General Outcomes
Transcript of General Outcomes
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General Outcomes: • Explain the nature of oxidation-reduction reactions
• Apply the principles of oxidation-reduction to electrochemical cells
Redox Reactions
Chapter 13
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Today’s Objectives:
1. Define oxidation and reduction operationally (historically) and theoretically
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Section 13.1 (pg. 558-567)
Today’s Agenda
1. Review Are You Ready p. 554 #1-13 (Prior Knowledge Activity)
2. Introduce Redox
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Electrochemical Reactions
• electrons are transferred
• Most common chemical change in living and non-living systems ▫ Example: metabolism, combustion, photosynthesis,
metal plating, fuel cells, biosensors and the production of metals from their ores (metallurgy)
▫ Technological applications were available long before there was a scientific understanding of the process
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Metallurgy
• Science and technology of extracting metals from naturally occurring compounds and adapting these metals for useful purposes
• Long history See technology timeline on Figure 1 – p. 558
▫ Example: Making samurai swords required a special fire to obtain the higher temperature needed – technological development used far before scientific understanding emerged
▫ With increasing technology humanity progressed from the Stone Age, through the Bronze Age and the Iron Age to our modern age.
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Metallurgy
• The term reduction was associated with producing metals from their compounds
▫ i.e. compounds being reduced to metals
• Reducing Agents promote the reduction of a metal compound to a metal element
▫ Example: CuS(s) + H2(s) Cu(s) + H2S(g) copper (II) sulfide is being reduced by the hydrogen reducing agent
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Metallurgy
• Corrosion (rusting of metals) returned metal to its natural state as a compound, therefore considered the opposite of metallurgy
• Eventually all reactions with oxygen were considered oxidation
• Chemists realized that other substances caused similar oxidation reactions and extended the term beyond reactions with oxygen
• Oxidizing Agent promotes the oxidation of a metal to produce a metal compound
▫ Example: Cu(s) + Br2(g) CuBr2(s)
The oxidation of copper by the bromine oxidizing agent
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Homework
• Practice Questions – p. 559 #1-6
• Case Study: Early Metallurgy - p. 560 #1-3
• Investigation 13.1 - p. 560 (predictions)
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Today’s Objectives:
1. Define half-reactions
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Section 13.1 (pg. 558-567)
Electron Transfer Theory
• Consider single replacement reactions ▫ One element replaces another element in a compound
• Reactions combined two half-reaction parts that represent what happens to each reactant
• half-reactions are balanced chemical equations (by mass and charge) that represent the loss or gain of electrons by a substance
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Electron Transfer Theory
• Atomic theory requires a loss of electrons when a reaction converts atoms (electrically neutral particles)
to ions in solution, and a gain of electrons for the opposite reaction
▫ Example: Corrode Zn with a HCl(aq) oxidizing agent
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
▫ The half-reactions are as follows:
Zn(s) Zn2+(aq)
+ 2e-
2H+(aq) + 2e-
H2(g)
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Figure 6 – p. 561
Recall that the net charge in chemical equations is zero
Reduction – Oxidation Reactions “REDOX”
• Chemical reaction in which electrons are transferred between entities
• Must have both the process of reduction and oxidation happening for the reaction to occur
▫ REDUCTION – electrons are gained by an entity
▫ OXIDATION – electrons are lost by an entity
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▫ How can you remember this?
“LEO the lion says GER”
Losing Electrons = Oxidation
Gaining Electrons = Reduction
Other mnemonic devices:
OIL RIG (Oxidation Is Loss, Reduction Is Gain)
ELMO (Electron Loss Means Oxidation)
Reduction – Oxidation Reactions “REDOX”
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Zn(s) Zn2+(aq)
+ 2e-
2H+(aq) + 2e-
H2(g)
▫ Where is oxidation and reduction occurring??
(OIL RIG)
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OXIDATION - lose electrons
REDUCTION - gain electrons
Electron Product
Electron Reactant
Reduction – Oxidation Reactions “REDOX”
• Examples of Redox Reactions:
Formation, decomposition, combustion, single replacement, cellular respiration, photosynthesis
NOT double replacement
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Sample Problem 13.1 (p. 563)
• Consider the production of silver metal from aqueous silver nitrate in the presence of copper
Cu(s) + AgNO3(aq) Cu(NO3)2(aq) + Ag(s)
• Write the balanced half-reactions: The silver ions obtain electrons by colliding with the copper atoms on the metal surface
Cu(s) Cu 2+ (aq) + 2e-
2 [Ag+(aq) + e- Ag (s)]
Where is oxidation and reduction occurring??
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Figure 7 – p. 562
OXIDATION
REDUCTION
electrons gained are equal to the
electrons lost
Sample Problem 13.1 (p. 563)
• Write the net-ionic equation by adding together the half-reactions and cancelling common terms
Cu(s) Cu 2+ (aq) + 2e-
2 Ag+(aq) + 2e- 2Ag (s)
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Figure 7 – p. 562
Cu(s) + 2 Ag+(aq) + 2 e- + Cu 2+ (aq) + 2 Ag(s) + 2 e-
Cu(s) + 2 Ag+
(aq) Cu 2+(aq) + 2 Ag(s)
Cancelled terms must be identical, including
states of matter
▫ Silver ions are reduced to silver metal by reaction with the copper metal reducing agent
▫ Simultaneously, copper metal is oxidized to copper(II) ions by reaction with silver ion oxidizing agent.
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Sample Problem 13.1 (p. 563)
Figure 7 – p. 562
Communication Example 1 – p. 563 Figure 8 – p. 563
▫ There are two methods for developing net ionic equations:
1) half-reaction method we just learned
OR
2) Using the net-ionic equation method from Chem 20
Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) (dissociate high solubility and ionic compounds)
Cu(s) + 2Ag + (aq) + 2 NO3
- (aq) Cu2+
(aq)+ 2NO3-(aq) + 2Ag(s) (cancel spectator ions)
2 Ag+(aq)+ Cu(s) 2 Ag(s) + Cu 2+ (aq) (Do we get the same net ionic reaction?? YES!)
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▫ There are two methods for developing net ionic equations:
1) half-reaction method we just learned
OR
2) Using the net-ionic equation method from Chem 20
Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) (dissociate high solubility and ionic compounds)
Cu(s) + 2Ag + (aq) + 2 NO3
- (aq) Cu2+
(aq)+ 2NO3-(aq) + 2Ag(s) (cancel spectator ions)
2 Ag+(aq)+ Cu(s) 2 Ag(s) + Cu 2+ (aq) (Do we get the same net ionic reaction?? YES!)
NON-IONIC
TOTAL IONIC
NET-IONIC
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Summary Electron Transfer Theory
p. 564
• A redox reaction is a chemical reaction in which electrons are transferred between entities
• The total number of electrons gained in the reduction equals the total number of electrons lost in the oxidation
• Reduction is a process in which electrons are gained by an entity
• Oxidation is a process in which electrons are lost by an entity
• Both reduction and oxidation are represented by balanced half-reaction equations.
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Reduction Oxidation
• Historically, the formation of a metal from its “ore” (or oxide)
▫ i.e. nickel(II) oxide is reduced by hydrogen gas to nickel metal
NiO(s) + H2(g) Ni(s) + H2O(l)
Ni +2 Ni 0
• A gain of electrons occurs (so the entity becomes more negative)
• Electrons are shown as the reactant in the half-reaction
REDOX Reactions…. so far
• Historically, reactions with oxygen
▫ i.e. iron reacts with oxygen to produce iron(III) oxide
4 Fe(s) + O2(g) Fe2O3(s)
Fe 0 Fe+3
• A loss of electrons occurs (so the entity becomes more positive)
• Electrons are shown as the product in the half-reaction
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Complex Half-Reactions
• Polyatomic ions and molecular compounds undergo more complicated REDOX processes
• Most of these reactions occur in an acidic or basic aqueous solution and
• Experimental evidence indicates that water molecules, H+ and OH – are important in these reactions
• Often called the “ion-electron” method
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• Reduction half-reaction for nitrous acid
▫ Consider the reactants and products in a partial equation, balancing all atoms other than oxygen and hydrogen
e– + H+(aq)
+ HNO2(aq) NO(g) + H2O(l)
▫ Add water molecules to balance the oxygen atoms and H+ to
balance hydrogen since available in an acidic solution
▫ Include electrons to balance the charge
▫ This balanced half-reaction equation represents a gain of electrons, i.e. the reduction of nitrous acid
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Sample Problem 13.2 (p. 565)
Communication Examples 2 & 3 – p. 566
• Half-reaction when copper metal is oxidized in a basic solution to form copper (I) oxide. ▫ Balance all atoms in the equation using water to balance oxygen,
H+ for hydrogen, and electrons for charge
2OH–(aq) + H2O(l) + 2Cu (s) Cu2O(s) + 2H+
(aq) + 2e– + 2OH–(aq)
▫ Since the solution is alkaline, add an equal amount of OH– as H+
to both sides of the equation, creating more water molecules while maintaining a balance of charge and mass
▫ Cancel common terms
2OH–(aq) + 2Cu (s) Cu2O(s) + H2O (aq) + 2e–
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Sample Problem 13.3 (p. 565)
2H2O (aq)
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Summary Writing Half-Reaction Equations
p. 567
• Write the chemical formulas fro the reactants and products.
• Balance all atoms, other than O and H.
• Balance O by adding water.
• Balance H by adding hydrogen ions.
• Balance charge on each side by adding electrons and cancel common terms.
For basic solutions only:
• Add hydroxide ions to both sides to equal the number of hydrogen ions present
• Combined hydrogen ions and hydroxide ions on the same side to form water. Cancel equal amounts of water from both sides.
Homework
• Practice Questions – p. 564 #7-11
• Practice Question – p. 566 #12
• Section 13.1 Review – p. 567 #1-7
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