Oxidation- Reduction Chemistry

OxidationReduction

Chemistry

Chem 36

Spring 2002

Definitions

n Redox reactions involve electron transfer:

Lose e- =

Oxidation

Cu (s) + 2 Ag+ (aq) ¡ú Cu2+ (aq) + 2 Ag (s)

Gain e- =

Reduction

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Half-Reactions

? Consider each process indivually:

Oxidation

Cu (s) ¡ú Cu2+ (aq) + 2 e-

Reduction

[Ag+ + e- ¡ú Ag (s) ] x 2

Overall:

Cu (s) + 2Ag+ (aq) ¡ú 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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2

Example

SO32- + H+ + MnO 4- ¡ú SO42- + Mn2+ + H2O

1. Write Skeleton Half-Reactions

Oxidation

SO32- ¡ú SO42-

Reduction

MnO4- ¡ú Mn2+

2. Mass Balance

SO32- + H2O ¡ú SO42- + 2H+

MnO4- + 8H+ ¡ú 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 ¡ú SO42- + 2H+ + 2eMnO4- + 8H+ + 5e- ¡ú Mn2+ + 4H2O

4. Combine!

[SO32- + H2O ¡ú SO42- + 2H+ + 2e-] x 5

[MnO4- + 8H+ + 5e- ¡ú Mn2+ + 4H2O] x 2

5SO32- + 5H2O + 2MnO4- + 16H + + 10e- ¡ú

5SO42- + 10H+ + 10e- + 2Mn2+ + 8H2O

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3

Example: Simplify and Verify

Collecting and cancelling gives:

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

Verify!

¨¹Mass Balance

¨¹Charge Balance

Done!

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In Basic Solution?

?First: Balance as though in acid

?Next:

¨¹Add OH - to both sides of reaction in an

amount equal to the amount of H+

¨¹Change: H+ + OH- to H2O

¨¹Verify!

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4

Example (Basic Solution)

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

In Acid Solution

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

+ 4OH -

+ 4OH -

Add OHMake Water

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

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Example: Simplify and Verify

Collecting and cancelling gives:

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

Verify!

¨¹Mass Balance

¨¹Charge Balance

Done!

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