1 / 21

Electrochemistry

Electrochemistry. Unit 13. Oxidation-Reduction Reactions. Now for a quick review. For the following reaction determine what is oxidized/reduced/reducing agent/OA. HIO 3 + H 2 SO 3  I 2 + H 2 SO 4 Now balance it. Electrochemical Cells.

baruch
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 Unit 13

  2. Oxidation-Reduction Reactions • Now for a quick review. For the following reaction determine what is oxidized/reduced/reducing agent/OA. • HIO3 + H2SO3 I2 + H2SO4 • Now balance it.

  3. Electrochemical Cells • Electrochemistry is the study of the interchange of chemical and electrical energy • The conversion between chemical energy and electrical energy is carried out in an electrochemical cell • Spontaneous redox reactions take place in a galvanic cell • Also known as voltaic cells • oxidation and reduction reactions kept separate • half-cells • requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons • Electrons flow from the oxidized substance to the reduced substance.

  4. Electrochemical Cells • The substances must be separated so electrons are forced to flow through a wire. • Electron flow will be short-lived unless a salt bridge is run between the substances to allow ion flow. • Otherwise, the reduced side becomes negatively charged and prevents the flow of electrons. • With a salt bridge negative ions flow toward the oxidized side preventing buildup of negative charge.

  5. Galvanic Cells • Oxidation occurs at the anode • Reduction occurs at the cathode

  6. Galvanic Cells • Electrons flow from one substance to another because of a difference in potential energy. • The potential energy difference exists as a result of a variation in electron affinity. • Electrons flow from higher to lower potential energy. • Toward the element with greater affinity. • Electrical potential is measured in volts. • In order for a reaction to be spontaneous, the value for the potential energy difference (volts) must be positive.

  7. Galvanic Cell • Voltage is calculated by comparing the potential of the two half-cells • Potential values for half-cells are determined by comparing to the reduction of H+(aq)  H2(g) • Standard conditions for measuring cell potential are : 25oC, 1 atm, 1 M • We represent the value for cell potential as E0 cell. • Example: • Calculate E0 cell for the Cu/Zn voltaic cell.

  8. Standard Reduction Potentials

  9. Galvanic Cells 1. Determine E0 cell for the following: Al(s) + Fe2+(aq) → Al3+(aq) + Fe(s) 2. Calculate the E0 cell for the cell with the following half-reactions: Cd(s)  Cd2+(aq) + 2e- Pb(s)  Pb2+(aq) + 2e-

  10. Free Energy Change and Cell Potential • Free energy change is related to cell potential by the equationΔGө = -nFEө • F = Faraday = 96,500 J/mol volt • 1 volt = 1 J/C; C = 1 coulomb ; 96,500 coulombs of charge in 1 mol of electrons • Electric current is measured in amperes, 1 ampere (amp) = 1 C/s

  11. Example: Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g) At standard conditions, ∆Gө = -454.4 kJ/mol; Eөcell= ? • ΔGө = -nFEө ; Eө = -ΔGө/nF; n = 2 mol e- from Mg to H+. Eө = (-454,400 J/mol) = 2.354 V 2 (96,500 J/mol volt)

  12. Example: Determine whether the potential of a cell involving an exothermic reaction whose entropy is increasing, will increase or decrease with a change in temperature. • ΔGө = -nFEө ; Eө = -ΔGө ; nF ΔG = ΔH – TΔS ; -ΔG = -ΔH + TΔS Eө = - ΔH + TΔS ; Now describe changes in Eө as T changes nFnF

  13. Example: A reaction which oxidizes Mg metal produces a current of 10.0 amps for 3.0 hours. How many grams of Mg2+ are produced? • Mg(s)  Mg2+ + 2 e- Therefore 1 mol of Mg will transfer 2 mol e- (n = 2)

  14. Galvanic Cells (non-standard) The conditions of E0cell include 1 M concentrations for both reactant and product, but when concentrations vary it has an effect on cell potential. N.b. you do not need to know the following derivation. ΔG = ΔG0 + RT ln Q -nFEcell = -nFE0cell + RT ln Q Ecell = E0cell - RT ln Q nF Ecell = E0cell – 2.303 (8.314)298.15)ln Q n(96,485) Ecell = E0cell - 0.0592 Vlog Q n This is called the Nernst Equation. You will not be asked to solve calculations with it, but you are expected to make predictions about the effect of concentration on Ecell and understanding the equation will help you to do that.

  15. Galvanic Cells (non-standard) • What is important to note looking at the equation is: • When Q < 1; [reactant] > [product]; log Q < 0; so Ecell > E0cell • When Q = 1; [reactant] = [product]; log Q = 0; so Ecell = E0cell • When Q > 1; [reactant] < [product]; log Q > 0; so Ecell < E0cell • This means that the larger the value for the ratio of [reactant] • [product] • The greater the cell potential. A voltaic cell consists of a Mn/Mn2+ half cell and a Cd/Cd2+ half cell. If the [Mn2+] is 1.4 M and the [Pb2+] is 0.50 M, will the cell potential be greater than, equal to, or less than E0cell?

  16. Reactivity Consider the reaction Zn(s) + CuSO4(aq). What will happen? We demonstrated using cell potentials that zinc will give electrons to copper(II) ions to reduce them to copper metal atoms. Would the following reaction occur: Cu(s) + ZnSO4(aq)? Why not? Zn is a more reactive metal than Cu. More reactive metals are stronger reducing agents. By comparing displacement reactions like these between different combinations of metals and their ions, we can build up a list of relative strengths of the metals as reducing agents. This is called a reactivity series and it allows us to be able to predict whether a particular reaction between a metal and ions of another metal would actually occur.

  17. Reactivity Series Here is a small part of the reactivity series of metals: Mg strongest reducing agent (most readily becomes oxidized) Al Zn Fe Pb Cu Ag weakest reducing agent (means Ag+ would be the best OA) Refer to the reactivity series given above to predict whether the following reactions will occur: no yes • ZnCl2(aq) + 2Ag(s) → 2AgCl(s) + Zn(s) • 2FeCl3(aq) + 3Mg(s) → 3MgCl2 + 2Fe(s)

  18. Reactivity Series • Use the following reactions to: • Deduce the reactivity series for the four metals (most to least • reactive) • Identify the strongest oxidizing and reducing agents • And use the series to determine which reaction(s) (a-c) would • spontaneously occur. Ni, Sn, Cu, Au (most) (least) Ni + Sn2+ → Ni2+ + Sn Sn + Cu2+ → Sn2+ + Cu RA = Ni; OA = Au3+ 3Cu + 2Au3+ → 3Cu2+ + 2Au a. Cu + Ni2+ b. Au + Sn2+ c. Ni + Au3+

  19. Reactivity Series For non-metals, like the halogens, reactivity refers to the ability to act as an oxidizing agent (non-metals are more likely to take electrons). Determine the reactivity series for the halogens and use it to predict which reaction(s) (a-c) would occur spontaneously. F2 most reactive (best OA) Cl2 Br2 I2 least reactive (I- is best RA) • F2 + 2Cl- b. I2 + 2Br- c. Cl2 + 2Br-

  20. Electrolysis • Electrolysis = using electrical energy to break a compound into its elements • Separate H2O into H2 and O2 • Cause a non-spontaneous reaction to occur • Often used to separate metals from ores • Non-spontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy • Requires direct current (battery)

  21. Electrolytic Cell electrolyte

More Related