Redox Reactions and Electrolysis Chapter 7 Redox Reactions and Electrolysis

What is a Redox Reaction? 7.1 What is a Redox Reaction?

A. Oxidation and Reduction Gain of oxygen 2Mg(s) + O2(g)  2MgO(s) Mg has been oxidized Loss of oxygen CuO(s) + H2(g)  Cu(s) + H2O(l) CuO has been reduced

B. Redox Reactions Redox Reactions – oxidation-reduction reactions take place together

A. Oxidation and Reduction Gain of oxygen 2Mg(s) + O2(g)  2MgO(s) Mg has been oxidized Removal of hydrogen from a compound (organic) Loss of oxygen CuO(s) + H2(g)  Cu(s) + H2O(l) CuO has been reduced Addition of hydrogen to a compound (organic)

Check-Up-#1- Pg. 107

Redox and Electron Transfer 7.2 Redox and Electron Transfer

Redox and Electron Transfer We can define oxidation and reduction with respect to electron transfer and changes in oxidation number

Oxidation and Reduction Gain of oxygen 2Mg(s) + O2(g)  2MgO(s) Mg has been oxidized Removal of hydrogen from a compound (organic) Oxidation Is Loss of electrons (OIL) Loss of oxygen CuO(s) + H2(g)  Cu(s) + H2O(l) CuO has been reduced Addition of hydrogen to a compound (organic) Reduction Is Gain (RIG)

A. Half-Equations We can divide a reaction into two half equations – show separate oxidation-reduction reactions 2Na(s) + Cl2(g)  2NaCl(s) Redox-Reaction Video

A. Half-Equations We can divide a reaction into two half equations – show separate oxidation-reduction reactions Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)

B. Balancing Half-Equations Use 2 half-reactions to write a balanced ionic equation - # electrons lost and gained in a redox reaction must be equal

Worked Example #1 – Pg. 108 Construct a balanced ionic equation for the reaction between nickel and iron (III) ions, from the half equations: Ni(s) → Ni2+ (aq) + 2e- Fe3+ (aq) + e- → Fe2+ (aq) Ni(s) → Ni2+ (aq) + 2e- 2(Fe3+ (aq) + e- → Fe2+ (aq)) 2Fe3+ (aq) + 2e- → 2Fe2+ (aq)) 2Fe3+ (aq) + 2e- → 2Fe2+ Ni(s) + 2Fe3+ (aq) → Ni2+ (aq) + 2Fe2+

Worked Example #2 – Pg. 108

Check-Up-#2 – Pg. 109

7.3 Oxidation Numbers

A. What are oxidation numbers? Oxidation Number – number given to each atom or ion in a compound which shows us its degree of oxidation (represents the charge an atom would have if e- were transferred completely to the atoms with the greater attraction for them in a given situation) +, -, 0 High (+) = more oxidized High (-) = more reduced

B. Oxidation Number Rules

B. Oxidation Number Rules

C. Applying the Oxidation # Rules Metal & Nonmetal

C. Applying the Oxidation # Rules Nonmetal & Nonmetal – most electronegative element given negative sign

C. Applying the Oxidation # Rules Compound Ions

Check-Up- #3 – Pg. 114

D. Redox and Oxidation Numbers We can define oxidation and reduction in terms of oxidation # changes of particular atoms during a reaction

Oxidation and Reduction Gain of oxygen 2Mg(s) + O2(g)  2MgO(s) Mg has been oxidized Removal of hydrogen from a compound (organic) Oxidation Is Loss of electrons (OIL) Increase in oxidation # Loss of oxygen CuO(s) + H2(g)  Cu(s) + H2O(l) CuO has been reduced Addition of hydrogen to a compound (organic) Reduction Is Gain (RIG) Decrease in oxidation #

D. Redox and Oxidation Numbers Sn + 4HNO3  SnO2 + 4NO2 + 2H2O

D. Redox and Oxidation Numbers Oxidizing Agent Reducing Agent Increases ox. # of another atom Deceases in ox. # Substance that gets reduced (gains e-) Decreases ox. # of another atom Increases in ox. # Substance that gets oxidized (loses e-)

Check-Up - #4 – Pg. 114

E. Naming Compounds Iron(II) Chloride Iron(III) Chloride

E. Naming Compounds Oxides of Nitrogen N2O NO NO2

E. Naming Compounds Nitrate Ions NaNO2 NaNO3

E. Naming Compounds Ions w/ Oxygen + 1 other element = end -ate Cl + oxygen = S + oxygen =

E. Naming Compounds Inorganic Acids w/ Oxygen = end in -ic H3PO3 = HClO4 =

Check-Up - #5 – Pg. 116 Sodium sulfate (IV) Sodium sulfate (VI) Iron(II) nitrate(V) Iron(III) nitrate(V) Iron(II) sulfate(VI) Copper(I) oxide Sulfuric(IV) acid Manganese(VII) oxide

Worked Example - #3 – Pg. 117 Sodium Chlorate (V)

Check-Up - #6 – Pg. 117

F. Balancing Chemical Equations Using Ox. # We can use ox. #s to balance equations involving redox reactions

Worked Example #4 – Pg. 117 Write a balanced chemical equation

Worked Example #5 – Pg. 117 Write a balanced chemical equation

Check-Up - #7 – Pg. 118

7.4 Electrolysis

Electrolytic Cells Electrolysis – decomposition of a compound into its elements by an electric current Used to extract metals which ore are high in reactivity series Used to produce non-metals such as chlorine and to purify some metals

Check-Up - #8 – Pg. 119

B. Redox Reactions in Electrolysis

Check-Up - #9 – Pg. 119

C. Extracting Aluminum 6O2-  3O2 + 12e- O2 reacts w/ hot carbon electrodes Oxidizes them to CO2 which escapes from cell Electrodes must be replaced periodically C. Extracting Aluminum Al3+ + 3e-  Al Reduced Dense, sinks to bottom

C. Extracting Aluminum Al2O3 molten via passing a high electric current through electrolyte (40,000 A) However, Al2O3 has a high MP = lots of E needed = $$$ Energy needed is lowered by dissolving Al2O3 in large amounts of cryolite (Na3AlF6) Lowers MP of Al2O3 to 970oC Improves electrical conductivity of electrolyte

Check-Up - #7 – Pg. 120

D. Electrolysis of Brine Concentrated aqueous solution of NaCl Obtained from seat water or dissolving rock salt in H2O Used to produce chlorine, hydrogen, and sodium hydroxide Ions in brine: Na+, Cl-, H+ (from H2O), OH- (from H2O)

2H+ + 2e-  H2 Na+ Cl- 2Cl-  Cl2 + 2e- H+ OH-

Check-Up - #11 – Pg. 122

E. The Electrolytic Purification of Copper Copper  smelting in furnace (water pipes) Not pure enough for electrical wiring because even small impurities reduce conductivity Need 99.99% pure copper Done through electrolysis

Anode decreases in thickness Cu  Cu2+ + 2e- Cu2+ + 2e-  Cu Cu atoms deposited on cathode Cu2+ Cathode increases in thickness When “full,” cathode is removed and replaced

Check-Up - #12 – Pg. 123