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Electrochemistry

Electrochemistry. Guaranteed to give you a jolt. Electrochemical cells. A chemical system in which oxidation and reduction can occur – often a single displacement reaction Zn + Cu +2  Zn +2 + Cu

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Electrochemistry

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  1. Electrochemistry Guaranteed to give you a jolt

  2. Electrochemical cells • A chemical system in which oxidation and reduction can occur – often a single displacement reaction Zn + Cu+2 Zn+2 + Cu • Oxidation reaction and reduction reaction are physically separated so that useable work can be obtained from the reaction

  3. Electrochemical cells • Voltaic cells (galvanic cells) – redox occurs spontaneously • Anode – where oxidation takes place (solid metal becomes aqueous positive ions) • Cathode – where reduction takes place (metal ions deposit as solid metal) • AN OX RED CAT

  4. Electrochemical cells Electrons flow from anode to cathode through wire cathode anode Salt bridge

  5. Electrochemical cells • Salt bridge – usually KNO3 or a semipermeable membrane– completes circuit by providing mobile ions • Conditions for spontaneity – one metal must lose electrons more easily than another (metals can be identical if one is warmer) p. 288 (activity series)

  6. Electric current • Rate of flow – amperes 1 amp = 1 coulomb/sec 1 coulomb = 1/96490 of a mole of electrons (6.24x1018e-) • Faraday’s number = 96490coul/mole • Potential – volts (joules/coulomb)

  7. Cell potential • Half cell potential (E) – the voltage contribution of a half reaction to the cell • Standard half cell potential (Eº) - voltage when solutions are 1M, gases are 1atm and temp. is 25ºC. • Half cell potentials are given as reductions relative to the reduction of H+ to H2, which is assigned 0 volts.

  8. Cell potential • Half cell potential is a measure of the tendency of a particle to gain electrons. • The cell potential is the sum of the two half cell potentials. • For oxidation, the sign of the reduction potential is reversed. • A positive cell potential means a spontaneous reaction, i.e. a galvanic cell.

  9. Cell potential • Example. Find the cell potential for a cell with a mercury/mercury (II) cathode and a lead/lead (II) anode. Standard reduction potentials: Hg+2 +2e- Hg Eº = 0.851V Pb+2 + 2e-  Pb Eº = -0.1262V • Lead is the anode, so it is oxidized (reduction equation is reversed)

  10. Cell potential Hg+2 +2e- Hg Eº = 0.851V Pb  Pb+2 + 2e-Eº = 0.1262V • Cell potential is the sum of the half cell potentials Hg+2 + Pb Pb+2 + Hg E = 0.851V + 0.1262V = 0.977V • Potential is intensive, so it’s not affected by coefficients

  11. Cell potential • A positive cell potential means a spontaneous reaction, i.e. a galvanic cell. • Galvanic cell calculations • Cell notation Zn|Zn+2║Cu+2|Cu anode salt bridge cathode

  12. Galvanic cell calculations • Vertical lines represent the barrier between two different states of matter. • Two different materials in the same part of the cell are separated by commas. H2,Pt|H+║Ag+|Ag+

  13. Galvanic cell calculations • Calculate the voltage of this cell: Mg|Mg+2║Au+3|Au Mg  Mg+2 + 2e- E º = +2.37V cathode: Au+3 + 3e- Au E º = +1.50V • Total cell voltage = 3.87V

  14. Current calculations • Extent of oxidation/reduction and amount of material oxidized or reduced is directly related to the number of electrons transferred • Moles e- = current x time / Faraday’s # = It/F F = 96485coul/mole

  15. Current calculations • A zinc anode with mass 2.30g is used in a copper/zinc cell. The cell has a current of 0.00140 amps. How long will the electrode last? • Solution: 2.30 g Zn is 0.0352 mole zinc. • Each mole zinc requires 2e-, so 0.0704 mole e- are required to completely oxidize the anode.

  16. Current calculations • There are 96,485 coulombs/mole, so 6790 coulombs are needed. 0.0704mol e- x 96485 coul/mol = 6790c • At 0.00140 coulombs/sec, the electrode will last 4.85x106 sec, or 1350 hours.

  17. Types of cells • Concentration cell – uses identical electrodes – potential difference due to concentration differences in the cell

  18. Nonstandard cells • Ecell = Eºcell – (RT/nF )lnQ • Q = reaction quotient • For the reaction aA + bB  cC + dD, Q = [C]c[D]d/[A]a[B]b • Find the voltage for a zinc/copper cell at 55ºC where the concentrations of zinc and copper are 0.10 and 0.75M respectively.

  19. Types of cells • Leclanché cell (dry cell) MnO2, water, NH4Cl in a paste around a graphite cathode, with a zinc anode

  20. Types of cells • Overall Leclanché cell equation: Zn+MnO2+NH4ClZnCl2+Mn2O3+NH3+H2O Balance it! Zn  Zn+2 + 2e- 2H+ + 2MnO2 + 2e-  Mn2O3 + H2O 2H++2MnO2+ZnZn+2+Mn2O3+H2O Add spectators (NH3 and Cl-): 2NH4Cl+2MnO2+ZnZnCl2+Mn2O3+H2O+2NH3

  21. Types of cells • Alkaline manganese cell – used in alkaline batteries – uses KOH as electrolyte • Zn+MnO2+H2OZn(OH)2+ Mn2O3

  22. Types of cells • Electrolytic cells – nonspontaneous redox reaction is forced to proceed by application of an electric current • Electrolysis of water – Hoffman apparatus 2H2O  2H2+ O2

  23. Electrolysis of aluminum oxide • Alumina (Al2O3) from bauxite is dissolved in molten cryolite (Na3AlF6).

  24. Electrolysis of aluminum oxide • The steel container is coated with carbon (graphite) and serves as the negative electrode (cathode). • Electrolysis of the alumina/cryolite solution gives aluminum at the cathode and oxygen at the anode. • Aluminum is more dense than the alumina/cryolite solution, and so falls to the bottom of the cell.

  25. Electrolysis of aluminum oxide • Aluminum can be tapped off the bottom as pure liquid metal. • The overall reaction is: 2Al2O3(l) 4Al(l) + 3O2(g) • Oxygen is discharged at the positive carbon (graphite) anode. • Oxygen reacts with the carbon anode to form carbon dioxide gas. The carbon anode slowly disappears as carbon dioxide and needs to be replaced regularly.

  26. Anodic protection

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