Acid-Base titration

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 +

4.4 Redox Reactions

Tracking electrons in oxidation reduction reactions.

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.

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.

Rules for assigning oxidation numbers.

1.

2.

3.

4.

5.

6.

7.

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 ). The oxidation state of a monatomic ion is the same as its charge.

 Fe 2+ oxidation number is 2+ In compounds, Group 1 is +1, Group 2 is +2, Fluorine is -1 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 ). Hydrogen is assigned an oxidation state of +1.  Metal hydrides are an exception In LiH, hydrogen has an oxidation state of -1. The sum of the oxidation states must be zero for an electrically neutral compound. For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.

Assigning Oxidation Numbers

  Cl 2 Sodium Metal  Lead IV  NaF     CaCl 2 H 2 SO 4 CrO 4 2 NO 3 -

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)

Balancing Redox equations

ClO 3 (aq) + I (aq)  Cl (aq)

The reaction occurs in an acidic solution.

+ I 2 (s) Split in to 2 half-reactions. An oxidation and a reduction.

Balancing ½ reactions

1.

2.

3.

4.

5.

Balance the atoms of the element being oxidized or reduced.

Balance the oxidation number by balancing electrons.

Balance the charge. (add H + if in acidic solution; add OH basic solution) if in Balance Hydrogen by adding H 2 O molecules.

Check to make sure that oxygen is balanced.

I - (aq)  I 2 (s) (in acidic solution) 1.

2.

3.

4.

5.

Balance the atoms of the element being oxidized or reduced.

Balance the oxidation number by balancing electrons. Oxidation ½ add electrons to the right side Balance the charge. (add H + if in acidic solution; add OH if in basic solution) Balance Hydrogen by adding H 2 O molecules.

Check to make sure that oxygen is balanced

ClO 3 - (aq)  Cl (aq) (in acidic solution) 1.

2.

3.

4.

5.

Balance the atoms of the element being oxidized or reduced.

Balance the oxidation number by balancing electrons. Reduction ½ adds e- to the left side.

Balance the charge. (add H + if in acidic solution; add OH if in basic solution) Balance Hydrogen by adding H 2 O molecules.

Check to make sure that oxygen is balanced

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.

The following reaction occurs in acidic solution:

Fe

2+ (aq)

+ MnO

4 (aq) 

Fe

3+ (aq)

+ Mn

2+ (aq)

The following reaction occurs in basic solution:

Cl

2 (g)

+ Cr(OH)

3 (s) 

Cl

- (aq)

+ CrO

4 2- (aq)