Transcript Document

Balance Redox Equations: Half Reaction
concentrate on electrons, balance each ½ rx. separately
Ex: Cr2O72- (aq) + Cl- (aq)  Cr3+ (aq) + Cl2 (g) IN ACID
Ox: Cl- (aq)  Cl2 (g)
Red: : Cr2O72- (aq)  Cr 3+ (aq)
Cr2O72- (aq)  2 Cr 3+ (aq)
2 Cl- (aq)  Cl2 (g)
1. Balance except
H and O
Cr2O72- (aq) + 14 H+(aq)  2 Cr 3+ (aq) + 7H2O(l)
2 Cl- (aq)  Cl2 (g)
2. Balance O with
H2O then add H+
3. Balance charge by 2 Cl- (aq)  Cl (g) + 2e2
add e- to side with
> # + charges,
X3
then X so # e- same
in both equations 6 Cl- (aq)  3Cl (g) + 6e2
Cr2O72- (aq) + 14 H+(aq) + 6e-  2 Cr 3+ (aq) + 7H2O(l)
4. Add ½ rx.
Cr2O72- (aq) + 14 H+(aq) + 6 Cl- (aq) + 6e-  2 Cr 3+ (aq) + 7H2O(l) + 3Cl2 (g) + 6eCr2O72- (aq) + 14 H+(aq) + 6 Cl- (aq)  2 Cr 3+ (aq) + 7H2O(l) + 3Cl2 (g)
Ex: Cr(OH)4- (aq) + ClO- (aq)  CrO42- (aq) + Cl- (aq) IN BASE
+3
+1
+6
-1
1. Balance except
H and O
2. Balance O with
H2O then add H+
Ox: Cr(OH)4- (aq)  CrO42- (aq)
Cr(OH)4- (aq)  CrO42- (aq) + 4H+(aq)
3. Balance charge by
add e- to side with
> # + charges,
then X so # e- same
in both equations
Cr(OH)4- (aq)  CrO42- (aq) + 4H+(aq) + 3e-
Red: ClO- (aq)  Cl- (aq)
ClO- (aq) + 2H+(aq)  Cl- (aq) + H2O(l)
ClO- (aq) + 2H+(aq) + 2e-  Cl- (aq) + H2O(l)
X3
X2
3ClO- (aq) + 6H+(aq) + 6e-  3Cl- (aq) + 3H2O(l)
2Cr(OH)4- (aq)  2CrO42- (aq) + 8H+(aq) + 6e4. Add ½ rx.
3ClO- (aq) + 6H+(aq) + 6e- + 2Cr(OH)4- (aq)  3Cl- (aq) + 3H2O(l) + 2CrO42- (aq) + 8H+(aq) + 6e2H+ (aq)
3ClO- (aq) + 2Cr(OH)4- (aq)  3Cl- (aq) + 3H2O(l) + 2CrO42- (aq) + 2H+(aq)
+2OH-(aq)
+2OH-(aq)
3ClO- (aq) + 2Cr(OH)4- (aq) + 2OH-(aq)  3Cl- (aq) + 5H2O(l) + 2CrO42- (aq)
BALANCING REDOX EQUATIONS – ½ RX METHOD
MASS AND CHARGE CONSERVATION
Ex: an oxidation reaction already balanced for O and H

[Cr(OH)4] -
oxid. # 3+
CrO4 2-
2-
O
Cr3+
1-
1O
O
H
4
4H +
+
Cr
O
O
complex ion
oxid # = 6+
MASS BALANCE (CONSERVATION)

+
CHARGE BALANCE (CONSERVATION)

4-
3+
+
4+
2-
2+
1-
SO ADD 3 e – to this side!

4-
3+
1-
+
+
2-
4+
1-
3-
Problem: Using the half-reaction methods, balance the following
equation showing the reaction of permanganate and ethanol to form
acetic acid. Assume the reaction takes place under ACIDIC
conditions.
MnO4- (aq) + C2H5OH (aq)  Mn2+ + CH3CO2H