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
2
1
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)
3
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
4
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
5
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
6
3
Example: Simplify and Verify
Collecting and cancelling gives:
5SO32- + 2MnO4- + 6H+ ¡ú 5SO42-+ 2Mn2+ + 3H2O
Verify!
¨¹Mass Balance
¨¹Charge Balance
Done!
7
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!
8
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
9
Example: Simplify and Verify
Collecting and cancelling gives:
5SO32- + 2MnO4- + 6H+ ¡ú 5SO42-+ 2Mn2+ + 3H2O
Verify!
¨¹Mass Balance
¨¹Charge Balance
Done!
10
5
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