Oxidation- Reduction Chemistry

OxidationReduction Chemistry

Chem 36 Spring 2002

Definitions

n Redox reactions involve electron transfer:

Lose e- = Oxidation Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)

Gain e- = Reduction

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1

Half-Reactions

? Consider each process indivually: Oxidation Cu (s) Cu2+ (aq) + 2 e-

Reduction [Ag+ + e- Ag (s) ] x 2 Overall: Cu (s) + 2Ag+ (aq) Cu2+ (aq) + 2Ag (s)

Oxidized (reducing agent)

Reduced (oxidizing agent)

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Balancing Redox Reactions

? The Half-Reaction Method Three Steps: 1. Determine net ionic equations for both half-reactions 2. Balance half-reactions with respect to mass and charge 3. Combine so as that electrons cancel

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Example

SO32-

+

H+

+

MnO

4

SO42-

+

Mn2+

+

H2O

1. Write Skeleton Half-Reactions

Oxidation

SO32- SO42-

Reduction

MnO4- Mn2+

2. Mass Balance

SO32- + H2O SO42- + 2H+

MnO4- + 8H+ Mn2+ + 4H2O

?Add H2O to side needing oxygen

?Add H+ to balance hydrogen

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Example: Continued

3. Charge Balance (use electrons)

SO32- + H2O SO42- + 2H+ + 2eMnO4- + 8H+ + 5e- Mn2+ + 4H2O

4. Combine!

[SO32- + H2O SO42- + 2H+ + 2e-] x 5 [MnO4- + 8H+ + 5e- Mn2+ + 4H2O] x 2

5SO32- + 5H2O + 2MnO4- + 16H+ + 10e- 5SO42- + 10H+ + 10e- + 2Mn2+ + 8H2O

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3

Example: Simplify and Verify

Collecting and cancelling gives:

5SO32- + 2MnO4- + 6H+ 5SO42-+ 2Mn2+ + 3H2O

Verify! ?Mass Balance ?Charge Balance

Done!

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In Basic Solution?

?First: Balance as though in acid ?Next:

?Add OH- to both sides of reaction in an amount equal to the amount of H+

?Change: H+ + OH- to H2O ?Verify!

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4

Example (Basic Solution)

Cr(OH)3 + OCl- + OH- CrO42- + Cl- + H2O

In Acid Solution

2Cr(OH)3 + 3OCl- 2CrO42- + 3Cl- + H2O + 4H+

+ 4OH-

Add OH-

+ 4OH-

Make Water

2Cr(OH)3 + 3OCl- + 4OH- 2CrO42- + 3Cl- + 5H2O

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Example: Simplify and Verify

Collecting and cancelling gives:

5SO32- + 2MnO4- + 6H+ 5SO42-+ 2Mn2+ + 3H2O

Verify! ?Mass Balance ?Charge Balance

Done!

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