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Electrochemistry The transfer of electrons provides a means for converting chemical energy to electrical energy, or vice versa. The study of the relationship.

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Presentation on theme: "Electrochemistry The transfer of electrons provides a means for converting chemical energy to electrical energy, or vice versa. The study of the relationship."— Presentation transcript:

1 Electrochemistry The transfer of electrons provides a means for converting chemical energy to electrical energy, or vice versa. The study of the relationship between electricity and chemical reactions is called electrochemistry. Electrochemical reactions are oxidation -reduction reactions. In redox reactions, electrons are transferred from one species (the reductant) to another (the oxidant). The two parts of the reaction are physically separated. The oxidation reaction occurs at the anode. The reduction reaction occurs at the cathode.

2 Balancing Redox Reactions - The Half-Reaction Method Half reaction method rules: 1.Write the unbalanced reaction. 2.Break the reaction into 2 half reactions: ―One oxidation half-reaction and ―One reduction half-reaction Each reaction must have complete formulas for molecules and ions. 3.Mass balance each half reaction by adding appropriate stoichiometric coefficients. To balance H and O we can add: ―H + or H 2 O in acidic solutions. ―OH - or H 2 O in basic solutions. 4.Charge balance the half reactions by adding appropriate numbers of electrons.  Electrons will be products in the oxidation half-reaction.  Electrons will be reactants in the reduction half-reaction. 5.Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half- reaction. 6.Add the two half reactions. 7.Eliminate any common terms and reduce coefficients to smallest whole numbers.

3 The Half-Reaction Method Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

4 Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. The Half-Reaction Method

5 In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH) 4 -, to chromate ions, CrO The hydrogen peroxide is reduced to hydroxide ions. Write and balance the net ionic equation for this reaction. The Half-Reaction Method

6 When chlorine is bubbled into basic solution, it forms hypochlorite ions and chloride ions. Write and balance the net ionic equation. This is a disproportionation redox reaction. The same species, in this case Cl 2, is both reduced and oxidized. The Half-Reaction Method

7 Stoichiometry of Redox Reactions Just as we have done stoichiometry with acid-base reactions, it can also be done with redox reactions. What volume of M KMnO 4 is required to oxidize 35.0 mL of M HCl? The balanced reaction is:

8 A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 20.0 mL of M potassium dichromate solution. Calculate the concentration of the iron (II) sulfate solution. The balanced equation is: Stoichiometry of Redox Reactions

9 There are two kinds electrochemical cells. Electrochemical cells containing nonspontaneous chemical reactions are called electrolytic cells. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells. 1.Galvanic cell (voltaic cell)— energy released during a spontaneous reaction (  G < 0) generates electricity 2.Electrolytic cell—consumes electrical energy from an external source to cause a nonspontaneous redox reaction to occur (  G > 0) 3.The cathode is negative in electrolytic cells and positive in voltaic cells. 4.The anode is positive in electrolytic cells and negative in voltaic cells. Electrochemistry

10 – Both types of cells contain two electrodes connected to an external circuit that provides an electrical connection between systems. Metallic Conduction. – When circuit is closed, electrons flow from the anode to the cathode; electrodes are connected by an electrolyte, which is an ionic substance or solution that allows ions to transfer between the electrodes, thereby maintaining the system’s electrical neutrality. Ionic Conduction. Electrochemistry

11 In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode). Voltaic cells consist of two half-cells which contain the oxidized and reduced forms of an element (or other chemical species) in contact with each other. Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires; creating a potential difference. A simple half-cell consists of: –A piece of metal immersed in a solution of its ions. –A wire to connect the two half-cells. –And a salt bridge to complete the circuit, maintain neutrality, and prevent solution mixing. Electrochemistry

12 Standard Potentials In a voltaic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between two electrodes in the electrochemical cell The flow of electrons in an electrochemical cell depends on 1.The identity of the reacting substances, 2.The difference in the potential energy of their valence electrons 3.The concentrations of the reacting species, and 4.The temperature of the system Electrochemistry The potential of the cell under standard conditions (1 M soln, 1 atm for gases, or a pure solids, or liquid) at 25ºC is called the standard cell potential, E º cell. All E º values are independent of the stoichiometric coefficients for the half-reactions.

13 Constructing a Cell Diagram Because voltaic cells are cumbersome to describe in words, a line notation called a cell diagram has been developed In a cell diagram –the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right; –phase boundaries are shown by single vertical lines; –the salt bridge, which has two phase boundaries, is shown by a double vertical line; Here’s a cell diagram for Zn/Cu cell: Zn(s)| Zn 2+ (aq, 1M) || Cu 2+ (aq, 1 M) | Cu(s) Anode Salt Bridge Cathode Electrochemistry

14 The Zinc-Copper Cell Cell components for the Zn-Cu cell are: 1.A metallic Cu strip immersed in 1.0 M copper (II) sulfate. 2.A metallic Zn strip immersed in 1.0 M zinc (II) sulfate. 3.A wire and a salt bridge to complete circuit The cell’s initial voltage is volts

15 Calculating Standard Cell Potentials The standard cell potential for a redox reaction, E º cell, is a measure of the tendency of the reactants in their standard states to form the products in their standard states—it is a measure of the driving force for the reaction (voltage) The standard cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half- reaction (E º cell = E º cathode – E º anode ). Calculations for the standard potential for the Zn/Cu cell represented by the cell diagram: Zn (s) | Zn 2+ (aq, 1 M) || Cu 2+ (aq, 1M) | Cu (s) Cathode: Cu 2+ (aq) + 2e –  Cu (s) E º cathode = 0.34 V Anode: Zn (s)  Zn 2+ (aq, 1M) + 2e – E º anode = V Overall: Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) E º cell = E º cathode – E º anode = 1.10 V Electrochemistry

16 The Copper - Silver Cell Cell components: 1.A Cu strip immersed in 1.0 M copper (II) sulfate. 2.A Ag strip immersed in 1.0 M silver (I) nitrate. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.46 volts.

17 The Copper - Silver Cell Compare the Zn-Cu cell to the Cu-Ag cell –The Cu electrode is the cathode in the Zn-Cu cell. –The Cu electrode is the anode in the Cu-Ag cell. Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is. Calculations for the standard potential for the Zn/Cu cell represented by the cell diagram: Cu (s) | Cu 2+ (aq, 1 M) || Ag + (aq, 1M) | Ag (s) Cathode: 2Ag + (aq) + 2e –  2Ag (s) E º cathode = 0.80 V Anode: Cu (s)  Cu 2+ (aq, 1M) + 2e – E º anode = – 0.34 V Overall: 2Ag (s) +Cu 2+ (aq)  2Ag + (aq) +Cu (s) E º cell = E º cathode – E º anode = 0.46 V

18 These experimental facts demonstrate that Cu 2+ is a stronger oxidizing agent than Zn 2+. –In other words Cu 2+ oxidizes metallic Zn to Zn 2+. Similarly, Ag + is is a stronger oxidizing agent than Cu 2+. – Because Ag + oxidizes metallic Cu to Cu 2+. If we arrange these species in order of increasing strengths, we see that: Electrochemistry

19 Easiest to reduce is Strongest Oxidizing Agent Ag + Cu 2+ Zn 2+ (find order using left hand side of table) Least likely to be reduced is Strongest Reducing Agent Zn Cu Ag (find order using right hand side of table) Because the half-reactions in the table are arranged in order of their E º values, the table can be used to predict the relative strengths of various oxidants and reductants Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it Any species on the right side of one half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it Electrochemistry

20 To measure relative electrode potentials, we must establish an arbitrary standard. That standard is the Standard Hydrogen Electrode (SHE). –The SHE is assigned an arbitrary voltage of … V The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potential for that half-reaction. Standard hydrogen electrode (SHE) — consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H +, which is in equilibrium with H 2 gas at a pressure of 1 atm at the Pt-solution interface Electrochemistry

21 Uses of Standard Electrode Potentials Electrodes that force the SHE to act as an anode are assigned positive standard reduction potentials. Electrodes that force the SHE to act as the cathode are assigned negative standard reduction potentials. Standard electrode (reduction) potentials tell us the tendencies of half-reactions to occur as written. For example, the half-reaction for the standard potassium electrode is:

22 The Zinc-SHE Cell For this cell the components are: 1.A Zn strip immersed in 1.0 M zinc (II) sulfate. 2.The other electrode is the Standard Hydrogen Electrode. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.76 volts.

23 The Zinc-SHE Cell The cathode is the Standard Hydrogen Electrode. –In other words Zn reduces H + to H 2. The anode is Zn metal. –Zn metal is oxidized to Zn 2+ ions. Calculations for the standard potential for the Zn/Cu cell represented by the cell diagram: Zn (s) | Zn 2+ (aq, 1 M) || H + (aq, 1M) | H 2(g) Cathode: 2H + (aq) + 2e –  H 2(g) E º cathode = 0.00 V Anode: Zn (s)  Zn 2+ (aq, 1M) + 2e – E º anode = V Overall: Zn (s) + H + (aq)  H 2(g) + Zn (s) E º cell = 0.76 V E º cell = E º cathode – E º anode = 0.00 – ( –.76) = 0.76 V

24 The Copper-SHE Cell The cell components are: 1.A Cu strip immersed in 1.0 M copper (II) sulfate. 2.The other electrode is a Standard Hydrogen Electrode. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.34 volts.

25 The Copper-SHE Cell In this cell the SHE is the anode –The Cu 2+ ions oxidize H 2 to H +. The Cu is the cathode. –The Cu 2+ ions are reduced to Cu metal. Calculations for the standard potential for the Zn/Cu cell represented by the cell diagram: Cu (s) | Cu 2+ (aq, 1 M) || H + (aq, 1M) | H 2(g) Cathode: 2H + (aq) + 2e –  H 2(g) E º cathode = 0.00 V Anode: Cu (s)  Cu 2+ (aq, 1M) + 2e – E º anode = V Overall: Cu (s) + 2H 2+ (aq)  H 2(g) + Cu (s) E º cell = 0.34 V E º cell = E º cathode – E º anode = 0.00 – ( –.34) = 0.34 V

26 Use standard electrode potentials to predict whether an electrochemical reaction at standard state conditions will occur spontaneously. In solution, will aqueous dichromate ions, Cr 2 O 7 2-, oxidize aqueous I - ions to aqueous iodine, I 2, or will aqueous I - ions oxidize aqueous Cr 3+ to aqueous dichromate ions, Cr 2 O 7 2- ? Steps for obtaining the equation for the spontaneous reaction. 1.Choose the appropriate half-reactions from a table of standard reduction potentials. 2.Write the equation for the half-reaction with the more positive E 0 value first, along with its E 0 value. 3.Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half- reaction and change the sign of the tabulated E 0. 4.Balance the electron transfer. 5.Add the reduction and oxidation half-reactions and their potentials. This produces the equation for the reaction for which E 0 cell is positive, which indicates that the forward reaction is spontaneous. Remember potentials are work functions and are not stoichiometric. Uses of Standard Electrode Potentials

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28 Electrode Potentials for Other Half-Reactions Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution? Follow the steps outlined in the previous slides. Note that E 0 values are not multiplied by any stoichiometric relationships in this procedure.

29 Electrode Potentials for Other Half-Reactions Will nitric acid, HNO 3, oxidize arsenous acid, H 3 AsO 3, in acidic solution? The reduction product of HNO 3 is NO in this reaction.

30 Commercial Galvanic Cells Galvanic cells can be self-contained and portable and can be used as batteries and fuel cells 1.A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. 2.A fuel cell is a galvanic cell that requires a constant external supply of one or more reactants in order to generate electricity. Electrochemistry

31 Two basic kinds of batteries 1. Disposable, or primary, batteries in which the electrode reactions are effectively irreversible and which cannot be recharged 2. Rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes; can be recharged by applying an electrical potential in the reverse direction, which temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell Major difference between batteries and galvanic cells is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass Electrochemistry

32 One example of a dry cell is flashlight and radio batteries. The cell’s container is made of zinc which acts as an electrode. A graphite rod is in the center of the cell which acts as the other electrode. The space between the electrodes is filled with a mixture of: 1.ammonium chloride, NH 4 Cl 2.manganese (IV) oxide, MnO 2 3.zinc chloride, ZnCl 2 4.and a porous inactive solid. The Dry Cell

33 Secondary Voltaic Cells Secondary cells are reversible, rechargeable. The electrodes in a secondary cell can be regenerated by the addition of electricity. –These cells can be switched from voltaic to electrolytic cells. One example of a secondary voltaic cell is the lead storage or car battery. The Lead Storage Battery In the lead storage battery the electrodes are two sets of lead alloy grids (plates). Holes in one of the grids are filled with lead (IV) oxide, PbO 2. The other holes are filled with spongy lead. The electrolyte is dilute sulfuric acid.

34 Provides the starting power in automobiles and boats; can be discharged and recharged many times The Lead Acid Storage Battery is an example of a very successful recylcing program. The anodes in each cell of this rechargeable battery are plates or grids of lead containing spongy lead metal, while the cathodes are similar grids containing powdered lead dioxide, PbO 2 The electrolyte is an aqueous solution of sulfuric acid The value of E º for such a cell is 2 V; connecting three cells in series produces a 6-V battery, and a typical 12-V car battery contains six of these cells connected in series. The Lead Storage Battery

35 Diagram of the lead storage battery.

36 Batteries Lithium-iodine battery –Water-free battery –Consists of two cells separated by a metallic nickel mesh that collects charge from the anodes –The anode is lithium metal, and the cathode is a solid complex of  2 –Electrolyte is a layer of solid Li  that allows Li + ions to diffuse from the cathode to the anode –Highly reliable and long-lived –Used in cardiac pacemakers, medical implants, smoke alarms, and in computers –Disposable

37 The Nickel-Cadmium (Nicad) Cell Nicad batteries are the rechargeable cells used in calculators, cameras, watches, etc. A water-based cell with a cadmium anode and a highly oxidized nickel cathode This design maximizes the surface area of the electrodes and minimizes the distance between them, which gives the battery both a high discharge current and a high capacity Lightweight, rechargeable, and high capacity but tend to lose capacity quickly and do not store well; also presents disposal problems because of the toxicity of cadmium

38 Fuel Cells A galvanic cell that requires an external supply of reactants because the products of the reaction are continuously removed Does not store electrical energy but allows electrical energy to be extracted directly from a chemical reaction Have reliability problems and are costly Used in space vehicles Hydrogen is oxidized at the anode. Oxygen is reduced at the cathode.

39 Standard electrode potentials, those compiled in appendices, are determined at thermodynamic standard conditions. Reminder of standard conditions. The Effect of Concentration on Cell Potential: The Nernst Equation

40 The value of the cell potentials change if conditions are nonstandard. The Nernst equation describes the electrode potentials at nonstandard conditions. The Nernst equation is: Substitution of the values of the constants into the Nernst equation at 25 o C gives: For this half-reaction: The corresponding Nernst equation is: The Nernst Equation

41 Substituting E 0 into the above expression gives: If [Cu 2+ ] and [Cu + ] are both 1.0 M, i.e. at standard conditions, then E = E 0 because the concentration term equals zero. The Nernst Equation

42 Calculate the potential for the Cu 2+ / Cu + electrode at 25 0 C when the concentration of Cu + ions is three times that of Cu 2+ ions. The Nernst Equation

43 Calculate the potential for the Cu 2+ /Cu + electrode at 25 0 C when the Cu + ion concentration is 1/3 of the Cu 2+ ion concentration. The Nernst Equation

44 Calculate the electrode potential for a hydrogen electrode in which the [H + ] is 1.0 x M and the H 2 pressure is 0.50 atmosphere. The Nernst Equation

45 The Nernst equation can also be used to calculate the potential for a cell that consists of two nonstandard electrodes. Calculate the initial potential of a cell that consists of an Fe 3+ /Fe 2+ electrode in which [Fe 3+ ]=1.0 x M and [Fe 2+ ]=0.1 M connected to a Sn 4+ /Sn 2+ electrode in which [Sn 4+ ]=1.0 M and [Sn 2+ ]=0.10 M. A wire and salt bridge complete the circuit. The Nernst Equation

46 Calculate the E 0 cell by the usual procedure.

47 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis A coulomb is the amount of charge that passes a given point when a current of one ampere (A) flows for one second. Charge (C) = current (A) * time (s) 1 amp = 1 coulomb/second Faraday’s Law states that during electrolysis, one faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent. –This corresponds to the passage of one mole of electrons through the electrolytic cell.

48 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes. Charge (Coulombs) = current (Amperes) * time (sec) moles e – = charge (Coulombs). 96,486 Coulombs /moles e - Moles Pd 2+ : moles of electrons

49 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Calculate the volume of oxygen (measured at STP) produced by the oxidation of water in the previous example.


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