2. Regents Warm-Up Solid samples of the element phosphorus can be white, black, or red in color. The variations in color are due to different (1) atomic masses (2) molecular structures (3) ionization energies (4) nuclear charges

3. Redox Half Reactions AIM: How do we identify oxidation/reduction half reactions? DO NOW: What is the oxidation number of chromium in the chromate ion, CrO4 2 – ? (1) +6 (3) +3 (2) +2 (4) +8

4. Redox Half Reactions Example – for the reaction Zn 0 (s) + Co +3 (aq)  Zn +2 (aq) + Co 0 (s) 1.Identify, write and balance the oxidation half reaction. 2.Identify, write and balance the reduction half reaction. 3.Balance the electrons that are transferred.

5. ID Oxidation Half Rxn 1.The oxidation: STEP ONE: find the species whose oxidation state INCREASES across the arrow. In this case it is the zinc. STEP TWO: write the species as a half reaction, check to balance mass ( equal number of each element on each side);Zn 0 (s)  Zn +2 (aq) STEP THREE: add electrons to the more positive side to balance charge Zn 0 (s)  Zn +2 (aq) + 2e - Here you add 2e- to the right cancel the 2+ to 0.

6. ID Reduction Half Rxn 2. The reduction: STEPS ONE: find the species whose oxidation state DECREASES across the arrow. In this case it is the cobalt. STEP TWO: write the species as a half reaction, check to balance mass ( equal number of each element on each side);Co 3+ (s)  Co 0 (aq) STEP THREE: add electrons to the more positive side to balance charge 3e- + Co 3+ (s)  Co 0 (aq) Here you add 3e- to the left cancel the 3+ to 0.

7. Balancing Half Rxns 3.Balance the electrons released from the oxidation with the electrons consumed in the reduction, make them equal Zn 0 (s)  Zn +2 (aq) + 2e- oxidation releases 2 mol of e- 3e- + Co 3+ (s)  Co 0 (aq) reduction consumes 3 mol of e-

8. Balancing Half Rxns In this example, 2 and 3 are factors of 6, therefore multiply the entire oxidation by 3, and the entire reduction by 2 such that each half reaction has 6 electrons. 6 is the LCM here. 3 Zn 0 (s)  3 Zn +2 (aq) + 6e- 6e- + 2 Co 3+ (s)  2 Co 0 (aq) the electrons now cancel across the arrow and the half reactions can be added.

9. Balancing Half Reactions The total overall rxn is: 3 Zn 0 (s)  3 Zn +2 (aq) + 6e- 6e- + 2 Co 3+ (s)  2 Co 0 (aq) _______________________________ 3Zn 0 (s) + 2 Co 3+ (s)  2 Co 0 (aq) + 3 Zn +2 (aq)

10. Practice a. PCl 3 + Cl 2 → PCl 5 b. Cu + 2AgNO 3 →Cu(NO 3 ) 2 + 2Ag c. CO 2 + 2LiOH → Li 2 CO 3 + H 2 O d. FeCl 2 + 2NaOH → Fe(OH) 2 + 2NaCl e. MnO 2 + 4HCl → Cl 2 + 2H 2 O + MnCl 2