Lecture 11
Aqueous Solutions and
Chemical Reactions II
REDOX Reactions in
Acid and Basic Solutions
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Acids
Acids donate protons in aqueous solutions
HCl (aq) Æ H+ (aq) + Cl- (aq)
H2SO4 (aq) Æ H+ (aq) + HSO4- (aq)
Strong acid like HCl and H2SO4 experience
~100% dissociation in polar solvents, just
like soluble ionic compounds.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Redox Reactions in Acidic
Solutions (I)
Reactions between metals and all strong acids
except Nitric Acid
H+ ions can oxidize many metals (X).
H+ (aq) + X (s) Æ H2 (g) + X a+ (aq)
H+(aq) pulls electrons out of the metal to form H2(g).
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Ex) Redox Reactions in Acidic
Solutions (I)
Ex) What happens when a solid piece of
magnesium is placed in a solution of
hydrochloric acid?
Mg (s) + 2 H+ (aq) Æ Mg2+ (aq) + H2 (g)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Redox Reactions in Acidic
Solutions (II)
Reactions between metals and Nitric Acid.
NO3- is a stronger oxidizing agent them H+.
Thus, NO3- will oxidize metals that H+ cannot.
NO3- (aq) + X (s) Æ NO (g) + X a+ (aq)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Ex1) Redox Reactions in Acidic
Solutions (II)
Ex1) A solid copper penny is placed in a
solution of nitric acid.
Cu (s) + NO3- (aq) Æ NO (g) + Cu2+(aq)
We need to balance using a special procedure.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Balancing REDOX Reactions
in Acidic Solutions
1) Write the two unbalanced half-reactions.
• Oxidation half-reaction
• Reduction half-reaction
For each of the half-reactions:
2) Balance all atoms except for O and H.
3) Balance for O by adding H2O molecules.
4) Balance for H by adding H+ ions.
5) Balance the charge by adding electrons.
6) Cross multiply to cancel e-.
7) Add half reactions and cancel things that are the same.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Reduction Half-Reaction
Cu (s) + NO3- (aq) Æ NO (g) + Cu2+(aq)
Step 1) NO3- (aq) Æ NO (g)
Step 2) NO3- (aq) Æ NO (g)
(Nitrogen is already balanced)
Step 3) NO3- (aq) Æ NO (g) + 2 H2O (l)
Step 4) NO3- (aq) + 4 H+ (aq) Æ NO (g) + 2 H2O (l)
Step 5) NO3- (aq) + 4 H+ (aq) + 3e- Æ NO (g) + 2H2O (l)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Oxidation Half-Reaction
Cu (s) + NO3- (aq) Æ NO (g) + Cu2+(aq)
Step 1) Cu (s) Æ Cu2+(aq)
Step 2) Cu (s) Æ Cu2+(aq) (copper is already balanced)
Step 3) Cu (s) Æ Cu2+(aq) (we don’t need to balance O’s)
Step 4) Cu (s) Æ Cu2+(aq) (we don’t need to balance H’s)
Step 5) Cu (s) Æ Cu2+(aq) + 2 e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 6) Cross multiply
Reduction half-reaction
[NO3- (aq) + 4 H+ (aq) + 3e- Æ NO (g) + 2 H2O (l)] x 2
= 2 NO3- (aq) + 8 H+ (aq) + 6e- Æ 2 NO (g) + 4 H2O (l)
Oxidation half reaction
[Cu (s) Æ Cu2+(aq) + 2 e-] x 3
= 3 Cu (s) Æ 3 Cu2+(aq) + 6 e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 7) Add the two ½ reactions
2NO3- (aq) + 8H+ (aq) + 6e- Æ 2NO (g) + 4H2O (l)
3Cu (s) Æ 3Cu2+(aq) + 6 e-
2NO3-(aq) + 3Cu(s) + 8H+(aq) Æ 2NO(g) + 3Cu2+(aq) + 4H2O(l)
The balanced redox reaction
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Strong oxidizing agents that act
in acidic solutions
NO3- (aq) Æ NO (g)
Nitrate)
Permanganate)
MnO4- (aq) Æ Mn2+(aq)
Dichromate)
Cr2O72- (aq) Æ Cr3+(aq)
All of the above oxidize p and d block metals, sulfite
ions, peroxides, and substances that have a lower (less
positive) oxidation state than usual, in acidic solutions.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Oxidation of Sulfite
SO32- Æ SO42The oxidation number of sulfur increases from
+4 to +6
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Oxidation of Hydrogen Peroxide
H2O2(aq) Æ O2(g)
The oxidation number on oxygen increases
from -1 to 0.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Substances with a lower
(less positive)
oxidation state than usual.
• In C2O42-, Carbon has an oxidations state of +3.
• As Carbon is in group 4A, it would rather have
an oxidation state of +4.
• It will oxidize (lose electrons) to raise its
oxidation state.
C2O42- (aq) Æ
CO2 (g)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Substances with a lower
(less positive)
oxidation state than usual.
Some non-metals can raise their oxidation states by
bonding with like elements.
This happens when hydrochloric acid is combined
with a strong oxidizing agent.
e.g.) Chlorine ion (Cl-)
2 Cl-(aq) Æ
Cl2(g)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
What if you have a choice of two
species that can be oxidized?
When compounds contain p-block or d-block
metals and non-metals with less positive than
usual oxidation states…
The metal will be oxidized!!
e.g. SnCl2
Sn2+ will be oxidized to Sn4+
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Ex2) Redox Reactions in Acidic
Solutions (II)
Ex2) An acidic solution containing tin (II) chloride
is mixed with a solution of potassium
permanganate.
KMnO4 (aq) + SnCl2 (aq) Æ
MnO4- (aq) + Sn2+(aq) Æ Mn2+(aq) + Sn4+(aq)
We need to balance using our special procedure.
If the question states that the solution is acidic or acidified,
you know you need to balance using this procedure.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Reduction Half-Reaction
MnO4- (aq) + Sn2+(aq) Æ Mn2+(aq) + Sn4+(aq)
Step 1) MnO4- (aq) Æ Mn2+(aq)
Step 2) MnO4- (aq)Æ Mn2+(aq) (Manganese already balanced)
Step 3) MnO4- (aq) Æ Mn2+(aq) + 4 H2O (l)
Step 4) MnO4-(aq) + 8 H+(aq) Æ Mn2+(aq) + 4 H2O (l)
Step 5) MnO4-(aq) + 8H+(aq)+ 5e- Æ Mn2+(aq)+4 H2O(l)
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Oxidation Half-Reaction
MnO4- (aq) + Sn2+(aq) Æ Mn2+(aq) + Sn4+(aq)
Step 1) Sn2+(aq) Æ Sn4+(aq)
Step 2) Sn2+(aq) Æ Sn4+(aq)
(tin is already balanced)
Step 3) Sn2+(aq) Æ Sn4+(aq)
(we don’t need to balance O’s)
Step 4) Sn2+(aq) Æ Sn4+(aq)
(we don’t need to balance H’s)
Step 5) Sn2+(aq) Æ Sn4+(aq) + 2 e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 6) Cross multiply
Reduction half-reaction
[MnO4-(aq) + 8H+(aq)+ 5 e- Æ Mn2+(aq) + 4 H2O(l)] x 2
= 2 MnO4-(aq) + 16 H+(aq)+ 10 e- Æ 2 Mn2+(aq) + 8 H2O(l)
Oxidation half reaction
[Sn2+(aq) Æ Sn4+(aq) + 2 e-] x 5
= 5 Sn2+(aq) Æ 5 Sn4+(aq) + 10 e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 7) Add the two ½ reactions
2 MnO4-(aq) + 16 H+(aq)+ 10e- Æ 2 Mn2+(aq) + 8 H2O(l)
5 Sn2+(aq) Æ 5 Sn4+(aq) + 10 e-
2 MnO4- + 5 Sn2+ + 16 H+Æ 2 Mn2+ + 5 Sn4+ + 8 H2O
The balanced redox reaction
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Basic Solutions
• Basic solutions have a high concentration
of hydroxide ions [OH-].
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Strong oxidizing agent that acts
in Basic Solutions
Permanganate)
MnO4- (aq) Æ MnO2 (aq)
Permanganate will oxidize p or d block metals, sulfite
ions, and substances that have a lower (less positive)
oxidation state than usual, in basic solutions.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Ex) Redox Reactions in Basic
Solutions
Ex) A basic solution containing magnesium sulfite
is mixed with a solution of potassium
permanganate.
KMnO4 (aq) + MgSO3 (aq) Æ
MnO4- (aq) + SO32-(aq) Æ MnO2(aq) + SO42-(aq)
We need to balance using our special procedure.
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Reduction Half-Reaction
MnO4- (aq) + SO32-(aq) Æ MnO2(aq) + SO42-(aq)
Step 1) MnO4- (aq) Æ MnO2(aq)
Step 2) MnO4- (aq) Æ MnO2(aq)
(Manganese is balanced)
Step 3) MnO4- (aq) Æ MnO2(aq) + 2 H2O
Step 4) MnO4- (aq) + 4 H+ Æ MnO2(aq) + 2 H2O
Step 5) MnO4- (aq) + 4 H+ + 3e- Æ MnO2(aq) + 2 H2O
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
The Oxidation Half-Reaction
MnO4- (aq) + SO32-(aq) Æ MnO2 (aq) + SO42-(aq)
Step 1) SO32-(aq) Æ SO42-(aq)
Step 2) SO32-(aq) Æ SO42-(aq)
(sulfur is already balanced)
Step 3) SO32-(aq) + H2O (l) Æ SO42-(aq)
Step 4) SO32-(aq) + H2O (l) Æ SO42-(aq) + 2H+(aq)
Step 5) SO32-(aq) + H2O (l) Æ SO42-(aq) + 2H+(aq) + 2e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 6) Cross multiply
Reduction half-reaction
[MnO4-(aq) + 4 H+ + 3e- Æ MnO2(aq) + 2 H2O] x2
= 2 MnO4-(aq) + 8 H+ + 6e- Æ 2 MnO2(aq) + 4 H2O
Oxidation half reaction
[SO32-(aq) + H2O (l) Æ SO42-(aq) + 2 H+(aq) + 2e-] x3
= 3 SO32-(aq) + 3 H2O (l) Æ 3 SO42-(aq) + 6 H+(aq) + 6e© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 7) Add the two ½ reactions
2 MnO4- + 8 H+ + 6e- Æ 2 MnO2 + 4 H2O
3 SO32- + 3 H2O Æ 3 SO42- + 6H+ + 6e-
2MnO4- + 3 SO32- + 2 H+ Æ 2MnO42- + 3 SO42- + H2O
For basic solutions there is one more step!
© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com
Step 8) Add OH- to get rid of H+
2MnO4- + 3 SO32- + 2 H+ Æ 2 MnO2 + 3 SO42- + H2O
+ 2 OH-
+ 2 OH-
2 MnO4- + 3 SO32- + H2O Æ 2 MnO2 + 3 SO42- + 2 OH© 2008 AP Chem Solutions. All rights reserved. www.apchemsolutions.com