Practice (from last lesson):
2. Consider Mn3+  MnO4-. Is Mn oxidized or reduced? Oxidized
Element
Mn
ON of Reactant
+3
ON of Product
+7
∆ON
+4
3.Consider 3SO2 + 3H2O + ClO3-  3SO42- + 6H+ + ClElement ON of Reactant ON of Product
∆ON
H
+1
+1
0
Cl
+5
-1
-6
O
-2
-2
0
S
+4
+6
+2
a) Which element is oxidized? S reduced? Cl
b) Which element gains electrons? Cl loses electrons? S
c) Which element is the RA? S OA? Cl
Practice:
4. All of the following are oxidation or reduction half-reactions. Find
the ∆ON of the element in which it changes and identify each as an
oxidation or reduction.
Element that
Oxidation or
Half-Reaction
∆ON
changes ON
reduction?
C2H5OH  CH3COOH
C
+2
Oxidation
Fe2O3  FeO
Fe
-1
Reduction
H3PO4  P4
P
-5
Reduction
CH3COOH  CH3COH
C
-1
Reduction
5. Circle the following which are redox reactions. Use the ∆ON to
determine the answer.
Unit 5 – Electrochemistry
5.5 Balancing Redox Reactions
Lesson Goals:
• balance the equation for a net ionic redox
reaction in acidic or basic solution
Homework /Next Class(es)?
Hebden pg 207 # 24
Date
Next Class
A) Balancing Redox Reactions (Acidic)
Consider the following reaction:
U4+ + MnO4-  Mn2+ + UO22+ (acidic)
Major Species
What is happening
to the ON’s?
Reactant
Product
Reduction or Oxidation?
U
U4+ = +4
UO22+ = +6
Oxidation
Mn
MnO4- = +7
Mn2+ = +2
Reduction
Try your best to balance this REDOX reaction (hint: Think 5.4)
Incorrect!!
2U4+ + MnO4-  Mn2+ + 2UO22+
But charges don’t balance!
Incorrect!! U4+ + MnO4- + 4H+ + 3e-  Mn2+ + UO22+ + 2H2O
Wait! Electrons cannot be shown…. It’s not a redox reaction
Correct Answer: 5U4+ + 2H2O + 2MnO4-  5UO22+ + 4H+ + 2Mn2+
It’s near impossible to balance like we normally do, EVEN with MOHE
A) Balancing Redox Reactions (Acidic)
In order to properly balance a REDOX REACTION, you must:
i) Break up reaction into two ½ reactions
ii) Balance with MOHE MOHE!
Example: U4+ + MnO4-  Mn2+ + UO22+ (acidic)
(U4+ + 2H2O  UO22+ + 4H+ + 2e- ) x 5
(MnO4- + 8H+ + 5e-  Mn2+ + 4H2O ) x 2
4
2
5U4++10 H2O+2MnO4-+16H++10e-5UO22++20 H++2Mn2++8H2O+10e-
(5)
5U4+ + 2H2O + 2MnO4-  5UO22+ + 4H+ + 2Mn2+
Practice (You Try)
1. Balance the following reaction:
SO2 + IO3-  SO42- + I2 (acidic solution)
(SO2 +2H2O  SO42- + 4H++ 2e- ) x 5
( 2IO3- + 12H+ + 10e-  I2 + 6H2O )
________________________________________________
5SO2 +10H2O + 2IO3- + 12H+ + 10e-  5SO42- + 20H++ 10e- + I2 + 6H2O
5SO2 + 2IO3- + 4H2O  5SO42- + 8H+ + I2
B) Balancing Redox Reactions (Basic)
Follow the redox balancing rules for balancing in acidic… BUT
convert to base by adding the water reaction to cancel H+
H2O  H+ + OHor
H+ + OH-  H2O
Example 2: Balance the following reaction in a basic solution:
SO2 + IO3-  SO42- + I2 (basic)
(from answer to practice #1)
5SO2 + 2IO3- + 4H2O  5SO42- + 8H+ + I2
+ 8OH-
+8OH-
8OH- + 5SO2 + 2IO3- + 4H2O  5SO42- + 8H2O + I2
8OH- + 5SO2 + 2IO3-  5SO42- + 4H2O + I2
C) Balancing Advanced Redox Reactions
• TYPE 1: Some redox reactions involve the REACTANTS self-reacting
in a reduction-oxidation reaction;
– E.g. Br2  Br- + BrO3-
• TYPE 2: Some redox reactions involve many different chemical
species and it can be hard to determine what is happening.
• -E.g. KMnO4 + H2S + H2SO4  K2SO4 + MnSO4 + S (acidic)
TYPE 1: Self-Redox
• Eg) Br2  Br- + BrO3- (acidic)
Half rx’s are:
[1]/[2] ( Br2 + 2e-  2 Br- ) x 5
+ + 10e2
+
12H
+
6H
O
Br2
2  BrO3
[3] 6Br2 +6H2O +10e-  10Br- + 2BrO3- + 12H+ + 10e[4]6Br2 + 6H2O  10Br- + 2BrO3- + 12H+
÷2
Overall Reaction
3Br2 + 3H2O  5Br- + BrO3- + 6H+
TYPE 2: Difficult Starting Point
Example 4:
KMnO4 + H2S + H2SO4  K2SO4 + MnSO4 + S (acidic)
(Step 1) In challenging balancing reactions, it can be useful to use
oxidation numbers to tell you what the two half reactions are.
Element
K
Mn
S
H
O
ON of
Reactant
+1
+7
-2 in H2S
+6 in SO42+1
-2
ON of
Product
+1
+2
0 in S
+6 in SO42+1
-2
∆ON
0
-5
+2
0
0
0
Reduction or
Oxidation?
None
Reduction
Oxidation
None
None
None
TYPE 2: Difficult Starting Point
Example 4:
KMnO4 + H2S + H2SO4  K2SO4 + MnSO4 + S (acidic)
(2) Next split reaction into its two simple half reactions
Oxidation Reaction:
Reduction Reaction:
( H2S
 S + 2H+ + 2e- ) x 5
Where did the “SO4” come from?
2 KMnO4 + 3 H2SO4 + 10H+ + 10e-  2 MnSO4 + K2SO4 + 8H2O
Where did the “K” go?
Balanced Redox Reaction:
2KMnO4 + 3H2SO4 + 5H2S  2MnSO4 + K2SO4 + 8H2O + 5S
(3) Add the missing chemical species into one, OR both half-reaction as needed.
(4) Continue balancing using MOHE
Practice:
2. Balance the following reactions:
a) S2- + ClO3- → Cl- + S (basic)
3 S2- + ClO3- + 3H2O → 3S + Cl- + 6OHAcidic / basic just means the conditions
are acidic/basic.
•
It doesn’t mean that the H+ / OHneeds to be a part of the reaction
b) ClO- → Cl- + ClO3- (basic)
3ClO- → 2Cl- + ClO3•
Homework /Next Class(es)?
Hebden pg 207 # 24
However it does mean that acidic
solutions shouldn’t have OH- and
basic solutions should not have H+
Date
Next Class