Chem 163 Final

Oxidation Reduction Reactions (Redox Reaction)

two 1/2 reactions, transfer of an e-

Oxidation

lose e-

* increases oxidation number

Oxidizing agent

oxidizes another species

* itself is reduced

Reduction

gain e-

* reducing oxidation number

Reducing agent

reduces another species

* itself is oxidized

1. How to assign oxidation number

the number of an atom is 0

* Fe(s) = 0
* Na (s) = 0

2. How to assign oxidation number

the oxidation number of a monatomic ion is its charge

* Na^+ = +1
* Cl^+ = -1
* Al^3+ = +3

3. How to assign oxidation number

the sum of the oxidation number in a compound is 0

* Na^+1 Cl^-1 = 0

4. How to assign oxidation number
* Polyatomic ions

the sum of the oxidation numbers is the charge of the ion

Assign the oxidation number of Mn in these two compounds

* Potassium Permanganate: KMnO4
* Potassium Manganate: K2MnO4

KMnO4 = +7

K2MnO4= +6

What is the Oxidation number of Cr in the dichromate ion?

* CrO7 ^2-

CrO7 ^2- = -2

Solid Zinc reacts with aq Silver (I) ion to produce aq Zinc (II) ion and solid silver

Zn(s) + Ag^+ (aq) → Zn^2+ (aq) + Ag(s)

* What is being oxidized?
* What is being reduced?
* Write the 1/2 reactions

Sn^2+ (aq) + Fe^3+ (aq) → Sn^4+ (aq) + Fe^2+ (aq)

* What is the balanced redox reaction

Balancing Redox Reactions in Acidic Solutions Rules

1. Write out the half reactions
2. Balance 1/2 reactions


1. ox 1/2 rxn
2. red 1/2 rxn: # of balance element being reduced
3. Balance oxygen by adding water
4. Balance hydrogen by adding H+ (aq)
5. Balance charge by adding e-
6. Add 1/2 reactions and simplify

\

Fe^-2 (aq) + NO3^- → Fe^3+ (aq) + NO(g)

Cr2O7^2- (aq) + S^2- (aq) → S(s) + Cr^3+ (aq)

Balancing Redox Reactions in Basic Reactions

1. Divide the reaction into half reactions
2. Balance the elements other than H and O
3. Balance the O atoms by adding H2O
4. Balance the H atoms by adding H+
5. Add OH- ions to BOTH SIDES to neutralize any H+
6. Combine H+ and OH- to make H2O
7. Simplify by cancelling out access H2O
8. Balance the charges by adding e=
9. Add the half reactions and simplify

Pb(OH)3^- (aq) + ClO^- (aq) → PbO2 (aq) + Cl^- (aq)

Pb(s) + NO3^- (aq) → Pb ^2+ (aq) + NO2 (aq)

Voltaic or Galvanic Cells

an electrochemical cell in which a spontaneous reaction generates an electric current

* consist of two half cell (1/2 cells)
* electrical connection between the two 1/2 cells
* Salt bridge

Voltaic Cell Notation

* ox 1/2 left (anode)
* red 1/2 right (cathode)
* phase boundary **I** (single vertical bar)
* salt bridge with two vertical bars II

The cell notation for a voltaic cells is as follows:

* Al(s) I Al^3+ (aq) II Cu^2+ I Cu(s)

Write the oxidation and reduction 1/2 reactions and balance then and add them together

Nernst Equation

The emf of a particular cell is 0.500 v. the cell reaction is :

* 2Al (s) + 3Cu^2+ (aq) → 2Al ^3+ (aq) + 3Cu (s)

Calculate the maximum electrical work of this cell is obtained from 1.00g aluminum