1. Chapter 20
Electrochemistry -study of the relationships between electricity and chemical reactions *remember redox *LEO *GER -loss or gain of e- will be determined by assigning oxidation # -page 828

2. -to have redox reaction, both oxidation and reduction must occur oxidation: -oxidation # of an element increases -element loses e- -compound adds O -compound loses H -half-reaction has e- as products reducing agent- substance that is oxidized

3. reduction: -oxidation # of an element dec
reduction: -oxidation # of an element dec. -element gains e- -compound loses O -compound gains H -half-reactions have e- as reactants oxidizing agent- substance that is reduced

4. Cd  loses e- = oxidized/reducing agent
Ex:
Cd(s) + NiO2(s) + 2H2O(ℓ)  Cd(OH)2(s) + Ni(OH)2(s)
Identify what is oxidized and reduced and the oxidizing and reducing agents.
Cd  loses e- = oxidized/reducing agent
NiO2 gains e- = reduced/oxidizing agent

5. Balancing Redox Reactions half-reactions- show either oxidation or reduction alone *shows where e- are lost or gained (must be =) *page 830

6. -redox reactions involve the transfer of e- from one substance to another -redox reactions have the potential to generate an electric current -in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring (half-cells)

7. voltaic/galvanic cell -a device in which chemical energy is changed to electrical energy (spon. reactions) *e- flowing through a wire and ions moving in solution both constitute an electrical current electrodes- solid metals connected by an external circuit which allow transfer of e- anode- electrode where oxidation occurs cathode- electrode where reduction occurs e- consumed here half-cell- each compartment of a galvanic cell page 836 figure 20.5

8. -one half-cell is the site of the oxid
-one half-cell is the site of the oxid. half-reaction and the other the site of the red. half-reaction salt bridge- used to separate two half-cells and allows ions to migrate and maintain electrical neutrality of the solutions (contains an electrolyte solution) *anions always migrate towards anode *cations migrate towards cathode *e- flow from anode (-) to cathode(+)

10. *e- flow spon. from anode to cathode b/c of difference in potential energy -potential energy of e- in anode is higher than in cathode (e- flow from anode to cathode) -difference in PE measured in volts (V) *1V = 1J/C(coulomb) -PE difference between the two electrodes is the cell potential (Ecell) *also called electromotive force (emf) *Ecell of any voltaic cell is +

11. -will look at cells under standard conditions -called standard reduction potential *measured values pg 841and 1064 Eºcell = Eºred(cathode) – Eºred(anode)

12. -standard reduction potential provides a measure of the tendency of reduction to occur *the more + E°red, the greater the tendency for reduction under standard conditions *when we assign an electrical potential to a half reaction we write the reaction as a reduction (e- always a reactant) *changing stoichiometry does not affect the value of std red potential

13. Free Energy -measure of the spontaneity of a process that occurs at constant temp and pressure ∆G = -nFE n = # of e- transferred from balanced equation F = Faraday’s constant (96485 C/mol or J/Vmol) E = standard cell potential (V) *-G and +E indicate spont. reaction -units are kJ/mol of reaction

14. -nonspon. redox reactions occur using electrical energy to drive them -called electrolysis reactions -take place in electrolytic cells -consists of two electrodes immersed either in molten salt or a solution -a battery acts as an e- pump, pushing e- into one electrode and pulling them from another -reduction occurs at cathode -oxidation at anode Page 860

15. -electrolysis can be used for electroplating (deposits a thin layer of one metal on another to improve beauty or resistance to corrosion) -stoichiometry of half-reactions shows how many e- are needed for electroplating Ex: Na+ + 1e-  Na *1mol of e- plates out 1mol of Na Ex: Cu2+ + 2e-  Cu *2mol of e- are required to produce 1mol of Cu

16. -quantity of charge passing through cell is measured in coulombs
-quantity of charge passing through cell is measured in coulombs *1 mol of e- = 96485C -a coulomb is the quantity of charge passing a point in a circuit in 1s when the current is 1ampere (A) *coulombs = amperes x seconds *1 Faraday = 96485C