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Electrochemistry

Electrochemistry. Other Applications of Redox Chemistry. Redox and the use of electrical cells. A cell is a device that is either used to pass electricity through a solution for purification or separation purposes, or a device that produces electricity, i.e. a battery.

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Electrochemistry

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  1. Electrochemistry Other Applications of Redox Chemistry

  2. Redox and the use of electrical cells A cell is a device that is either used to pass electricity through a solution for purification or separation purposes, or a device that produces electricity, i.e. a battery. Cells in which we supply the current are called electrolysis cells. or electrolytic cells. The process of passing a current through something is called electrolysis.

  3. Cells that produce electrical current are called electrochemical cells, Galvanic cells, or voltaic cells. These types of cells are behind the technology of batteries that we use everyday. There are two types, wet and dry cell batteries. Car batteries are wet cell, and flashlights use dry cell batteries.

  4. Electrolysis- passing of a current through a solution or molten compound. • The apparatus used to accomplish electrolysis is an electrolysis cell, which causes a redox reaction to occur. • Terminology- • Anode- oxidation occurs here in all cells • Cathode- reduction always occurs here • Electrons move through metal and ions transport charge by conduction

  5. Electrolysis cell

  6. Solutions of salts will always behave the same when being electrolyzed. All Cu (II) salts will yield Cu+2 ions that will reduce to Cu metal. Cu+2 + 2e-1 Cu. • All Bromide salts will yield Br-1 which is oxidized to Br2. 2 Br-1  Br2 + 2 e-1

  7. When electrolysis occurs in solution, the reactions are sometimes difficult to predict because the water may be oxidized or reduced. An example of this is potassium nitrate in water. • Generally when a group I or II metal is in the solution, the water will be reduced because the metal would react violently with water. 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq)

  8. Stoichiometry of Electrolysis • Redox involves e-, and current is the flow of e-. Charge is defined as (current) (time) • Charge is measured in coulombs, and: • C = (amperes) (time) • The charge of one mole of electrons is called a Faraday and is equal to 9.65 x 104 C This allows us to convert current into stoichiometric amounts of metals electroplated by the cell

  9. How many grams of Cu(s) are deposited on the cathode of a cell if 2.00 amps runs through the Cu+2 solution for 20.0 minutes? • Find total charge, as that can be related to electrons and thus to moles of copper. • 20 min= 1200 sec x 2.00 amps= 2400C Cu+2 + 2 e- Cu (s) X g Cu = 2400 C x 1 mol e-1 x 1 mol Cu x 63.55 g Cu 9.65x 104C 2 mol e- 1 mol Cu 0.790 g Cu

  10. How much time would be required to deposit 0.500 g of Ni(s) on an object using a 3.00 A current in a cell reducing Ni+2? • Ni+2 + 2 e- Ni(s) • X C = .500g Ni x 1 mol Ni x 2 mol e-x 9.65x 104 C 58.69 g Ni 1 mol Ni 1 mol e- 1644 C = (3.00 A)(x sec) 1644/3.00 = x sec = 548 sec or 9.13 min

  11. Find the current required to deposit 0.500 g of Cr metal from a solution of Cr+3 in a period of 1.00 hr. Cr+3 +3 e- Cr (s) X C = .500 g Cr x 1 mol Cr x 3 mol e- x 9.65x 104 C 52.00 g Cr 1 mol Cr 1 mol e- 2.784 x 103 C = (x A)(3600 sec) 2.784 x 103C/3600 sec = 0.77 A

  12. Galvanic cells- producing electricity

  13. Galvanic cells are a way to harness the electrical potential that exists and drives spontaneous redox reactions. • Oxidation at anode but negative charge • Reduction at cathode but pos charge • Electrons flow from one to the other due to potential difference. We must isolate the cells to harness energy. We must have either a salt bridge or porous plate between the cells to allow for ions to move and counteract charge build-up.

  14. Voltaic cell

  15. In each cell, there is a redox reaction occurring. • Cu+2 + 2 e- Cu (s) red cathode Zn  Zn+2 + 2e- oxidation anode • Each of these reactions has a certain electrical potential and the potential of the entire cell will be the sum of both reactions. • Ecell = Eox + Ered The potentials for the reactions are given on a table of standard reduction potentials.

  16. The table of reduction potentials lists the reduction potentials for various reactions. We use this table to determine the voltages for our reactions. • Cu+2 + 2 e- Cu (s) Eº = 0.34 V Zn  Zn+2 + 2e- Eº = 0.76 V • Zn + Cu+2 Cu + Zn+2 Eºcell= 1.10V • This is the amount of voltage produced by the cell in the diagram.

  17. When using the reduction potential tables, one of the species must be oxidized in our cell, so it will be the reverse reaction of what appears on the table. Since the reaction is reversed, the potential listed on the table will have the opposite sign from the potential listed on the table. • Zn  Zn+2 + 2e- Eº = 0.76 V • On the table Zn+2 + 2e-  Zn Eº = -0.76V • We reverse the reaction, so potential becomes +0.76 V.

  18. NO matter what reduction potential table you use, the species with the more positive reduction potential will be reducing in your reaction, and the species with the more negative reduction potential will be forced to oxidize, and its potential will be reversed. With the tables you will have, the species higher on the table will reduce, and the species lower on the table will reverse. The more positive the reduction potential, the stronger that species is as an oxidizing agent.

  19. Arrange the following in order of increasing strength as oxidizing agents. MnO4- (in acid), Sn+2, Al+3, Co+3, and Ag+ in their standard states. For electrochem, standard states are 298 K, 1 M solutions and 1 atm for gases. Al+3 < Sn+2<Ag+<MnO4-1<Co+3

  20. Cell Notation • To describe voltaic cells, a standard method of notation has been adopted. This standard cell notation for a cell involving Al oxidizing to Al+3 and Cu+2 reducing to Cu would be as follows: • Al | Al+3 || Cu+2 | Cu • The single lines indicate a boundary between phases and the double lines indicate the salt bridge or barrier between the cells. The anode (oxidation) is on the left and the cathode (reduction) is on the right.

  21. Al | Al+3 || Cu+2 | Cu • Al  Al+3 + 3e- Eº = 1.66 V • Cu+2+ 2e-  Cu Eº = 0.34 V • Eºcell = 2.00 V • Even though the balanced equation for this reaction would be 2 Al + 3 Cu+2 2 Al+3 + 3 Cu , The standard potential will only be 2.00 V as calculated from the table, the potentials will not be stoichiometric.

  22. Spontaniety of Galvanic cells • Wmax = Wele = - (n)(F)(Ecell); work leaves cell so its negative ΔG = Wmax so, ΔG = - (n)(F)(Ecell) Since ΔG must be negative for spontaneous process, Ecell must be positive for Rxn spontaniety.

  23. Ecell and Keq • ΔG = - R T ln K • - n F Ecell = - R T ln K = 2.303(RT) log K • Ecell = 0.0591 V (log K) n F n If the cell potential of a cell involving 2 moles of electrons is 0.15 v, find Keq for the reaction. 0.15 = 0.0591 log K 2 log K = 5.07 : K = 1.19 x 105

  24. ΔG , K and Ecell • ΔG K Ecell Rxn Negative >1 pos spont 0 =1 0 at EQ Pos <1 neg nonspont (rev spont)

  25. Which acid will oxidize Hg  Hg2+2, HCl or HNO3? • 2 Hg(l)  Hg2+2 + 2e- Eº = -0.85 v • 2H+ + 2e-  H2(g) E º = 0.0 v • Ecell = -0.85 v not spont 6 Hg(l)  3Hg2+2 + 6e- Eº = -0.85 v 2NO3-1 + 8H+ + 6e-1  2NO + 4 H2O Eº= 0.96 v Ecell = 0.11 v HNO3 spont.

  26. Find Keq for the reaction: • Sn(s)|Sn+2 || Cu+2|Cu+1 • Sn  Sn+2 + 2e- Eº = +0.14 v • 2Cu+2 + 2 e-  2Cu+1 Eº = + 0.15 v • Eº cell = 0.29 v • 0.29 v = 0.0591 log K 2 9.81 = log K K = 6.5 x 109

  27. Concentration and EMF When solutions are 1.0 M and gases are at 1 atm, Ecell = Ecellº. As reactants are consumed, Ecell decreases, until equilibrium is established and the battery is dead, Ecell = 0. To find Ecell at any point during the reaction, when concentrations are not 1 M, we use the Nernst Equation Ecell = Ecellº - 0.0591 V log Q n (From ΔG = ΔGº + RT ln Q)

  28. Zn(s)|Zn+2|| Cu+2|Cu(s) • Zn(s) + Cu+2 Zn+2 + Cu(s) Ecell = 1.10 V - 0.0591V log [Zn+2] 2 [Cu+2] If Zn+2/Cu+2 is < 1, Ecell will be greater than Ecellº. If Zn+2/Cu+2 > 1, then Ecellwill be less than Ecellº.

  29. Co(s)|Co+2 || Fe+2|Fe(s) • If [Co+2] = 0.15 and [Fe+2] = 0.68, find Ecell and determine if the reaction will proceed spontaneously. • Co(s) + Fe+2 Co+2 + Fe(s) • Co(s)  Co+2 +2e- Eº = 0.28 v Fe+2 + 2e- Fe(s) Eº = -0.44 v E°cell = -0.16 v Ecell = -0.16v - 0.0591v log [Co+2] 2 [Fe+2] Ecell = -0.16 + log 0.22 = -0.16 + 0.019 = - 0.14 V Not spontaneous

  30. Zn(s) + 2H+(aq,?M)  Zn+2(aq,1 M) + H2(g,1atm) Given that the Ecell = 0.54v, find H+and pH of the solution. Zn(s)  Zn+2 + 2e- E° = 0.76 v = E°cell 2H+ + 2e-  H2(g) E° = 0.0 v 0.54 = 0.76 v - 0.0591v log [Zn+2][H2] 2 [H+]2 -0.22 = -0.0591 log 1 2 [H+]2 7.4 = -2 log [H+] log [H+] = -3.7; [H+] = 2.00 x 10-4 ; pH = 3.70

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