1 / 64

Electrochemistry

Electrochemistry. Importance of Electrochemistry starting your car use a calculator listen to a radio corrosion of iron preparation of important industrial materials Electrochemistry - the study of the interchange of chemical and electrical energy. Galvanic Cells. Redox reactions

clifford-bryant
Download Presentation

Electrochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electrochemistry • Importance of Electrochemistry • starting your car • use a calculator • listen to a radio • corrosion of iron • preparation of important industrial materials • Electrochemistry - the study of the interchange of chemical and electrical energy

  2. Galvanic Cells • Redox reactions • electron transfer reactions • oxidation - loss of electrons, i.e., an increase in oxidation number • reduction - gain electrons, i.e., a decrease in oxidation number

  3. Galvanic Cells • Ex: 8 H+ + MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3 + 4 H2O • If MnO4- and Fe+2 are present in the same solution, the electrons are transferred directly when the reactants collide. • No useful work is done. • Chemical energy is released as heat. • How can the energy be harnessed? • Separate the half reactions. • Have the electron transfer go through a wire, which can then be directed through a motor or other useful device

  4. Galvanic Cells • Galvanic cell • Uses a spontaneous redox reaction to produce current to do work • A device in which chemical energy is changed to electrical energy • Also known as a voltaic cell

  5. Galvanic Cells • Requirements for a Galvanic Cell • Half reactions are in separate compartments • Compartments are connected by a salt bridge • a porous disk connection • a salt bridge (U bridge) containing a strong electrolyte held in a jello-like matrix

  6. Galvanic Cells • Why a salt bridge? • Without a salt bridge, the reaction will not go • Current starts, but stops due to charge buildup in both compartments • one side becomes negatively charged as electrons are added • one side becomes positively charged as electrons are lost • Creating a charge separation requires a large amount of energy

  7. Galvanic Cells • Reaction in a galvanic cell occurs at each of the electrodes • Be able to label the anode, cathode, direction of electron flow, and direction of ion flow

  8. Galvanic Cells • Cell Potential • The oxidizing agent “pulls” electrons towards itself from the reducing agent • The “pull” or driving force on the electrons is the cell potential, Ecell, aka, the electromotive force (emf) of the cell. • Measured in volts, V • 1 V = 1 Joule/Coulomb

  9. Galvanic Cell • Cell Potential • Measure with a voltmeter

  10. Standard Reduction Potentials • Predict cell potentials if the half cell potentials are known • add half cell potentials to get cell potential • Eox + Ered = Ecell • There is no way to measure half cell potentials

  11. Standard Reduction Potentials • So how do we get these half cell potentials if they can’t be measured directly? • Measure a cell potential, assigning one reaction to be the “zero” • 2H+ + 2 e---> H2Ered = 0.00 V

  12. Standard Reduction Potentials • So for 2H+ + Zn --> H2 + Zn+2 the voltage is found to be 0.76 V • If 2 H+ + 2 e- --> H2Ered = 0.00 V and Ecell = Eox + Ered, then Eox = 0.76 - 0.00 V = 0.76 V • So we can say Zn --> Zn+2 + 2 e- Eox = 0.76 V

  13. Standard Reduction Potentials • Combine other half reactions with half reactions with known half cell potentials to complete the table of Standard Reduction Potentials

  14. Standard Reduction Potential • Things to know about standard reduction potentials: • Eovalues correspond to solutes at 1 M and all gases at 1 atm • When a half reaction is reversed, the sign of Eo is reversed • When a half reaction is multiplied by an integer, Eo remains the same!! Do not multiply Eoby any number!

  15. Standard Reduction Potentials • The cell potential is always positive for a galvanic cell Given Fe+2 + 2e- --> Fe Ered = -0.44 V MnO4- + 5 e- + 8 H+--> Mn+2 + 4H2O Ered = 1.51 V • Reverse the reaction that will result in an overall cell potential that is positive, i.e., reverse the rxn involving iron

  16. Standard Reduction Potentials • Find the Eo for the galvanic cell based on the reaction Al+3 + Mg --> Al + Mg+2 • Sketch this cell and label the anode, cathode, direction of electron and ion flow, and indicate what reaction occurs at each electrode

  17. Standard Reduction Potentials • Calculate Eo and sketch the galvanic cell based on the (unbalanced) reaction: MnO4- + H+ + ClO3---> ClO4- + Mn+2 + H2O • Balance the reaction

  18. Galvanic Cells • Line Notation • Used to describe electrochemical cells • Anode components are listed on the left • Cathode components are listed on the right • Separate half cells with double vertical lines: ll • Indicate a phase difference with a single vertical line: l

  19. Galvanic Cells • Ex: Mg(s) l Mg+2(aq) ll Al+3(aq) l Al(s) is the line notation for the example in slide # 15 • For slide #16, the reactants and products are all ions, and so cannot act as an electrode • an inert electrode is needed - use Pt • Pt(s) l ClO3-(aq),ClO4- ll MnO4-(aq),Mn+2(aq) l Pt(s)

  20. Cell Potential, Electrical Work, and Free Energy • Relationship between free energy and cell potential • after some serious derivations: DGo = -nFE o • n = number of moles of electrons • F = the charge on 1 mole of electrons = Faraday = 96,485 Coulombs/1 mole e- • this equation provides an experimental means to obtain DGo. • Cell potential is directly related to DGo. • Confirms that a galvanic cell runs in a direction that results in a positive E o, because a positive E o means a negative DGo, which means the reaction is spontaneous

  21. Cell Potential and Free Energy • Calculate DGo for the reaction Cu+2 + Fe --> Cu + Fe+2 Is this reaction spontaneous?

  22. Cell Potential and Free Energy • Predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au+3 solution.

  23. Dependence of Cell Potential on Concentration • The standard reduction potentials can be used to tell us the cell potential for cells under standard conditions (all concentrations = 1 M) • What will happen to the cell potential when the concentration is not 1 M?

  24. Dependence of Cell Potential on Concentration • Cu(s) + 2Ce+4(aq) --> Cu+2(aq) + 2 Ce+3(aq) • Under standard conditions, Eo = 1.36 V. What will the cell potential be if [Ce+4] is greater than 1.0 M? • Use Le Chatelier’s principle. • Increasing the [Ce+4] (when the [Ce+4] > 1 M), then the forward reaction is favored, so the driving force on the electrons is greater, so the cell potential is greater

  25. Dependence of Cell Potential on Concentration 2 Al(s) + 3Mn+2(aq) --> 2 Al+3(aq) + 3 Mn(s) Eo = 0.48 V • Predict whether Ecell is larger or smaller than Eocell a. [Al+3] = 2.0 M, [Mn+2] = 1.0 M b. [Al+3] = 1.0 M, [Mn+2] = 3.0 M

  26. Concentration Cells • Cell potential depends on concentration • Make a cell (a concentration cell) in which both compartments contain the same components, but at different concentrations • Usually has a small cell potential • The difference in concentration produces the cell potential

  27. Concentration Cells • Given a cell in which both compartments contain aquesous AgNO3 • Ag+ + e- ---> Ag Eo = 0.80 V • Left side: 0.10 M Ag+ • Right side: 1.0 M Ag+ • What will happen?

  28. Concentration Cells • There will be a positive cell potential due to the difference in Ag+ concentrations • To reach equilibrium, the driving force is to equalize the Ag+ concentration. • This can be done if the 1.0 M Ag+ could be reduced and the 0.10 M Ag+ could be increased. • The Ag in the 0.10 M Ag+ compartment will dissolve while the Ag in the 1.0 M Ag+ compartment will increase in mass.

  29. The Nernst Equation DG = DGo + RT ln Q DGo= -nFEo • -nFE = -nFEo + RT lnQ • E = Eo - RT ln Q Nernst equation nF • The Nernst equation gives the relationship between cell potential and the concentrations of the cell components

  30. The Nernst Equation • @ 25oC, the Nernst equation can be written as: E = Eo - 0.0592 log Q n • Use the Nernst equation to calculate the cell potential when one or more of the components are not in their standard states

  31. The Nernst Equation • For 2 Al(s) + 3 Mn+2 --> 2 Al+3 + 3 Mn(s) • Calculate E when [Mn+2] = 0.50 M and [Al+3] = 1.50 M • Applying Le Chatelier’s principle, we would predict that the reverse reaction would be favored, so E should be less than Eo.

  32. The Nernst Equation • The cell potential calculated by the Nernst equation is the maximum cell potential before any current flows • As an galvanic cell discharges, the concentrations of the components will change, so E will change over time • A cell will spontaneously discharge until equilibrium has been reached

  33. The Nernst Equation • Equilibrium has been reached when Q = K, so Ecell = 0 • A dead battery is one in which the cell reaction has reached equilibrium • There is no driving force for electrons to be pushed through the wire DG = 0 at equilibrium, meaning the cell no longer has the ability to do work

  34. The Nernst Equation • Describe the cell based on the following half reactions and concentrations: VO2+ + 2 H+ + e- --> VO+2 + H2O Eo = 1.00 V Zn+2 + 2 e- --> Zn Eo = -0.76 V T = 25o C [VO2+] = 2.0 M [H+] = 0.50 M [VO+2] = 1.0 x 10-2M [Zn+2] = 1.0 x 10-1 M

  35. Calculating the Keq for Redox Reactions • For a cell at equilibrium, Ecell = 0, and Q = K Ecell = Eocell - 0.0592 log Q n 0 = Eocell - 0.0592 log Keq n Eocell = 0.0592 log Keq n log Keq = nEocell @ 25oC 0.0592

  36. Batteries • A battery is • a galvanic cell • a group of galvanic cells connected in series • cell potentials of the individual cells add up to give the total battery cell potential • a source of direct current

  37. Batteries • Lead Storage Battery • can function for several years under temperature extremes from -30oF to 120oF • 12 Volt storage battery made of six cells • anode = Pb • cathode = Pb coated with PbO2 • electrolyte solution = H2SO4 solution

  38. Batteries Pb + HSO4- --> PbSO4 + H+ + 2 e- PbO2 + HSO4- + 3 H+ + 2 e- --> PbSO4+ 2 H2O ___________________________________________________________ Pb + PbO2 + 2HSO4- + 2 H+ --> 2 PbSO4 + 2 H2O • PbSO4 adheres to the electrodes • As the cell discharges, the sulfuric acid is consumed. The condition of the battery can be determined by measuring the density of the solution. As the sulfuric acid concentration decreases, the density decreases.

  39. Batteries • The lead storage battery can be recharged because the products adhere to the electrodes, so the alternator can force current through the battery in the opposite direction and reverse the reaction • Even though the battery can be recharged, physical damage from road shock and chemical side reactions (e.g. electrolysis of water)eventually cause battery failure

  40. Batteries • Dry Cell Battery • invented more than 100 years ago by George Leclanche • Acid Version • anode: Zn • cathode: Carbon rod in contact with a moist paste of solid MnO2, solid NH4Cl, and carbon • 2NH4+ + 2 MnO2 + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O • produces 1.5 V

  41. Batteries • Alkaline dry cell • anode: Zn • cathode: carbon rod in contact with a moist paste of solid MnO2, KOH or NaOH, and carbon • 2MnO2 + H2O + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O • the alkaline dry cell lasts longer because the zinc doesn’t corrode as fast under basic conditions

  42. Batteries • Nickel-Cadmium battery • Cd + NiO2 + 2 OH- + 2 H2O --> Cd(OH)2 + Ni(OH)2 + 2 OH- • This battery can be recharged because, like the lead storage battery, the products adhere to the electrodes.

  43. Fuel Cells • Fuel Cell • a galvanic cell for which the reactants are continuously supplied • used in the U.S. space program • based on the reaction of hydrogen and oxygen to form water: • 2 H2(g) + O2(g) --> 2H2O(l) • not exactly a portable power source as this cell weighs about 500 pounds

  44. Corrosion • Corrosion is the return of metals to their natural states, i.e., the ores from which they are obtained. • Corrosion involves oxidation of the metal • Corroded metal loses its structural integrity and attractiveness • Metals corrode because they oxidize easily • Metals commonly used for decorative and structural purposes have less positive reduction potentials than oxygen gas

  45. Corrosion • Noble Metals • copper, gold, silver, and platinum are relatively difficult to oxidize, hence the term noble metals

  46. Corrosion • Some metals form a protective oxide coating, preventing the complete corrosion of the metal • Aluminum is the best example, forming Al2O3, which adheres to, and protects the aluminum • Copper forms an external layer of copper carbonate, known as patina • Silver forms silver tarnish which is silver sulfide • Gold does not corrode in air

  47. Corrosion of Iron • Important to control the corrosion of iron because it is so important as a structural material • The corrosion of iron is not a direct oxidation process (iron reacting with oxygen), but is actually an electrochemical reaction.

  48. Corrosion of Iron • Steel has a nonuniform surface • These nonuniform areas are where iron can be more easily oxidized (the anode regions) than at other regions (the cathode regions) • anode: Fe--> Fe+2 + 2 e- (the electrons flow through the steel to the cathode) • cathode: O2 + 2 H2O + 4 e- --> 4OH- • The Fe+2 formed in the anodic regions travel through the moisture to the cathodic regions. There, the Fe+2 reacts with oxygen to form rust, which is hydrated iron (III) oxide, Fe2O3.nH2O

  49. Corrosion of Iron • For iron to corrode, moisture must be present to act as a salt bridge between the anode and cathode regions. • Steel does not rust in dry air • Salt accelerates the rusting process • Cars rust faster where salt is used on roads to melt ice and snow • Salt on the moist surfaces increases the conductivity of the aqueous solution formed

  50. Corrosion Prevention • Apply a protective coating • apply paint or metal plating • Chromium or tin are often used to plate steel because they form protective oxides • Zinc can be used to coat steel, a process called galvanizing • Fe--> Fe+2 + 2 e-Eo = 0.44 V • Zn --> Zn+2 + 2 e-Eo = 0.76 V • Zinc, then, is more likely to be oxidized than iron. It acts as a protective coating because it will react with the oxygen preferentially, so the oxygen and the iron don’t come into contact. The zinc is sacrificed.

More Related