1. Objective: Determine the equivalence point. Equivalence point n OH - = n H + If 25.00mL of 0.0800M NaOH is needed to react with 10.00 mL of HCl. What is the molarity of HCl? 1. Write the reaction 2. Use M 1 V 1 =M 2 V 2 In this case the formula could be written M OH - V OH - = M H + V H + Acid-Base titration

2. 4.4 Redox Reactions Tracking electrons in oxidation reduction reactions.

3. Oxidation-Reduction Reactions (Red-Ox) A type of electrochemistry with reactions that involve a transfer of electrons between atoms. The charge or oxidation number changes Ex: Fe 2+  Fe 3+ + 1 e - In this red-ox ½ reaction the Iron (II) lost an electron to become Iron (III). The electron does not just “go into space” so there must be a companion ½ reaction that gains an electron to make the reaction balance.

4. Oxidation Number The oxidation number of an element in a molecule is the charge that it would have if all shared electrons were assigned to the more electronegative elements in their bonds. Draw the Lewis diagram for water Assign the bonding electrons to the more E n atoms. Determine what the charge would be.

5. Rules for assigning oxidation numbers. 1. The oxidation state of an element is zero.  including all elemental forms of the elements (N 2, P 4, S 8, O 2, O 3 ). 2. The oxidation state of a monatomic ion is the same as its charge.  Fe 2+ oxidation number is 2+ 3. In compounds, Group 1 is +1, Group 2 is +2, Fluorine is -1 4. Oxygen is usually assigned an oxidation state of -2 in its covalent compounds.  Exceptions to this rule include peroxides (compounds containing the group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H 2 O 2 ). 5. Hydrogen is assigned an oxidation state of +1.  Metal hydrides are an exception In LiH, hydrogen has an oxidation state of -1. 6. The sum of the oxidation states must be zero for an electrically neutral compound. 7. For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.

6. Assigning Oxidation Numbers Cl 2 Sodium Metal Lead IV NaF CaCl 2 H 2 SO 4 CrO 4 2- NO 3 -

7. Red-ox reactions A spontaneous red-ox reaction can be used to perform electrical work.  Fuel Cells  Primary voltaic cells (alkaline batteries)  Storage Cells (car batteries)

8. Balancing Redox equations ClO 3 - (aq) + I - (aq)  Cl - (aq) + I 2 (s) The reaction occurs in an acidic solution. Split in to 2 half-reactions. An oxidation and a reduction.

9. Balancing ½ reactions 1. Balance the atoms of the element being oxidized or reduced. 2. Balance the oxidation number by balancing electrons. 3. Balance the charge. (add H + if in acidic solution; add OH - if in basic solution) 4. Balance Hydrogen by adding H 2 O molecules. 5. Check to make sure that oxygen is balanced.

10. I - (aq)  I 2 (s) (in acidic solution) 1. Balance the atoms of the element being oxidized or reduced. 2. Balance the oxidation number by balancing electrons. Oxidation ½ add electrons to the right side 3. Balance the charge. (add H + if in acidic solution; add OH - if in basic solution) 4. Balance Hydrogen by adding H 2 O molecules. 5. Check to make sure that oxygen is balanced

11. ClO 3 - (aq)  Cl- (aq) (in acidic solution) 1. Balance the atoms of the element being oxidized or reduced. 2. Balance the oxidation number by balancing electrons. Reduction ½ adds e- to the left side. 3. Balance the charge. (add H + if in acidic solution; add OH - if in basic solution) 4. Balance Hydrogen by adding H 2 O molecules. 5. Check to make sure that oxygen is balanced

12. Combining ½ reactions Combine so that the number of e- cancel. Multiply by coefficients to accomplish this. Add together and eliminate what appears on both sides.

13. The following reaction occurs in acidic solution: Fe 2+ (aq) + MnO 4 - (aq)  Fe 3+ (aq) + Mn 2+ (aq)

14. The following reaction occurs in basic solution: Cl 2 (g) + Cr(OH) 3 (s)  Cl - (aq) + CrO 4 2- (aq)

15. Balancing ½ reactions Simple  Fe 2+ (aq)  Fe 3+ (aq) + 1 e - Complex When elements other than those getting oxidized or reduced take part in the reaction.  MnO 4 - (aq)  Mn 2+ (aq)