Balance Redox Equations: Half Reaction

1
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. Balance O with H2O then add H
2 Cl- (aq) ? Cl2 (g)
2 Cl- (aq) ? Cl2 (g) 2e-
3. Balance charge by add e- to side with
gt charges, then X so e- same in
both equations
Cr2O72- (aq) 14 H(aq) 6e- ? 2 Cr 3 (aq)
7H2O(l)
X3
6 Cl- (aq) ? 3Cl2 (g) 6e-
4. Add ½ rx.
Cr2O72- (aq) 14 H(aq) 6 Cl- (aq) 6e- ? 2
Cr 3 (aq) 7H2O(l) 3Cl2 (g) 6e-
Cr2O72- (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
6
1
-1
Ox Cr(OH)4- (aq) ? CrO42- (aq)
Red ClO- (aq) ? Cl- (aq)
1. Balance except H and O
2. Balance O with H2O then add H
ClO- (aq) 2H(aq) ? Cl- (aq) H2O(l)
Cr(OH)4- (aq) ? CrO42- (aq) 4H(aq)
3. Balance charge by add e- to side with
gt charges, then X so e- same in
both equations
Cr(OH)4- (aq) ? CrO42- (aq) 4H(aq) 3e-
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) 6e-
4. Add ½ rx.
3ClO- (aq) 6H(aq) 6e- 2Cr(OH)4- (aq) ?
3Cl- (aq) 3H2O(l) 2CrO42- (aq) 8H(aq) 6e-
2H (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)
2
BALANCING REDOX EQUATIONS ½ RX METHOD MASS AND
CHARGE CONSERVATION
Ex an oxidation reaction already balanced for O
and H
Cr(OH)4 - ? CrO4 2-
4H
MASS BALANCE (CONSERVATION)
CHARGE BALANCE (CONSERVATION)
?

4
3
4 -
2 -
2
1 -
SO ADD 3 e to this side!

?

3 -
4 -
2 -
4
3
1 -
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