation; they are good reducing agents). Any reduction reaction in Table 18.1 is spontaneous when paired with the reverse of the reaction listed below it. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-20

21 EXAMPLE 18.5 Predicting Spontaneous Redox Reactions and Sketching Electrochemical Cell Without calculating predict whether each of the following redox reactions is spontaneous. If the reaction is spontaneous as written, make a sketch of the electrochemical cell in which the reaction could occur. If the reaction is not spontaneous as written, write an equation for the spontaneous direction in which the reaction would occur and sketch the electrochemical cell in which the spontaneous reaction would occur. In your sketches, make sure to label the anode (which should be drawn on the left), the cathode, and the direction of electron flow. Fe (s) + Mg 2+ (aq) Fe 2+ (aq) + Mg (s) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-21

22 EXAMPLE 18.5 continued Fe (s) + Mg 2+ (aq) Fe 2+ (aq) + Mg (s) Solution oxidation will occur (in the reverse) Mg 2+ (aq) + 2 e Mg (s) E = 2.37 V reduction will occur Fe 2+ (aq) + 2 e Fe (s) E = 0.45 V E Mg 2+ /Mg is more negative than E Fe 2+ /Fe Fe 2+ attracts e more than Mg 2+ Fe 2+ (aq) + Mg (s) Fe (s) + Mg 2+ (aq) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-22

EXAMPLE 18.5 continued Mg Mg 2+ Fe 2+ Fe FIGURE 18.

24 Predicting Whether a Metal Will Dissolve in Acid 2 H + (aq) + 2 e H 2 (g) Zn (s) Zn 2+ (aq) + 2 e 2 H + (aq) + Zn (s) H 2 (g) + Zn 2+ (aq) When zinc is immersed in hydrochloric acid, the zinc is oxidized, forming ions that become solvated in the solution. Hydrogen ions are reduced, forming bubbles of hydrogen gas. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-24

18.7 Batteries: Using Chemistry to Generate Electricity Dry-Cell Batteries Oxidation (Anode) Zn (s) Zn + (aq) + 2 e Reduction (Cathode) 2 MnO 2 (s) + 2 NH 4+ (aq) + 2 e Overall Mn 2 O 3 (s) + 2 NH 3

25 18.7 Batteries: Using Chemistry to Generate Electricity Dry-Cell Batteries Oxidation (Anode) Zn (s) Zn + (aq) + 2 e Reduction (Cathode) 2 MnO 2 (s) + 2 NH 4+ (aq) + 2 e Overall Mn 2 O 3 (s) + 2 NH 3 (aq) + H 2 O(l) Zn (is) + 2 MnO 2 (s) + 2 NH 4+ (aq) FIGURE Dry-Cell Batteries (a) common dry cell Zn + (aq) + Mn 2 O 3 (s) + 2 NH 3 (aq) + H 2 O(l) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-25

26 Oxidation (Anode) Zn (s) + 2 OH (aq) Zn(OH) 2 (s) + 2 e Reduction (Cathode) 2 MnO 2 (s) + 2 H 2 O(l) + 2 e 2 MnO(OH) (s) + 2 OH (aq) Overall Zn (s) + 2 MnO 2 (s) + 2H 2 O(l) 2 MnO(OH) (s) + Zn(OH) 2 (s) + H 2 O(l) FIGURE Dry-Cell Batteries (b) Alkaline Battery Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-26

28 Oxidation (Anode) Pb (s) + HSO 4 (aq) PbSO 4 (s) + H + (aq) + 2 e Reduction (Cathode) PbO 2 (s) + HSO 4 (aq) + 3 H + (aq) + 2 e PbSO 4 (s) + 2 H 2 O(l) Overall Pb (s) + PbO 2 (s) + 2 HSO 4 (aq) + 2 H + (aq) 2 PbSO 4 (s) + H 2 O(l) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-28

29 Other Rechargeable Batteries The Nickel Cadmium (NiCad) Battery Oxidation (Anode) Cd (s) + 2 OH (aq) Cd(OH) 2 (s) + 2 e Reduction (Cathode) 2 NiO(OH) (s) + 2 H 2 O (l) + 2 e 2 Ni(OH) 2 (s) + 2 OH (aq) Overall Cd (s) + 2 NiO(OH) (s) + 2 H 2 O (l) Cd(OH) 2 (s) + 2 NiO(OH) 2 (s) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-29

18.8 Electrolysis: Driving Nonspontaneous Chemical Reactions With Electricity FIGURE

34 Summarizing, Characteristics of Electrochemical Cell Types: In all electrochemical cells: Oxidation occurs at the anode. Reduction occurs at the cathode. In voltaic cells: The anode is the source of electrons and has a negative charge (anode ). The cathode draws electrons and has a positive charge (cathode ). In electrolytic cells: Electrons are drawn away from the anode, which must be connected to the positive terminal of the external power source (anode ). Electrons are forced to the cathode, which must be connected to the negative terminal of the power source (cathode ). Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-34

Predicting the Products of Electrolysis anion oxidized at anode cation reduced at cathode

36 Mixtures of Cations Na + (aq) + e Na (s) E = 2.71 V K + (aq) + e Na (s) E = 2.92 V or Anions Cl 2 (g) + 2 e 2 Cl (aq) E = V Br 2 (g)+ 2 e 2 Br (aq) E = V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-36

37 Aqueous Solutions Anode (Oxidation) O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = V E = V (ph = 7) Cathode (Reduction) 2 H 2 O (l) + 2 e H 2 (g) + 2 OH E = 0.83 V E = 0.41 V (ph = 7) E cell = E cat E an = 0.41 V (+0.82 V) = 1.23 V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-37

38 Pure water is a poor conductor of electrical current, but the addition of an electrolyte allows electrolysis to take place, producing hydrogen and oxygen gas in a stoichiometric ratio. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-38

40 Aqueous Solutions Reduction 2 Na + (aq) + 2 e 2 Na (s) E = 2.71 V Reduction 2 H 2 O (l) + 2 e H 2 (g) + 2 OH E = 0.41 V (ph 7) Oxidation I 2 (s) + 2 e 2 I (aq) E = +0.54V Oxidation O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = V (ph 7) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-40

42 Overpotential Reduction 2 Na + (aq) + 2 e 2 Na (s) E = 2.71 V Reduction 2 H 2 O (l) + 2 e H 2 (g) + 2 OH E = 0.41 V (ph 7) Oxidation Cl 2 (s) + 2 e 2 Cl (aq) E = 1.36 V Oxidation 2 H 2 O (l) O 2 (g) + 4 H + (aq) + 4 e E = V (ph 7) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-42

Stoichiometry of Electrolysis FIGURE 18.

18.9 Corrosion: Undesirable Redox Reactions O 2 (g) + 4 H 2 O (l) + 4 e 4 OH (aq) E = +0.40 V O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = +1.

44 18.9 Corrosion: Undesirable Redox Reactions O 2 (g) + 4 H 2 O (l) + 4 e 4 OH (aq) E = V O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = V Aluminum is stable because its oxide forms a protective film over the underlying metal, preventing further oxidation. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-44

A scratch in paint often allows the underlying iron to rust. Fe 2+ (g) + 2 e Fe (s) E = 0.45 V O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = +1.

45 A scratch in paint often allows the underlying iron to rust. Fe 2+ (g) + 2 e Fe (s) E = 0.45 V O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = V 4 Fe 2+ (aq) + O 2 (g) + (4+ 2n H 2 O) 2 Fe 2 O 3 (s) n H 2 O + 8 H + (aq) E = V Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-45

This principle underlies the use of sacrificial electrodes to prevent the corrosion of iron.

47 Preventing Corrosion If a metal more active than iron, such as magnesium or aluminum, is in electrical contact with iron, the metal rather than the iron will be oxidized. This principle underlies the use of sacrificial electrodes to prevent the corrosion of iron. In galvanized nails, a layer of zinc prevents the underlying iron from rusting. The zinc oxidizes in place of the iron, forming a protective layer of zinc oxide. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-47

BC Transit had 20 hydrogen fuel cell buses in service in Whistler, BC, for

49 18.1 Pulling the Plug on the Power Grid 2 H 2 (g) + O 2 (g) 2 H 2 O (g) The energy produced by this fuel cell can power an entire house. BC Transit had 20 hydrogen fuel cell buses in service in Whistler, BC, for the 2010 Winter Olympic games. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-49

50 18.2 Balancing Oxidation Reduction Equations reducing oxidizing agent agent Ca (s) + 2 H 2 O (aq) Ca(OH) 2 (aq) + H 2 (g) oxidation reduction oxidation corresponds to an increase in oxidation state and reduction corresponds to a decrease in oxidation state Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-50

51 EXAMPLE 18.1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Balance the redox equation: Fe 2+ (aq) + MnO 4 (aq) Fe 3+ (aq) + Mn 2+ (aq) Solution oxidation reduction Step 1 Assign oxidation states to all atoms and identify the substances being oxidized and reduced. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-51

52 EXAMPLE 18.1 continued Fe 2+ (aq) + MnO 4 (aq) Fe 3+ (aq) + Mn 2+ (aq) oxidation reduction oxidation Fe 2+ (aq) MnO 4 (aq) reduction Fe 3+ (aq) Mn 2+ (aq) Step 2 Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-52

53 EXAMPLE 18.1 continued 8 H + (aq) + Fe 2+ (aq) MnO 4 (aq) Fe 3+ (aq) Mn 2+ (aq) + 4 H 2 O (l) Step 3 Balance each half-reaction with respect to mass in the following order: Balance all elements other than H and O. Balance O by adding H 2 O Balance H by adding H + Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-53

54 EXAMPLE 18.1 continued Fe 2+ (aq) Fe 3+ (aq) + e 5 e + 8 H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4 H 2 O (l) Step 4 Balance each half-reaction with respect to charge by adding electrons. (Make the sum of the charges on both sides of the equation equal by adding as many electrons as necessary.) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-54

55 EXAMPLE 18.1 continued Fe 2+ (aq) Fe 3+ (aq) + e x 5 5 e + 8 H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4 H 2 O (l) x 1 5 Fe 2+ (aq) 5 Fe 3+ (aq) + 5 e 5 e + 8 H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4 H 2 O (l) Step 5 Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-55

56 EXAMPLE 18.1 continued 5 Fe 2+ (aq) 5 Fe 3+ (aq) + 5 e 5 e + 8 H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4 H 2 O (l) 5 Fe 2+ (aq) + 8 H + (aq) + MnO 4 (aq) 5 Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O (l) Step 6 Add the two half-reactions together, canceling electrons and other species as necessary. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-56

57 EXAMPLE 18.1 continued 5 Fe 2+ (aq) + 8 H + (aq) + MnO 4 (aq) 5 Fe 3+ (aq) + Mn 2+ (s) + 4 H 2 O (l) Reactants Products 5 Fe 5 Fe 8 H 8H 1 Mn 1 Mn 4 O 4 O +17 charge +17 charge Step 6 Verify that the reaction is balanced both with respect to mass and with respect to charge. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-57

58 EXAMPLE 18.2 Half-Reaction Method of Balancing Aqueous Redox Equations in Basic Solution Balance the equation occurring in basic solution: I (aq) + MnO 4 (aq) I 2 (aq) + MnO 2 (s) Solution Similar to acid solution except add a step to neutralize H + with OH oxidation reduction Step 1 Assign oxidation states to all atoms and identify the substances being oxidized and reduced. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-58

59 EXAMPLE 18.2 continued I (aq) + MnO 4 (aq) I 2 (aq) + MnO 2 (s) oxidation reduction oxidation I (aq) MnO 4 (aq) reduction I 2 (aq) MnO 2 (s) Step 2 Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-59

60 EXAMPLE 18.2 continued 2 I (aq) I 2 (aq) 4 OH (aq) + 4 H + (aq) + MnO 4 (aq) 4 H 2 O (l) MnO 2 (s) + 2 H 2 O (l) + 4 OH (aq) Step 3 Balance each half-reaction with respect to mass in the following order: Balance all elements other than H and O. Balance O by adding H 2 O Balance H by adding H + Neutralize H + by adding OH (to both sides equally) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-60

61 EXAMPLE 18.2 continued 2 I (aq) I 2 (aq) + 2 e 3 e + 4 H 2 O (l) + MnO 4 (aq) MnO 2 (s) + 2 H 2 O (l) + 4 OH (aq) Step 4 Balance each half-reaction with respect to charge by adding electrons. (Make the sum of the charges on both sides of the equation equal by adding as many electrons as necessary.) Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-61

62 EXAMPLE 18.2 continued 2 I (aq) I 2 (aq) + 2 e x 3 3 e + 4 H 2 O (l) + MnO 4 (aq) MnO 2 (s) + 2 H 2 O (l) + 4 OH (aq) x 2 6 I (aq) 3 I 2 (aq) + 6 e 6 e + 8 H 2 O (l) + 2 MnO 4 (aq) 2 MnO 2 (s) + 4 H 2 O (l) + 8 OH (aq) Step 5 Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-62

63 EXAMPLE 18.2 continued 6 I (aq) 3 I + 6 e 2 (aq) 4 6 e + 8 H 2 O (l) + 2 MnO 4 (aq) 2 MnO 2 (s) + 4 H 2 O (l) + 8 OH (aq) 6 I (aq) + 4 H 2 O (l) + 2 MnO 4 (aq) 3 I 2 (aq) + 2 MnO 2 (s) + 8 OH (aq) Step 6 Add the two half-reactions together, canceling electrons and other species as necessary. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-63

64 EXAMPLE 18.2 continued 6 I (aq) + 4 H 2 O (l) + 2 MnO 4 (aq) 3 I 2 (aq) + 2 MnO 2 (s) + 8 OH (aq) Reactants Products 6 I 6 I 8 H 8H 2 Mn 2 Mn 12 O 12 O 8 charge 8 charge Step 6 Verify that the reaction is balanced both with respect to mass and with respect to charge. Chemistry, 1Ce Copyright 2014 Pearson Canada Inc. slide 18-64

Oxidation-Reduction Review. Electrochemistry. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions. Sample Problem.

Oxidation-Reduction Review. Electrochemistry. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions. Sample Problem. 1 Electrochemistry Oxidation-Reduction Review Topics Covered Oxidation-reduction reactions Balancing oxidationreduction equations Voltaic cells Cell EMF Spontaneity of redox reactions Batteries Electrolysis