 Anything that uses batteries: › Cell phones › Game boys › Flash lights › Cars  Jewelry—electroplating

 The study of the interchange of chemical and electrical energy  All electrochemistry reactions are oxidation- reduction reactions.

 A substance that is oxidized loses electrons  A substance that is reduced gains electrons; a reducing agent loses electrons.  LEO says GER  Processes must occur together.

 An oxidizing agent causes something else to be oxidized, so it is reduced (gains electrons).  A reducing agent causes something else to be reduced, so it is oxidized (loses electrons).

 Redox reactions are often broken into two half reactions, one showing the oxidation and the other showing the reduction.  FeCl 2 + Ca  Fe + CaCl 2 › Fe 2+ + Ca  Fe + Ca 2+  Fe 2+  Fe (reduction)  Ca  Ca 2+ (oxidation)

 When the two ions are in the same solution, the electrons are transferred directly from one to the other in a collision.  In order to harness the energy, a flow of electrons (current) must be created.

 A U-tube filled with an electrolyte or a porous disk that allows ions to flow  The purpose of a salt bridge is to prevent the build-up of charge that would stop the flow of electrons.

 Disk with small openings that allows ions to flow back and forth  Like a salt bridge—prevents the build-up of ions on one side of the cell

 A device in which chemical energy is changed to electrical energy  Oxidation occurs at the anode; reduction occurs at the cathode.  An ox; red cat

 Cell potential—the “pull” or driving force on electrons (emf)  Unit = volt  Potentials are calculated by using a standard reference electrode: Hydrogen

 Values for emf at 25 o C and 1M concentration for solutions/1 atm pressure for gases  Table—p. 833

 All reactions are reduction potentials.  One reaction must be reversed to show oxidation.  The sign must change for the reversed reaction.  Coefficients do not matter.  Cell runs spontaneously to produce positive cell potential.

 A galvanic reaction is based on the following reaction: Al 3+ + Mg  Al + Mg 2+  Give the balanced cell reaction and find the cell potential.

 Shorthand to represent a galvanic cell  Mg(s)IMg 2+ (aq)IIAl 3+ (aq)IAl(s)  Anode components (oxidation) on left; cathode components (reduction) on right

1. Cell potential and balanced cell reaction 2. Direction of electron flow 3. Designation of anode & cathode 4. Nature of each electrode and ions present in each compartment

 Completely describe the galvanic cell based on the following half reactions: › Ag + + e -  Ag E = 0.80 v › Fe 3+ + e -  Fe 2+ E = 0.77 V

 w = -q E  w = work  q = quantity of charge transferred ( coulombs/mole electrons)  E = cell potential

 A cell has maximum potential of 2.50 V. If the actual voltage is 2.10 V, how much work could be done by the flow of 1.33 mol of electrons?

 Think of the free energy as the energy that does the work.  If w = -q E, then also  G = -q E  For q you may use nF (n = number of moles of electrons x Faraday's constant.)

 Is this reaction spontaneous? Cu 2+ + Fe  Cu + Fe 2+  G = -q E = -nF E

 Gives relationship between cell potential and concentration of cell components:   G =  G o + RTlnQ  Remember  G = -nF E  So, -nF E = -nF E* + RTlnQ  At 25 o C, E = E* ( /n) log Q

 At 25 o C: E = E* - {(.0591/n) log Q} › E = Cell potential › E* = Standard potential › n = number of moles of electrons › Q = reaction quotient When equilibrium is reached,  G = 0 (Battery is dead.)

 Opposite of galvanic  Uses electrical energy to produce chemical change (electrolysis)

 Stoichiometry problems  1ampere (A) = 1 Coulomb/s  1 mole e - = Coulombs  Current & time  quantity of charge in coulombs  moles of electrons  moles of metal  grams of metal

 How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g of sliver metal?

 Using a current of 1.00 x 10 6 A for 2.00 hours, what mass of aluminum can be produced from aluminum oxide?