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)


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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


• 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


• 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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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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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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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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General Outcomes

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