1 / 27

Electrochemistry

Electrochemistry. Chapter 20 Brown-LeMay. Review of Redox Reactions. Oxidation - refers to the loss of electrons by a molecule, atom or ion - LEO goes Reduction - refers to the gain of electrons by an molecule, atom or ion – GER

ulf
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 Chapter 20 Brown-LeMay

  2. Review of Redox Reactions • Oxidation- refers to the loss of electrons by a molecule, atom or ion - LEO goes • Reduction - refers to the gain of electrons by an molecule, atom or ion – GER • Chemical reactions in which the oxidation state of one or more substances changes are called oxidation-reduction reactions (or redox reactions)

  3. Zn(s)+ 2H+(aq) Zn2+(aq)+ H2(g) • Zn = 0, Zn2+ = +2 (LEO) reducing agent H+ = +1, H2 = 0 (GER) oxidizing agent • Thus, the oxidation number of both the Zn(s) and H+(aq) change during the course of the reaction, and so, this must be a redox reaction • Review balancing redox-equations

  4. Balancing Redox by Half Reactions • Half reactions are a convenient way of separating oxidation and reduction reactions. • Balance the titration of acidic solution of Na2C2O4 (colorless) with KMnO4(deep purple) • 1st Write the incomplete ½ reactions • MnO4(aq)  Mn2+(aq)is reduced (pale pink) • C2O4(aq)  CO2(g) is oxidized

  5. MnO4-(aq)  Mn2+(aq) + 4H2O C2O42-(aq)  2CO2(g) • Then bal H by adding H+ 8H+ +MnO4-(aq)  Mn2+(aq) + 4H2O C2O42-(aq)  2CO2(g) • finish by balancing the e’s • For the permanganate 7+ left and 2+ on the right 5e- + 8H+ +MnO4(aq)  Mn2+(aq) + 4H2O • On the oxalate 2- on the right and o on the left C2O42-(aq)  2CO2(g) +2e-

  6. 2(5e- + 8H+ +MnO4-(aq)  Mn2+(aq) + 4H2O) 5(C2O42-(aq)  2CO2(g) +2e-) 10e- + 16H+ + MnO4-(aq)  2Mn2+(aq) + 8H2O 5C2O42-(aq)  10CO2(g) +10e- 16H++2MnO4- +5C2O42-  2Mn2+(aq)+8H2O + 10CO2

  7. Balancing Eq in Basic Solution • The same method is used but OH- is added to neutralize the H+ used. • The equation must again be simplified by canceling the terms on both sides of the equation.

  8. Spontaneous redox reactions may be use to perform electrical work Voltaic or galvanic cells are devices that electron transfer occurs in an external circuit. Voltaic Cells

  9. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • Zn0 is spontaneously oxidized to Zn2+ • Cu2+ is spontaneously reduced to Cu0 • Oxidization half reaction at the anode Zn(s) Zn2+(aq) + 2e- • Reduction half reaction at the cathode Cu2+(aq) + 2e- Cu(s)

  10. As oxidization occurs, Zn is converted to Zn2+ and 2e-. • The electrons flow toward the cathode, where they are used in the reduction reaction • We expect the Zn electrode to lose mass • Electrons flow from the anode to cathode so the anode is negative and the cathode is positive

  11. Electrons cannot flow through the solution; they have to be transported though an external wire. • Anions and cations move through a porous barrier or salt bridge • Cations move into the cathodic compartment to neutralize the excess negatively charged ions (cathode: Cu2+ + 2e- Cu, so the counter ion of Cu is in excess)

  12. A build up of excess charge is avoided by movement of cations and anions through the salt bridge

  13. The anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by the oxidation. • Molecular View - “Rules” of voltaic cells * at the anode electrons are products oxidization occurs * at the cathode electrons are reactants reduction occurs

  14. Cell EMF “the driving force” • Reactions are spontaneous because the cathode has a lower electrical potential energy than the anode • Potential difference: difference in electrical potential measured in volts • One volt is the potential difference required to impart one joule (J) of energy to a charge of one coulomb (C) • 1V = 1 J/C

  15. Cell EMF “the driving force” • Electromotive force (emf) is the force required to push electrons though the external circuit • Cell potential: Ecell is the emf of a cell • Ecell > 0 for a spontaneous reaction • For 1 molar, 1 atm for gases at 250C(standard conditions), the standard emf (standard cell potential) = E0cell

  16. Standard Reduction Potentials& Cal. Cell Potentials • Standard Reduction Potentials E0red are measured relative to a standard • The emf of a cell is E0cell = E0red(cathode) – E0red(anode) • The standard hydrogen electrode is used as the standard (standard hydrogen electrode) (SHE) • 2H+(aq,1M)+2e- H2(g,1atm) E0cell = 0 V

  17. The standard hydrogen electrode • The (SHE) is assigned a potential of zero • Consider the following half reaction Zn(s) Zn2+(aq) + 2e- • We can measure E0cell relative to the SHE • In the cell the SHE is the cathode • It cons of a Pt electrode in a tube – 1M H+sol • H2 is bubbled through the tube • E0cell = E0red(cathode)-E0red(anode) • 0.76V = 0V-E0red(anode) • Therefore E0red(anode) = -0.76V

  18. The standard hydrogen electrode • Standard electrode potentials are written as reduction reactions • Zn2+(aq) (aq,1 M)+ 2e- Zn(s) E0 = -0.76V • Since the reduction potential is negative in the presence of the SHE the reduction of Zn2+is non-spontaneous • However the oxidization of Zn2+is spontaneous with the SHE

  19. Standard reduction potential • The standard reduction potential is an intensive property • Therefore, changing the stoichiometric coefficient does not affect E0red • 2Zn2+(aq) + 4e- 2Zn(s) E0red= -0.76 V

  20. E0red > 0 are spontaneous relative to the SHE E0red < 0 are non- spontaneous relative to the SHE The larger the difference between E0red values the larger the E0cell The more positive the E0cell value the greater the driving force for reduction

  21. Oxidizing and Reducing Agents • Consider the table of standard reduction potentials • We use the table to determine the relative strength of reducing and oxidizing agents • The more positive the E0red the stronger the oxidizing agent (written as the reactant) • The more negative the E0red the stronger the reducing agent (written as the product)

  22. Oxidizing and Reducing Agents • We can use tables to predict if one reactant can spontaneously oxidize or reduce another • Example F2 can oxidize H2 or Li Ni2+ can oxidize Al(s) Li can reduce F2

  23. Spontaneity of Redox Reactions E0cell = E0red(red process) – E0red(oxid process) • Consider the reaction • Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s) • The standard cell potential is • E0cell = E0red (Ag+/Ag) – E0red(Ni2+/Ni) • E0cell = (0.80 V) - (-0.28) • E0cell = 1.08 V the value indicates the reaction is spontaneous

  24. EMF and free energy change • Delta G = -nFE • where delta G is the change in free energy • n = the number of moles of electrons transferred • F = Faraday’s constant • E = emf of the cell

  25. EMF and free energy change • F = 96,500 C/mole- = 96,500 J/(V)(mole-) • Since n and F are positive, if Delta G < 0, then E > 0 and the reaction will be spontaneous. • Effect of concentration on cell EMF the cell is function until E=0 at which point equilibrium has been reached and the cell is “dead” The point at which E=0 is determined by the concentrations of the species involved in the redox reaction

  26. Walter Nernst (Nobel Prize 1920) • Nernst Equation • G = G0 + RTlnQ • -nFE = -nFE0 + RTlnQ • Solve the equation for E give the Nernst Eq • E = E0 - RT/nF lnQ • Or for base 10 log E = E0- 2.3RT/nF logQ

  27. The nernst eq at 298K = E = E0 – 0.0592/n log Q • Consider if you may • Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • If [Cu2+] = 5.0M and [Zn2+] = 0.05 M • Ecell= 1.10 V – 0.0592/2 log 0.05/5 =1.16 V • Cell emf and chemical equilibrium • Log K = nE0/0.0592 thus if we know cell emf, we can calc the equilibrium constant

More Related