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 Oxidizer: Oxidizes another species, it gets reduced. Also called oxidizing agent or oxidant

Reducing agent or reductant: Reduces another species, gives up electrons, it is oxidized Many reactions are oxidation-reduction (redox) reactions I 2 O 5(s) + 5CO (g)  I 2(s) + 5CO 2(g) 3NO 2(g) + H 2 O (l)  2HNO 3(aq) + NO (g)

Balancing Redox Reactions

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 H 2 O. 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.

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

Half Reactions Zn (s)  Zn e - Cu e -  Cu (s) Zn electrode  mass because Zn (s)  Zn e - Cu electrode  mass because Cu e -  Cu (s)

Zn CuSO 4 Cu 2+ SO 4 2- Zn 2+ Cu 2+

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!

Salt bridge Won’t react Pt electrode but will conduct e - Which electrode is (+) ? V.M. (-) Anode e- e-  e- e-  e- e-  (+) Which reaction at anode? Which reaction at cathode?

Salt bridge: can’t just build up a charge in the cell, keep neutrality by ions in salt bridge. Cu cathode V.M. (-) Zn anode e- e-  e- e-  e- e-  (+) Solution of ZnSO 4 Zn 2+ Cu 2+ Solution of CuSO 4 NO 3 - Na + For the reaction Flow of e - allows the reaction to occur

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 = E cell = cell voltage Measured in volts

Standard emf = standard cell potential, E cell Standard conditions: 1M concentration for reactants and products and 1atm pressure when gases are used. *Standard Electrode Potentials E cell = E ox + E red

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

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

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 =  G 0 + RT ln Q Q = reaction quotient like equilibrium expression but not at equilibrium (ion product)

Well! Since  G = -nFE -nFE = -nFE 0 + RT ln Q Solve for E NOW you can measure E and determine concentration of reactant or product

? emf ofat 25°C [Al 3+ ] = 4.0x10 -3 M [I - ] = 0.010M step 1)what’s E 0 Al  Al e - I 2  2I - + 2e - step 2)R = J/K·mol F = 96,500 n = 6 V E 0 = Find values for all variables and constants

Q = [Al 3+ ] 2 [I - ] 6 = (4.0x10 -3 ) 2 (0.010) 6 = 1.6x log Q = step 3) Plug and Chug

Equilibrium Constants When E = 0, Q = K c The cell is no longer active NO NET REACTION Substitute in Nernst Equation at 298K Can calculate K c from E 0 for cell.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

amps x seconds = coulombs ex. current 60.0A 4.00x10 3 s How much Mg can be formed from Mg 2+ 1) coulombs = (4.00x10 3 s)(60.0A) = 2.4x10 5 coulombs

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

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