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21-1 OXIDATION-REDUCTION REACTIONS Oxidation State/Numbers Redox Reactions Balancing Redox Reactions Chapters 4.4; 21.1 Silberberg

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OXIDATION-REDUCTION REACTIONS

• Oxidation State/Numbers• Redox Reactions• Balancing Redox Reactions• Chapters 4.4; 21.1 Silberberg

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Chapter 21

Electrochemistry: Chemical Change and Electrical Work

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Electrochemistry: Chemical Change and Electrical Work

21.1 Redox Reactions and Electrochemical Cells

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Goals & Objectives

• See these Learning Objectives on page 170 & 894.

• Understand these Concepts:• 4.8-10; 21.1-3.

• Master these Concepts:• 4.7-10; 21.1

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Oxidation-Reduction Reactions

• Reactions in which substances undergo changes in oxidation number are called oxidation-reduction reactions or simply redox reactions.

• Oxidation and reduction always occur simultaneously in chemical reactions.

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Basic Concepts

• Oxidation– an algebraic increase in oxidation number– the process in which electrons are– (or appear to be) lost.

• Reduction– an algebraic decrease in oxidation number– the process in which electrons are– (or appear to be) gained.

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Pneumonic

• LEO the lion says GER

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Basic Concepts

• Oxidizing agents– substances that gain electrons and oxidize

other substances. Oxidizing agents are always reduced.

• Reducing agents– substances that lose electron and reduce

other substances. Reducing agents are always oxidized.

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Assigning Oxidation Numbers

• Use a set of arbitrary rules to assign the oxidation number of each element in a compound or ion.

• Start with two simple rules:– 1. The oxidation number of an element

uncombined with another element is zero.– 2. The sum of the oxidation numbers of all

the atoms in a species is equal to its total charge.

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Assigning Oxidation Numbers

• Substance Oxidation Number(ON)• H2 0

• O2 0

• P4 0

• Ni 0• Fe3+ +3• Br-1 -1

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Assigning Oxidation Numbers

• Substance Oxidation Number(ON)• Co2+ +2• S2- -2• O2

2- -1

• H-1 -1

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Assigning Oxidation Numbers

• Additional Rules– The oxidation number of H is +1 in

combination with nonmetals and -1 in combination with metals.• HCl ON of H = +1• NaH ON of H = -1

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Assigning Oxidation Numbers

– The oxidation number of oxygen is -2 in most of its compounds.• MgO ON of O = -2• Na2SO4 ON of O = -2

• H2O ON of O = -2 ON of H = +1

• H2O2 ON of H = +1 ON of O = -1

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Assigning Oxidation Numbers

– Group 1 metals in all their compounds exhibit an oxidation number of +1• NaCl ON of Na = +1• LiOH ON of Li = +1

– Group 2 metals in all their compounds exhibit an oxidation number of +2• MgO ON of Mg = +2• BaSO4 ON of Ba = +2

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Assigning Oxidation Numbers

– Group 13 metals in most of their compounds exhibit an oxidation number of +3.• Al2O3 ON of Al = +3

• GaF3 ON of Ga = +3

– Group16 elements in compounds with metals and ammonium ion exhibit an oxidation number of -2• H2S ON of S = -2

• (NH4)2Se ON of Se = -2

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Assigning Oxidation Numbers

– Halogens (Group 17) in all their compounds exhibit an oxidation number of -1 except when in combination with oxygen or another halogen higher in the group.• NaCl ON of Cl = -1• AlCl3 ON of Cl = -1

• HClO ON of Cl = +1• HClO2 ON of Cl = +3

• HClO3 ON of Cl = +5

• HClO4 ON of Cl = +7

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Assigning Oxidation Numbers

• Assign an oxidation number to the underlined element in each of the following:

• NaNO2 H3PO4

• K2Sn(OH)6 KO2

• SO3-2 HCO3

-1

• Cr2O7-2 NH4

+1

• HC2H3O2

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Overview of Redox Reactions

Oxidation is the loss of electrons and reduction is the gain of electrons. These processes occur simultaneously.

The oxidizing agent takes electrons from the substance being oxidized. The oxidizing agent is therefore reduced.

The reducing agent takes electrons from the substance being oxidized. The reducing agent is therefore oxidized.

Oxidation results in an increase in O.N. while reduction results in a decrease in O.N.

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Figure 21.1 A summary of redox terminology, as applied to the reaction of zinc with hydrogen ion.

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) +10 0+2

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Redox Reactions

• Redox reactions must involve a change in oxidation number.– H2 + Cl2 = 2HCl

– Zn + HNO3 = Zn(NO3)2 + NO2 + H2O

– Ag+1 + Cl-1 = AgCl

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Half-Reaction Method for Balancing Redox Reactions

The half-reaction method divides a redox reaction into its oxidation and reduction half-reactions. - This reflects their physical separation in electrochemical cells.

This method does not require assigning O.N.s.

The half-reaction method is easier to apply to reactions in acidic or basic solutions.

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Steps in the Half-Reaction Method

• Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species.

• Balance the atoms and charges in each half-reaction.– First balance atoms other than O and H, then O, then H.– Charge is balanced by adding electrons (e-) to the left side in

the reduction half-reaction and to the right side in the oxidation half-reaction.

• If necessary, multiply one or both half-reactions by an integer so that – number of e- gained in reduction = number of e- lost in oxidation

• Add the balanced half-reactions, and include states of matter.

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Balancing Redox Reactions in Acidic Solution

Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)

Step 1: Divide the reaction into half-reactions.

Cr2O72- → Cr3+

I- → I2

Step 2: Balance the atoms and charges in each half-reaction.

Cr2O72- → 2Cr3+

Balance atoms other than O and H:

Balance O atoms by adding H2O molecules:

Cr2O72- → 2Cr3+ + 7H2O

For the Cr2O72-/Cr3+ half-reaction:

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Balance H atoms by adding H+ ions:

14H+ + Cr2O72- → 2Cr3+ + 7H2O

Balance charges by adding electrons:

6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O

Balance atoms other than O and H:

There are no O or H atoms, so we balance charges by adding electrons:

For the I-/I2 half-reaction:

2I- → I2

2I- → I2 + 2e-

This is the reduction half-reaction. Cr2O72- is reduced, and is the

oxidizing agent. The O.N. of Cr decreases from +6 to +3.

This is the oxidation half-reaction. I- is oxidized, and is the reducing agent. The O.N. of I increases from -1 to 0.

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Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e- lost in the oxidation equals the number of e- gained in the reduction.

6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O

3(2I- → I2 + 2e-)

The reduction half-reaction shows that 6e- are gained; the oxidation half-reaction shows only 2e- being lost and must be multiplied by 3:

6I- → 3I2 + 6e-

Step 4: Add the half-reactions, canceling substances that appear on both sides, and include states of matter. Electrons must always cancel.

6I- → 3I2 + 6e-

6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)

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Ion-Electron Method

• Use the ion electron method to write and balance the net ionic equation for the following:

• Sn2+ + Br2 = Sn4+ + Br-1

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Ion-Electron Method

• In acid solutions use H2O to balance O and H+ to balance H.

• Use the ion-electron method to write and balance the net ionic equation for the following:– Fe2+ + Cr2O7

2- = Fe3+ + Cr3+ (acidic)

– Zn + NO3-1 = Zn+2 + N2 (acidic solution)

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Balancing Redox Reactions in Basic Solution

An acidic solution contains H+ ions and H2O. We use H+ ions to balance H atoms.

A basic solution contains OH- ions and H2O. To balance H atoms, we proceed as if in acidic solution, and then add one OH- ion to both sides of the equation.

For every OH- ion and H+ ion that appear on the same side of the equation we form an H2O molecule.

Excess H2O molecules are canceled in the final step, when we cancel electrons and other common species.

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Sample Problem 21.1 Balancing a Redox Reaction in Basic Solution

PROBLEM: Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution:

MnO4-(aq) + C2O4

2-(aq) → MnO2(s) + CO32-(aq) [basic solution]

PLAN: We follow the numbered steps as described in the text, and proceed through step 4 as if this reaction occurs in acidic solution. Then we add the appropriate number of OH- ions and cancel excess H2O molecules.

SOLUTION:

Step 1: Divide the reaction into half-reactions.

MnO4- → MnO2 C2O4

2- → CO32-

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Sample Problem 21.1

Step 2: Balance the atoms and charges in each half-reaction.

Balance atoms other than O and H:

Balance O atoms by adding H2O molecules:

MnO4- → MnO2 C2O4

2- → 2CO32-

MnO4- → MnO2 + 2H2O 2H2O + C2O4

2- → 2CO32-

Balance H atoms by adding H+ ions:

4H+ + MnO4- → MnO2 + 2H2O 2H2O + C2O4

2- → 2CO32- + 4H+

Balance charges by adding electrons: 3e- + 4H+ + MnO4

- → MnO2 + 2H2O

[reduction]

2H2O + C2O42- → 2CO3

2- + 4H+ + 2e-

[oxidation]

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Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e- lost in the oxidation equals the number of e- gained in the reduction.

6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O x 2

6H2O + 3C2O42- → 6CO3

2- + 12H+ + 6e- x 3

Step 4: Add the half-reactions, canceling substances that appear on both sides.

6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O

6H2O + 3C2O42- → 6CO3

2- + 12H+ + 6e-

2MnO4- + 2H2O + 3C2O4

2- → 2MnO2 + 6CO32- + 4H+

2 4

Sample Problem 21.1

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2MnO4- + 2H2O + 3C2O4

2- + 4OH- → 2MnO2 + 6CO32- + 4H2O

Sample Problem 21.1

Basic. Add OH- to both sides of the equation to neutralize H+, and cancel H2O.

2MnO4- + 2H2O + 3C2O4

2- + 4OH- → 2MnO2 + 6CO32- + [4H+ + 4OH-]

2

Including states of matter gives the final balanced equation:

2MnO4-(aq) + 3C2O4

2-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

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Ion-Electron Method

• In basic solution, balance the equation as if it were acidic and then neutralize the H+ with OH-1 to produce water.

• Use the ion-electron method to write and balance the net ionic equation for the following:

• Br2 = Br-1 + BrO3-1 (basic solution)

• S-2 + Cl2 = SO4 -2 + Cl-1 (basic solution)

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