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Chapter 20

Chapter 20. Electrochemistry. Oxidation States electron bookkeeping * NOT really the charge on the species but a way of describing chemical behavior. Oxidation: Loss of electrons Reduction: Gain of electrons

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Chapter 20

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  1. Chapter 20 Electrochemistry

  2. Oxidation States electron bookkeeping * NOT really the charge on the species but a way of describing chemical behavior. Oxidation: Loss of electrons Reduction: Gain of electrons Oxidizer: Oxidizes another species, it gets reduced. Also called oxidizing agent or oxidant

  3. Reducing agent or reductant: Reduces another species, gives up electrons, it is oxidized Many reactions are oxidation-reduction (redox) reactions I2O5(s) + 5CO(g) I2(s) + 5CO2(g) 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) -2 0 +4 -2 +5 -2 +2 -2 +1 +5 -2 +2 -2 +4 -2 +1

  4. Balancing Redox Reactions

  5. The following steps summarize the procedure that we use to balance an oxidation-reduction equation by the method of half-reactions when the reaction occurs in acid solution: • Divide the reaction into two complete half-reactions, one for oxidation and the other for reduction. • Balance each half-reaction • First, balance the elements other than H and O • Next, balance the O atoms by adding H2O. • Then, balance the H atoms by adding H+ • Finally, balance the charge by adding e- to the side with the greater positive charge • Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other. • Add the two half-reactions and simplify where possible by canceling species appearing on both sides of the equation. • Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

  6. Voltaic Cells E released from redox reaction can be used to perform electrical work. anode = electrodes where oxidation occurs cathode = electrode where reduction occurs reduce the cat population

  7. Half Reactions Zn(s) Zn2+ + 2e- Cu2+ + 2e-  Cu(s) Zn electrode  mass because Zn(s) Zn2+ + 2e- Cu electrode  mass because Cu2+ + 2e-  Cu(s)

  8. Zn Cu2+ Zn2+ Cu2+ CuSO4 Cu2+ SO42- Zn2+

  9. Cell EMF Electromotive force E is favorable for e- to flow from anode to cathode e-’s at higher potential E in Zn electrode than in Cu electrode What do you know about E? It tends to dissipate. Lower E state is favorable!

  10. Which reaction at anode? Which reaction at cathode? Won’treact Pt electrode but will conduct e- e- e- e- V.M. Anode (-) (+) Salt bridge Which electrode is (+) ?

  11. For the reaction Flow of e- allows the reaction to occur e- e- e- Cu cathode Zn anode V.M. (-) (+) NO3- Na+ Salt bridge: can’t just build up a charge in the cell, keep neutrality by ions in salt bridge. Solution of ZnSO4 Solution of CuSO4 Zn2+ Cu2+

  12. The Potential Difference between two electrodes is measured in volts 1V = 1J/C (energy/unit charge) Volts = Electrical pressure or electromotive force The driving force for reaction to take place. Potential difference = emf = Ecell = cell voltage Measured in volts

  13. Standard emf = standard cell potential, Ecell Standard conditions: 1M concentration for reactants and products and 1atm pressure when gases are used. *Standard Electrode Potentials Ecell = Eox + Ered

  14. Emf and Free Energy G = free energy Energy produced by a reaction that can be used for work G is a measure of spontaneity G = -nFE

  15. N = number of moles of e-transferred E = emf of cell +E = spontaneous -G = spontaneous

  16. Concentration and Cell EMF EMF we have looked at so far has been at standard condition. i.e. 1 molar solutions ? What if you change concentration A guy named Nernst worked it out for you! G = G0 + RT ln Q Q = reaction quotient like equilibrium expression but not at equilibrium (ion product)

  17. Well! Since G = -nFE -nFE = -nFE0 + RT ln Q Solve for E NOW you can measure E and determine concentration of reactant or product

  18. ? emf of at 25°C [Al3+] = 4.0x10-3M [I-] = 0.010M step 1)what’s E0 Al  Al3+ + 3e- I2 2I- + 2e- step 2)R = 8.314 J/K·mol F = 96,500 n = 6 V +1.66 +0.536 E0 = 2.196 Find values for all variables and constants

  19. Q = [Al3+]2[I-]6 = (4.0x10-3)2(0.010)6 = 1.6x10-17 log Q = -16.8 step 3) Plug and Chug

  20. Equilibrium Constants When E = 0, Q = Kc The cell is no longer active NO NET REACTION Substitute in Nernst Equation at 298K Can calculate Kc from E0 for cell. Just assume equilibrium and replace E with 0 and Q with Kc

  21. amps x seconds = coulombs ex. current 60.0A 4.00x103 s How much Mg can be formed from Mg2+ 1) coulombs = (4.00x103s)(60.0A) = 2.4x105 coulombs

  22. Quantitative Aspects of Electrolysis Fe2+ + 2e- Fe 1 mol of Fe2+ needs 2 mol of e- to make 1 mol Fe 1/2 mol Fe2+ need 1 mol e- to make 1/2 mol Fe

  23. Al3+ + 3e- Al For 1 mol Al to form 1 mol Al3+ it must give up 3 mol e-. For 1/2 mol Al to form 1/2 mol Al3+ it must give up 3/2 mol e-. NOW, Can I measure electron flow? You betcha! e-

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