Chem 1011 – Intersession 2011
Class #3
11-May-11
1
Class 3:
Balancing Redox Reactions
• Sec 18.2 – Balancing Redox Reactions
▫ Half reaction method of balancing aqueous redox
equations in either acidic or basic solutions
• But first! Last time, on Chem 1011...
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Oxidizing / Reducing Agents
•
Oxidizing Agent
▫
causes something to be oxidized, by taking electrons
from it. When it takes these electrons, it undergoes
reduction (since it takes the electrons for itself).
Oxidizing agents gain electrons and become
reduced.
▫

Oxidizing agent refers to the entire substance where
the atom causing the oxidation is found.
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Oxidizing / Reducing Agents
•
Reducing Agent
▫
causes something to be reduced, by giving electrons
to it. When it gives these electrons it undergoes
oxidation (since it has lost its own electrons).
Reducing agents lose electrons and become
oxidized.
▫

Reducing agent refers to the entire substance where
the atom causing the reduction is found.
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
4
Problems
• For the following reactions, write the net ionic
equation and the half-reactions. For each of the
half reactions, state which is the oxidation halfreaction, which is the reduction half-reaction,
what is the oxidizing agent and what is the
reducing agent.
▫ (a)
Mg(s) + FeSO4(aq) → Fe(s) + MgSO4(aq)
▫ (b)
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
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Solution for (b)
• Total Ionic Equation:
Cu(s) + 2 Ag+(aq) + 2 NO3–(aq) → 2 Ag(s) + Cu2+(aq) + 2 NO3–(aq)
Spectator Ions
• Net Ionic Equation:
Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
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Oxidation Half Reaction
Cu(s) → Cu2+(aq) + 2 e–
Oxidation #: 0
+2
•
Cu loses 2 electrons
Notice
▫
▫
•
Increase in oxidation number ; loss of electrons
This is an oxidation
Reducing agents lose electrons and become
oxidized
▫
Therefore, Cu(s) is the reducing agent.
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
7
Reduction Half Reaction
Oxidation #:
•
Ag gains an electron
Ag+(aq) + e– → Ag(s)
+1
0
Notice
▫
▫
•
Reduction in oxidation number; gain of electrons
This is a reduction
Oxidizing agents gain electrons and become
reduced
▫
Therefore, Ag+(aq) is the oxidizing agent.
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Solution for (b)
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
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Combustion
• Combustion reactions are also redox!
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
–4 +1
0
+4 –2
+1 –2
Reduction
Oxidation
• Oxidation and reduction must occur simultaneously.
• Can’t have an oxidation without a reduction! (and vice versa)
• If an atom loses electrons, another atom must take them.
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
10
Balancing Redox Reactions
•
Balancing redox reactions by inspection can be
challenging because we need to balance for both
mass (atoms of each type) and charge.
•
Redox reactions taking place in aqueous solution can
be balanced by applying a special procedure knows
as the half-reaction method of balancing.
•
The steps involved differ slightly for reactions taking
place in acidic and basic solutions. We will look at
both cases.
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Half-Reaction Method
• Example – balance the following equation in
acidic solution:
SO32–(aq) + MnO4–(aq) → SO42–(aq) + Mn2+(aq)
1. Assign all oxidation numbers and identify your redox
pairs (oxidation and reduction):
SO32–(aq) + MnO4–(aq) → SO42–(aq) + Mn2+(aq)
+4 –2
+7 –2
+6 –2
+2
Reduction
Oxidation
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Half-Reaction Method
2. Separate into two half reactions
▫
Oxidation:

▫
SO32–(aq) → SO42–(aq)
Reduction:

MnO4–(aq) → Mn2+(aq)
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
13
Half-Reaction Method
3. Balance both equations in the following order:
a. All atoms other than O and H
Oxidation:
SO32–(aq) → SO42–(aq)
Reduction:
MnO4–(aq) → Mn2+(aq)
(both are already balanced in that regard)
b. Balance for oxygen by adding H2O(l) with the
appropriate coefficient
Oxidation:
Reduction:
SO32–(aq) + H2O(l) → SO42–(aq)
MnO4–(aq) → Mn2+(aq) + 4 H2O(l)
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Half-Reaction Method
3. Finish balancing the atoms...
c. Balance for hydrogen by adding H+(aq) with the
appropriate coefficient (Remember: this is an acidic
solution, after all!)
Oxidation:
Reduction:
SO32–(aq) + H2O(l) → SO42–(aq) + 2 H+(aq)
MnO4–(aq) + 8 H+(aq)→ Mn2+(aq) + 4 H2O(l)
It’s important to balance for atoms using this order!
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Half-Reaction Method
4. Balance for charge
▫
Add the number of electrons necessary to equate
the total charge on both sides
Oxidation:
SO32–(aq) + H2O(l) → SO42–(aq) + 2 H+(aq) + 2 e–
Reduction: MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
16
Half-Reaction Method
5. Multiply half reactions to obtain the lowest
common multiple for the electrons
Oxidation:
5 x [SO32–(aq) + H2O(l) → SO42–(aq) + 2 H+(aq) + 2 e–]
= 5 SO32–(aq) + 5 H2O(l) → 5 SO42–(aq) + 10 H+(aq) + 10 e–
Reduction:
2 x [MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)]
= 2 MnO4–(aq) + 16 H+(aq) + 10 e– → 2 Mn2+(aq) + 8 H2O(l)
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Half-Reaction Method
6. Re-Combine the half reactions, being sure that
all electrons cancel, as well as any other like
species.
Ox:
Red:
5 SO32–(aq) + 5 H2O(l) → 5 SO42–(aq) + 10 H+(aq) + 10 e–
2 MnO4–(aq) + 16 H+(aq) + 10 e– → 2 Mn2+(aq) + 8 H2O(l)
Total: 5 SO32–(aq) + 2 MnO4–(aq) + 6 H+(aq) → 5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l)
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Half-Reaction Method
7. Verify that everything is balanced (atoms and
total charge)
5 SO32–(aq) + 2 MnO4–(aq) + 6 H+(aq) → 5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l)
Reactants
Products
•
•
•
•
•
•
•
•
•
•
5S
2 Mn
23 O
6H
Charge = –6
Balancing Redox Equations
5S
2 Mn
23 O
6H
Charge = –6
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Chem 1011 – Intersession 2011
Class #3
11-May-11
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Problem
• Balance the following redox equation in acidic
solution
Cr2O72–(aq) + HNO2(aq) → Cr3+(aq) + NO3–(aq)
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Basic Solution
• The procedure is identical, just add ―step 8‖
▫ Step 8 is the conversion from acidic to basic
• We know that H+(aq) is indicative of an acid – so
neutralize it with base, OH–(aq)
5 SO32–(aq) + 2 MnO4–(aq) + 6 H+(aq) → 5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l)
• There are 6 H+(aq) ions on the left – neutralize
that with 6 OH–(aq) ions
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Basic Solution
5 SO32–(aq) + 2 MnO4–(aq) + 6 H+(aq) → 5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l)
+ 6 OH–(aq)
+ 6 OH–(aq)
6 H2O(l)
• Since the equation was balanced to begin with, if we add
6 OH– (aq) to the left hand side, we must add 6 OH–(aq) to
the right hand side to maintain the balance
• The addition of OH–(aq) to H+(aq) produces H2O(l)
• So, now our equation is:
5 SO32–(aq) + 2 MnO4–(aq) + 6 H2O(l) →
5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l) + 6 OH–(aq)
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
22
Basic Solution
• Since we just created more H2O(l), look to see if you
can cancel any
▫ Since there are 6 on the left and 3 on the right, we
can cancel out 3 of these
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5 SO32–(aq) + 2 MnO4–(aq) + 6 H2O(l) →
5 SO42–(aq) + 2 Mn2+(aq) + 3 H2O(l) + 6 OH–(aq)
• So, in basic solution, the balanced redox equation is:
5 SO32–(aq) + 2 MnO4–(aq) + 3 H2O(l) → 5 SO42–(aq) + 2 Mn2+(aq) + 6 OH–(aq)
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Check it over!
•
Like before, verify that all of the atoms are
balanced, as well as the charge
5 SO32–(aq) + 2 MnO4–(aq) + 3 H2O(l) → 5 SO42–(aq) + 2 Mn2+(aq) + 6 OH–(aq)
Reactants
Products
•
•
•
•
•
•
•
•
•
•
5S
2 Mn
26 O
6H
Charge = –12
5S
2 Mn
26 O
6H
Charge = –12
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Alternatively
• There is another way to approach a redox equation
in basic solution, if you choose to do so
▫ Instead of waiting until the end for ―Step 8,‖ you can
neutralize the H+(aq) with OH—(aq) right after you added
it when balancing the half reactions
▫ Follow the same procedure:
 You must add the same amount of OH—(aq) to both sides
to maintain the balance
 When this forms water, cancel what you can
Balancing Redox Equations
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Chem 1011 – Intersession 2011
Class #3
11-May-11
25
Problem
•
Write the balanced equation for this reaction in
basic solution.
CN–(aq) + MnO4–(aq) → OCN–(aq) + MnO2(s)
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Coming up at 11
•
Sec 9.5 – Covalent Bonding: Lewis Structures
▫
▫
▫
Single Covalent Bonds
Double and Triple Covalent Bonds
Covalent Bonding: Models and Reality
Balancing Redox Equations
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