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ELECTROCHEMISTRY

ELECTROCHEMISTRY. 1 Using chemical reactions to produce electricity (Electrochemical cell) 2 Using electrical energy to cause chemical reactions (Electrolytic Cell). Electrochemical, Galvanic, or Voltaic cells. Using redox reactions to produce electricity. V. Anode

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ELECTROCHEMISTRY

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  1. ELECTROCHEMISTRY 1 Using chemical reactions to produce electricity (Electrochemical cell) 2 Using electrical energy to cause chemical reactions (Electrolytic Cell)

  2. Electrochemical, Galvanic, or Voltaic cells Using redox reactions to produce electricity V Anode Oxidation Electrons lost Cathode Reduction Electrons gained electrolyte electrolyte Salt bridge

  3. Electrochemical cell example Zn 2+ + 2e- Zn Eo = -.34 V Cu2+ + 2e- Cu Eo=.76 V Eo is the potential of a substance to gain electrons from H+ (reduction potential) at standard conditions (1M, 1 atm., 25oC) The cell with the lowest Eo is oxidized, the equation and voltage are reversed. Zn  Zn2+ + 2e- Eo = .34 V anode (oxidation) Cu2+ + 2e- Cu Eo= .76 V cathode ( reduction) Cu2++Zn  Cu + Zn2+ Eo= 1.10 V cell reaction

  4. Tracing the “life” of the voltaic cell 2 electrons are lost from a Zn atom producing a zinc ion in solution. Zn  Zn2+ + 2e- 2 negatively charged ions enter the electrolyte solution from the salt bridge to keep the electrical neutrality. The 2 electrons from zinc collect on the copper electrode surface, attracting a copper ion from the solution. Cu2+ + 2e- Cu 2 positive charges enter the electrolyte solution from the salt bridge to keep the electrical neutrality. As the process continues, the zinc gets smaller, the copper electrode gets larger, the voltage drops. When the electrons stop moving from zinc to copper, the voltage is zero and the reaction is at equilibrium. E= 0 => at equilibrium

  5. Try this example: Fe/ Fe3+ (1M) // Ag+ (1M)/Ag Standard cell notation shows the anode metal/ anode electrolyte solution//(salt bridge)/cathode electrolyte solution/ cathode metal Fe3+ + 3e- Fe Eo = -.036V (smallest must be reversed) Ag+ + e- Ag Eo = .80V (multiply by 3* to balance) 3Ag+ + Fe  3Ag + Fe3+ Eo = .836V *Notice the Ag cell equation was tripled, but the Eo was not!

  6. If Eo is negative, the cell is not spontaneous as written, but is spontaneous in the reverse direction. Does it seem there might be a connection between cell potential and DGo ?

  7. What happens if the concentrations are not = 1M? The Nernst equation relates cell potential (E), reactant concentrations, and the Eo. E = Eo –(nF/RT)ln Q -nF/RT can be expressed as .0592/n …. n = number of electrons lost or gained in the reaction. E = Eo – (.0592/n) log Q ; Q= [Ox]/[red]

  8. Try this example: Find the potential of a cell made from Ni/ Ni2+ (.01M) and Cd/Cd2+(2.0M) Ni2+ + 2e- Ni Eo= -.23V Cd2+ + 2e- Cd Eo= -.40V reverse and add Cd + Ni2+ Cd2+ + Ni Eo = .17 V E = .17V – (.0592/2 )log [2M/.01M] E = .10 V

  9. How about this cell? Cu/ Cu2+ (.002M) // Cu2+ (.50M)/Cu How can electrons flow from Cu to Cu? There are more cations in the “right” cell than the “left”, the system is not in equilibrium until the [Cu2+ ] is the same in both cells. So oxidation occurs in the .002M cell( increasing [Cu2+ ]in the “left” cell) and reduction occurs in the .50M cell (decreasing the [Cu2+ ]in the “right” cell) E = 0.00 –(.0592/2) log[.002/.5] ; E= .0709 V

  10. Electrolytic Cells …. Reactions caused by application of electrical energy. Faraday’s law of electrolytic cells: The quantity of chemical change is directly related to the quantity of charge added to the cell. Electroplating …. Using external source of electrons to reduce cations onto the surface of an electrode. Electrolysis …. Separation of compounds using electrical energy

  11. Electroplating example How many g of Ag will be plated onto a steel spoon if a current of .750 amps runs for 30.0 minutes? Amp = ampere = coulombs/sec = charge/time= current F = Faraday = 96, 500 coul/mol of e- Mass = (I)t(gfm)/(96500 x n) (.750 coul/sec)1800sec(108 g/mol Ag) = 1.51 g Ag (96500 coul/ mol e-)(1 mol e-/ mol Ag)

  12. Electrolysis of water Passing an electric current through a solution of water and sulfuric acid or other salts not containing Cl-, Br-, I- causes water to decompose into hydrogen and oxygen. (These halides oxidize to halogens) Anode reaction 2H2O  O2 +4H+ + 4e- Cathode reaction 4H2O + 4e-  2H2 + 4OH- Net rxn 2H2O  2H2 + O2

  13. Try this example: Find the volume of hydrogen produced at STP when a current of 1.00 amp is run through a water sulfuric acid solution for 2.00 minutes. 1.00 coul/sec(120sec) = .00125 mol of electrons 96485coul/ mol e- 2 mol e- / mol H2 (previous slide) .00125 mol e- / (2 mol e- / mol H2) = .000625 mol H2 V = nRT/P ; V= .014 L H2

  14. The DGo ,Eo , K connection When E = 0 a cell is at equilibrium and DG = 0 DGo = -nFEo = -(RT) ln K : ln(K) = e-nFE/-RT Find the DGo and K of a Zn/Zn2+(1M) //Cu2+ (1M)/Cu cell Cell rxn Zn + Cu2+ Zn2+ + Cu Eo= 1.10 V DGo = -2 mol e- (96500coul/ mol e- )1.10 J/coul= -212,300J K = e(-212300 / -[8.314 x 298] = 1.64 x 1037

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