Chapter 20

 Involves the transfer or flow of electrons in a chemical reaction  This flow of electrons results in changes of charges (aka oxidation states) and the production of electricity

 Oxidation state = number assigned to an atom within an element, ion, or compound that indicates its “charge”  Rules for assigning “total” oxidation states 1. Total charge of elements = 0 2. Total charge of polyatomic ions = charge 3. Total charge of compounds = 0

 Rules for assigning individual atomic oxidation states 1. Metal cations = charge from the P.T 2. Hydrogen = +1 (in almost all cases) 3. Oxygen = -2 (in almost all cases) 4. Most electronegative element (upper right hand corner of P.T.) will be negative

1. Br 2 Br: _____ 2. Au(IO 3 ) 3 Au: ____I: _____O: _____ 3. K 2 Cr 2 O 7 K: ____Cr: ____O: _____

4. H 2 SH: _____S: ____ 5. MgCO 3 Mg: ____C: ____O: ____ 6. PO 4 -3 P: ____O: ____

4. H 2 SH: 1+S: MgCO 3 Mg: 2+ C: 4+O: PO 4 -3 P: 5+O: 2- Checkpoint: Of the three examples, how many did you get correct?

 Electrons are being lost  The atom’s oxidation state becomes more positive. (losing negative charge) ◦ Li  Li +1 + e - ◦ N -3  N + 3e - ◦ Zn  Zn +2 +2e - P O xidation S I T I V E

 Electrons are being gained  The atom’s oxidation state becomes more negative. (adding negative charge) ◦ Cl + e -  Cl -1 ◦ Mg e -  Mg ◦ Co +3 +e -  Co +2 N R E duction G A T I V E

 “Leo” the Lion goes “Ger” ◦ L = Losing G = Gaining ◦ E= ElectronsE = Electrons ◦ O = Oxidation R = Reduction  “Oil Rig” ◦ O = OxidationR = Reduction ◦ I = Is I = Is ◦ L = Losing G = Gaining

 Assign oxidation states and determine if oxidation, reduction, or neither is taking place 1. SO 3 -2  SO CaO  Ca 3. CrO 4 -2  Cr 2 O 7 -2

1. SO 3 -2  SO 4 -2 S:4+  6+oxidation 2. CaO  CaCa: 2+  0reduction 3. CrO 4 -2  Cr 2 O 7 -2 Cr: 6+  6+neither  Checkpoint: Of the three examples, how many did you get correct?

 The word “Redox” comes from the 2 words: reduction and oxidation because both occur simultaneously ◦ One atom is gaining electrons (reduction) ◦ One atom is losing electrons (oxidation)  A “Redox” reaction is one in which there is a flow or transfer of electrons

1. FeCl 3 (aq) + SnCl 2 (aq)  SnCl 4 (aq) + FeCl 2 (aq) 2. MnO 2 (s) + HCl (aq)  MnCl 2 (aq) + Cl 2 (g) + H 2 O (l)

1. Fe +3  Fe  +2 reduction; gain e - Sn +2  Sn  +4 oxidation; lose 2e - 2. Mn +4  Mn  +2 reduction; gain 2e - 2Cl -1  Cl 2 -1  0 oxidation; lose 2e -

 Identify and label parts of a galvanic (electrochemical) cell  Calculate the cell potential

 A reaction in which oxidation and reduction occurs.  An electric current is created by the flow of electrons lost then gained.

 Complete Redox Reaction: Oxidation and reduction occurs simultaneously  Half Reaction : Either oxidized or reduced onlyAND includes the electrons in the reaction.

1. 2 HCl  H 2 + Cl 2  2 H +  H 2 +2  0 reduction  2 Cl -  Cl 2 -2  0 oxidation Half Reactions  2 H + + 2e -  H 2  2Cl -  Cl 2 + 2e -

2. Co + Fe +3  Co +2 + Fe +2  Co  Co +2 (0  +2) oxidation  Fe +3  Fe +2 (+3  +2) reduction

2. Co + Fe +3  Co +2 + Fe +2 Half Reactions  Co  Co e -  Fe +3 + e -  Fe +2

3. Al + H +1  Al +3 + H 2  Al  Al +3 (0  +3) oxidation  2H +1  H 2 ( +2  0) reduction

3. Al + H +1  Al +3 + H 2 Half Reactions  Al  Al e - oxidation  2H + + 2e -  H 2 reduction

 Electrochemical cell: a system of materials that connects 2 half reactions to produce an electric current through transfer of electrons

 Electrode: conducting material (usually metal) that comes in contact with the aqueous solution.  Anode (-): Electrode where oxidation occurs ◦ source of negative electrons in the cell.  Cathode (+): Electrode where reduction occurs ◦ attracts the negative electrons

A Negative O D E CA+HODECA+HODE Electrons flow from Anode to C athode (alphabetical!)

◦ Reduction: CathodeAnode: Oxidation RED CATAN OX

Zn + CuSO 4  ZnSO 4 + Cu Half Reactions Zn  Zn e - (oxidation) Cu e -  Cu (reduction)

 Voltage: A measurement of the overall “pull” on electrons.  Standard Reduction Potential, E o : The relative “strength” of an element’s pull on electrons ◦ The more positive value, the more pull on electrons, therefore the more potential to undergo reduction ◦ Compared to a Hydrogen standard (Defined as 0.00 V)

 Half Cell: The cell in which only a half reaction occurs  Cell Potential (E cell o ) : Comparison of voltages between two half cells (electrodes) ◦ The more positive the value, the more likely the reaction will occur (spontaneous)

1. Determine half reactions 2. Find red. Pot. For each half 3. If necessary, flip the potential for the oxidation reaction. 4. Add half cell potentials to find total cell potential (E o cell )

1. Calculate E° Cell in the full redox reaction CuCl 2 + Zn ----> ZnCl 2 + Cu

2. Calculate E° Cell when the following half cells are connected.  Fe e -  Fe E°: V  Ag +1 + e -  Ag E°: V

3. Calculate E° Cell when the following half cells are connected.  Hg e -  Hg E°: V  Cr e -  Cr E°: V

 Voltaic or Galvanic Cell: reaction is spontaneous (+V) ◦ Produces an electrical current ◦ Used in batteries  Electrolytic Cell: reaction is not spontaneous (-V) ◦ Must be connected to a power source to react ◦ Electroplating, Electrolysis of Water

 Calculate the E cell for each reaction and then determine whether the rxn is spontaneous 1. Cr (s) + Ni +2 (aq)  Cr +3 (aq) + Ni (s)

2. Br 2(g) + I - (aq)  Br - (aq) + I 2(s)

1. Assign oxidation states 2. Remove spectator ions 3. Split into half reactions. Add e - to half reactions to balance charges. 4. Balances charges with H + ions and oxygen with H 2 O 5. Multiply half reactions to get equal e - 6. Combine to write whole reaction

Zn  Zn e - 2V e -  2V +4 Zn (s) + 2V +5 (aq)  2V +4 (aq) + Zn +2 (aq)

Zn +2  Zn e - 2e _ + F 2  2F - Zn +2 (aq) + F 2 (g)  Zn +4 (aq) + 2 F -1 (aq)

Cu  Cu e - 2Ag +1 +2e -  2Ag Cu (s) + 2Ag +1 (aq)  Cu +2 (aq) + 2Ag (s)

2(x3)Cl -  3Cl 2 + 2(x3)e - 14H + + 6e - + Cr 2 O 7 -2  2Cr H 2 O 14H + (aq) + 6Cl - (aq) + Cr 2 O 7 -2 (aq)  3Cl 2 (g) + 2Cr +3 (aq) + 7H 2 O (l)

Cl 2 + 2e -  2Cl - 2I -  I 2 + 2e - Cl 2 (g) + 2I - (aq)  I 2 (s) + 2Cl - (aq)

5S -2  5S + 2(x5)e - 8(x2)H + + 5(x2)e - + 2MnO 4 -1  2 Mn (x2)H 2 O 5S -2 (aq) + 16H + (aq) + 2MnO 4 -1 (aq)  2Mn +2 (aq) + 5S (s) + 8H 2 O (l)