Download Lecture 03B - Balancing Redox

Daniel L. Reger
Scott R. Goode
David W. Ball
www.cengage.com/chemistry/reger
Lecture 03B (Chapter 18,
sections 18.1, 18.2)
Balancing Redox Reactions
Oxidation and Reduction
• Oxidation is the loss of electrons by a chemical
process.
• When sodium forms a compound, Na+ is formed.
Sodium is oxidized.
• Reduction is the gain of electrons by a chemical
process.
• When Cl- ions are formed from elemental chlorine,
chlorine is reduced.
Oxidation-Reduction (“Redox”)
• An oxidation-reduction reaction, or redox
reaction, is one in which electrons are transferred
from one species to another.
• In every redox reaction, at least one species is
oxidized and at least one species is reduced.
• 2Na(s) + Cl2(g) → 2NaCl(s) is a redox reaction
because Na is oxidized and Cl is reduced.
Oxidizing and Reducing Agents
• An oxidizing agent is the reactant that accepts
electrons, causing an oxidation to occur.
• The oxidizing agent is reduced.
• A reducing agent is the reactant that supplies
electrons, causing a reduction to occur.
• The reducing agent is oxidized.
• In the reaction of sodium with chlorine, Na is the
reducing agent and Cl2 is the oxidizing agent.
Half-reactions
• In a half-reaction, either the oxidation or reduction
part of a redox reaction is given, showing the
electrons explicitly.
• Half-reactions emphasize the transfer of electrons
in a redox reaction.
• For 2Na(s) + Cl2(g) → 2NaCl(s) :
• Na → Na+ + 1eoxidation half-reaction
• Cl2 + 2e- → 2Clreduction half-reaction
Oxidation States
• The oxidation state is the charge on the monatomic
ion, or the charge on an atom when the shared
electrons are assigned to the more electronegative
atom.
• Electron pairs shared by atoms of the same
element are divided equally.
• In CaCl2, an ionic compound:
• calcium has an oxidation state of +2.
• chlorine has an oxidation state of -1.
Rules for Determining Oxidation Numbers
Rule 1: The O.N. of an atom in its (uncombined) elemental state is zero.
Exs. Al(0), O2(0), H2(0), Br2(0), Na(0)
Rule 2: +The O.N. 2+
of a monatomic
ion2-is equal-to the charge on the ion.
3+
Exs. Na (+1), Mg (+2), Al (+3), O (-2), Br (-1)
Rule 3: Atoms in polyatomic ions or molecular compounds usually have O.N.s identical
to the charges they would have as ions.
Exceptions:
-Hydrogen (except as H2) is usually +1, but sometimes -1 (ex. H-Ca-H)
-Oxygen (except as O2) is usually -2, but can be -1 when –O-O- bonds exist (peroxides)
-Halogens usually -1, unless combine with more EN element (dichlorine monoxide, ClO-Cl, Cl is +1; F is always -1)
Rule 4: The algebraic sum of all O.N.s of all atoms in a neutral compound or
polyatomic ion is equal to the net charge.
- For neutral compounds, the sum of the O.N.s is 0
- For a charged polyatomic ion, the sum of the O.N.s is equal to the charge.
Assigning Oxidation Numbers
• Assign oxidation numbers for each atom in K2CrO4.
• K = +1 by rule 2.
• O = -2 by rule 5.
• Cr = +6 by rule 6.
• 2(+1) + 6 + 4(-2) = 0, the overall charge on the
substance.
Assigning Oxidation Numbers
K2CO3
CH2O
NO3
-
2(O.N of K)
+
(O.N. of C)
+
3(O.N. of O)
= 0
2(+1)
+
?
+
3(-2)
= -4
2(+1)
+
(+4)
+
3(-2)
= 0
2(O.N. of H)
+
(O.N. of O)
= 0
2(+1)
+
(-2)
= 0
+
(-2)
= 0
(O.N of C)
+
(0)
+
2(+1)
(O.N of N)
+
3(O.N. of O)
= -1
+
3(-2)
= -6
+
3(-2)
= -1
(+5)
Test Your Skill
• Assign oxidation numbers to each atom in the
following substances.
(a) PF3 (b) CO
(c) NH4Cl
Test Your Skill
• Assign oxidation numbers to each atom in the
following substances.
(a) PF3 (b) CO
(c) NH4Cl
• Answers:(a) P = +3, F = -1
(b) C = +2, O = -2
(c) N = -3, H = +1, Cl = -1
Redox Rxns Determined by Change in O.N.
S(s)
+
(O.N of uncombined
element = 0)
O2(g) 
(O.N of uncombined
element = 0)
SO2(g)
(O.N. of S) + 2(O.N. of O) = 0
(+4)
+ 2(-2)
= 0
Sulfur is oxidized, Oxygen is reduced
4Al(s)
(O.N of uncombined
element = 0)
+
3O2(g) 
(O.N of uncombined
element = 0)
2Al2O3(s)
2(O.N. of Al) + 3(O.N. of O) = 0
+ 3(-2)
= -6
2(+3)
 Al is oxidized (O.N. increased by 3), Al is reducing agent
 Oxygen is reduced (O.N. decreases by 2), O is oxidizing agent
+ 3(-2)
= 0
Redox e- Transfer: Examples
SO2(g) + H2O(l)  H2SO3(aq)
S
+4
O
-2
H
+4
-2
-2
+1
+1
4Fe(s) + 3O2 (g)  2Fe2O3(s)
Balancing Net Ionic Equations Using Redox
Half-reactions
Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(aq)
With more complex net ionic equations, it is
not always easy to see how the equation
should be balanced.
We have another set of steps to address
this!!!!
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 1: Write the unbalanced net ionic equation
Step 2: Determine which atoms are oxidized and reduced, and write the two unbalanced
half-reactions
Step 3: Balance both half-reactions for all atoms except H, O
Step 4: Balance each half-reactions for O
- Add H2O to side with less O
- Add H+ to side with less H
Step 5: Balance each half-reactions for charge
- Add electrons to side with greater positive charge
- Multiply half-reaction as needed so the electron count is the same on both sides
Step 6: Add the two half-reactions
- Cancel electrons and other species (spectator ions) that appear on both sides
Step 7: Confirm that equation balanced for atoms and charge
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 1: Write the unbalanced net ionic equation
Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(aq)
Step 2: Determine which atoms are oxidized and reduced,
and write the two unbalanced half-reactions
Cl is oxidized (loses e-) from -1 to 0  Reducing agent
Cr is reduced (gains e-) from +6 to +3  Oxidizing agent
Oxidation half-rxn
2 Cl-(aq)  Cl2(aq)
Reduction half-rxn
Cr2O72-(aq) 2 Cr3+(aq)
Step 3: Balance both half-reactions for all atoms except
H, O
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 4: Balance each half-reactions for O, H
- Add H2O to side with less O
- Add H+ to side with less H
Oxidation half-rxn
2Cl-(aq)  Cl2(aq)
Reduction half-rxn
Cr2O72-(aq)  2Cr3+(aq) +7H20 (l)
Cr2O72-(aq) + 14H+ (aq)  2Cr3+(aq) + 7H20 (l)
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 5: Balance each half-reactions for charge
- Add electrons to side with greater positive charge
- Multiply half-reaction as needed so the electron
count is the same on both sides
Oxidation half-rxn
2Cl-(aq)  Cl2(aq) + 2 eReduction half-rxn
Cr2O72-(aq) + 14H+ (aq)  2Cr3+(aq) + 7H20 (l)
Cr2O72-(aq) + 14H+ (aq) + 6 e-  2Cr3+(aq) + 7H20 (l)
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 5: Balance each half-reactions for charge
- Add electrons to side with greater positive charge
- Multiply half-reaction as needed so the electron
count is the same on both sides
Oxidation half-rxn
3 X [2Cl-(aq)  Cl2(aq) + 2e-] =
Reduction half-rxn
6Cl-(aq)  3Cl2(aq) + 6e-
Cr2O72-(aq) + 14H+ (aq) + 6e-  2Cr3+(aq) + 7H20 (l)
Balancing Net Ionic Equations Using Redox
Half-reactions (under acidic conditions)
Step 6: Add the two half-reactions
- Cancel electrons and other species (spectator ions)
that appear on both sides
6Cl-(aq)  3Cl2(aq) + 6e(Ox)
Cr2O72-(aq) + 14H+ (aq) + 6e-  2Cr3+(aq) + 7H20 (l) (Red)
Cr2O72-(aq) + 14H+ (aq) + 6Cl-(aq)  3Cl2(aq) + 2Cr3+(aq) + 7H20 (l)
Step 7: Confirm that equation balanced for atoms and
charge
Balancing Redox Reactions under basic
conditions
• In basic solutions, follow all of the same steps. When
finished, add an equivalent amount of OH- to each
side of the reaction to eliminate any H+ by combining
them to make H2O.
Test Your Skill
• Balance the following equation in acid solution:
• Cr2O72- + C2H5OH → Cr3+ + CO2
Test Your Skill
• Balance the following equation in acidic solution:
• Cr2O72- + C2H5OH → Cr3+ + CO2
• Answer: 2Cr2O72- + C2H5OH + 16H+
→ 4Cr3+ + 2CO2 + 11H2O
Test Your Skill
• Balance the following equation in basic solution:
• Zn + ClO- → Zn(OH)42- + Cl-
Test Your Skill
• Balance the following equation in basic solution:
• Zn + ClO- → Zn(OH)42- + Cl-
• Answer: Zn + ClO- + H2O + 2OH→ Zn(OH)42- + Cl-