Download BALANCING REDOX REACTIONS

DAY #2
Date: ___________
PAGE #1
SCH 4U
Unit: Electrochemistry
BALANCING REDOX REACTIONS
Simple REDOX reaction equations can be balanced by inspection or trial and error as we have
done for all reactions
However, REDOX reactions are often shown without some of the components (the focus is on
the reduced and oxidized components)
Therefore, other methods must be used to determine the true electron flow
TWO METHODS of BALANCING:
A
B
Balance by Oxidation Number
Balance by Half-Reaction Method
A
Balancing REDOX EQUATIONS by OXIDATION NUMBER
The total increase in oxidation number for a particular atom/ion must equal the total decrease
in oxidation number of another atom/ion
Many redox reactions happen in acidic or basic environments, which helps us “fill in the
ingredients”
IN AN ACIDIC SOLUTION
ClO3- (aq)
Example #1:
STEP #1:
+
Cl- (aq)
I2 (aq)
Assign oxidation numbers and identify those changing.
+5 -2
ClO3- (aq)
+
0
-1
Cl- (aq)
I2 (aq)
IO3- (aq)
+
OXIDATION
+
+5 –2
IO3- (aq)
REDUCTION
STEP #2:
Using the change in oxidation numbers, write the number of electrons transferred per
atom.
Cl:
I:
STEP #3:
+5
0
—>
—>
-1
+5
Therefore
Therefore
6 e- / Cl
5 e- / I
Using chemical formulas, determine the number of electrons transferred per reactant
(Use reaction subscripts for this)
Cl:
I:
+5
0
—>
—>
-1
+5
Therefore
Therefore
6 e- / Cl
5 e- / I
and
and
6 e- / ClO310 e- / I2
DAY #2
Date: ___________
PAGE #2
SCH 4U
Unit: Electrochemistry
STEP #4:
Calculate simplest whole number coefficients for the reactants that will balance the
total number of electrons transferred. Balance reactants and products.
** Common Factor
5 ClO3- (aq) +
So
STEP #5:
=
30 e- **
5 Cl- (aq)
3 I2 (aq)
6 IO3- (aq)
+
Since in acidic solution, balance “O” atoms using H2O (l), then “H” using H+ (aq)
5 ClO3- (aq) + 3 I2 (aq)
3 H2O (l) +
5 Cl- (aq) + 6 IO3- (aq) + 6 H+ (aq)
IF BASIC SOLUTION:
STEP #6:
Add OH- (aq) to both sides to equal H+ (aq)
STEP #7:
Combine OH- (aq) and H+ (aq) (to form water) and rationalize H2O (l) amounts.
Example #2: Basic Solution
STEP #1:
-5 +1 –2 +1
CH3OH (aq) +
STEP #2 & #3:
STEP #4:
STEP #5:
2 H2O (l) +
REDUCTION
+7
-2
+4 -2
MnO4- (aq)
CO3
OXIDATION
and
6 e- / CH3OH
Mn : +7
+6
C : -2
Therefore
Common Factor: 6 eSo
CH3OH (aq) +
6 MnO4- (aq)
CH3OH (aq)
2-
+ 6 MnO4- (aq)
(aq)
+
+4 Therefore
+6
–2
MnO42- (aq)
lose 6 e- / C
1 e- / MnO4-
gain 1 e- / Mn
and
CO32- (aq)
+
6 MnO42- (aq)
CO32- (aq)
+ 6 MnO42- (aq) + 8 H+ (aq)
If this reaction were in an acidic solution, you would now be finished. However for a basic solution, we add
enough OH- (aq) to both sides to equal the number of H+ (aq) present. The hydrogen and hydroxide ions on
the same side of the equation are then combined to form water.
STEP #6:
8 OH- (aq) + 2 H2O (l) +
CH3OH (aq)
+ 6 MnO4- (aq)
CO32- (aq) + 6 MnO42- (aq) + 8 H+ (aq) + 8 OH- (aq)
DAY #2
Date: ___________
PAGE #3
SCH 4U
Unit: Electrochemistry
Example #2: Basic Solution
(continued)
Finally, rationalize the total number of H2O (l) on both sides. In this case, the water on the reactant side can
be cancelled by also removing 2 H2O (l) from the product side, leaving the 6 extra water in the final equation.
STEP #7:
8 OH- (aq) + CH3OH (aq)
+ 6 MnO4- (aq)
CO32- (aq) + 6 MnO42- (aq) + 6 H2O (l)
PAGE 668 Q# 2, 3
Writing Half-Reaction Equations
Example #1:
Nitrous acid can be reduced in acidic solution to form nitrogen monoxide gas.
STEP #1:
Write chemical reaction.
HNO2
(aq)
NO (g)
STEP #2:
Balance N (atoms other than O or H)
HNO2
(aq)
NO (g)
STEP #3:
Balance O with H2O (l)
HNO2
(aq)
NO (g) + H2O (l)
STEP #4:
Balance H with H+ (aq)
H+ (aq) + HNO2
(aq)
NO (g) + H2O (l)
STEP #5:
Balance charge with e-
e- + H+ (aq) + HNO2
(aq)
NO (g) + H2O (l)
STEP #6:
State if it is an oxidation or reduction
REDUCTION (HNO2 is gaining e-)
DAY #2
Date: ___________
PAGE #4
SCH 4U
Unit: Electrochemistry
Writing Half-Reaction Equations (continued)
Example #2:
Copper metal can be oxidized in basic solution to copper (I) oxide.
STEP #1:
Write chemical reaction.
Cu
STEP #2:
Balance Cu (atoms other than O or H)
2 Cu
STEP #3:
Balance O with H2O (l)
STEP #4:
Balance H with H+ (aq)
H2O (l) + 2 Cu
(s)
Cu2O (aq) + 2 H+ (aq)
STEP #5:
Balance charge with e-
H2O (l) + 2 Cu
(s)
Cu2O (aq) + 2 H+ (aq) + 2 e-
STEP #6:
Since basic solution
STEP #7:
Combine to form water 2 OH- (aq) + H2O (l) + 2 Cu (s)
STEP #8:
Finally cancel waters
Example #3:
Cu2O (aq)
(s)
Cu2O (aq)
(s)
H2O (l) + 2 Cu
2 OH- (aq) + H2O (l) + 2 Cu
2 OH- (aq) + 2 Cu
(s)
(s)
(s)
Cu2O (aq)
Cu2O (aq) + 2 H+ (aq) + 2 e- + 2 OH- (aq)
Cu2O (aq) + 2 H2O (l) + 2 eCu2O (aq) + H2O (l) + 2 e-
Aqueous permanganate ions are converted to solid manganese (IV) oxide in
basic solution. Write the correct half reaction.
DAY #2
Date: ___________
PAGE #5
SCH 4U
Unit: Electrochemistry
B
Balancing REDOX EQUATIONS by HALF-REACTION METHOD
Again, many REDOX reactions occur in acidic/basic solutions, therefore H2O (l) , H+ (aq) and
OH- (aq) are used for balancing
Example #1:
Balance the following equation by half reaction method if it were found in an
acidic solution.
Mn2+ (aq)
STEP #1:
+
HBiO3 (aq)
Bi3+ (aq)
+
MnO4- (aq)
Separate the skeleton equation into the start of two half-reaction equations.
a)
Mn2+ (aq)
MnO4- (aq)
b)
HBiO3 (aq)
Bi3+ (aq)
STEP #2:
Balance each half-reaction equation. (See writing half-reactions)
a)
4 H2O (l)
b)
5 H+ (aq) + 2 e-
STEP #3:
Multiply each 1/2 reaction by simple whole numbers to balance e- lost and gained
+
Mn2+ (aq)
MnO4- (aq) + 8 H+ (aq) + 5 e-
+
Bi3+ (aq) + 3 H2O (l)
COMMON FACTOR
+
HBiO3 (aq)
=
10
2 Mn2+ (aq)
2 MnO4- (aq) + 16 H+ (aq) + 10 e-
a) x 2
8 H2O (l)
b) x 5
25 H+ (aq) + 10 e- +
STEP #4:
Add two half reactions canceling e– and anything else that is the same on reactant vs.
product side
a) x 2
8 H2O (l)
b) x 5
25 H+ (aq) + 10 e- +
+
5 HBiO3 (aq)
2 Mn2+ (aq)
5 HBiO3 (aq)
9 H+ (aq) + 2 Mn2+ (aq) + 5 HBiO3 (aq)
5 Bi3+ (aq) + 15 H2O (l)
2 MnO4- (aq) + 16 H+ (aq) + 10 e5 Bi3+ (aq) + 15 H2O (l)
5 Bi3+ (aq) + 2 MnO4- (aq) + 7 H2O (l)
DAY #2
Date: ___________
PAGE #6
SCH 4U
Unit: Electrochemistry
B
Balancing REDOX EQUATIONS by HALF-REACTION METHOD (continued)
Example #2:
Balance the following equation by half reaction method if it were found in an
basic solution.
Cr(OH)3 (aq) +
STEP #1:
IO3- (aq)
CrO4 2- (aq) +
I- (aq)
Separate the skeleton equation into the start of two half-reaction equations.
a)
b)
STEP #2:
Balance each half-reaction equation. (See writing half-reactions)
a)
b)
STEP #3:
Multiply each 1/2 reaction by simple whole numbers to balance e- lost and gained
COMMON FACTOR
=
a) x ____
b) x ____
STEP #4:
Add two half reactions canceling e– and anything else that is the same on reactant vs.
product side
a)
b)
PAGE 673 Q# 6,7