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Redox reaction: A reaction that can be split apart
into an oxidation step and a reduction step.
Example: 2 Na + Cl2
2 NaCl
Can be split apart as:
Na
Na+ + e- oxidation
1561
Redox reaction: A reaction that can be split apart
into an oxidation step and a reduction step.
Example: 2 Na + Cl2
2 NaCl
Can be split apart as:
Na
Na+ + e- oxidation
2e- + Cl2
2 Cl-
reduction
1562
Redox reaction: A reaction that can be split apart
into an oxidation step and a reduction step.
Example: 2 Na + Cl2
2 NaCl
Can be split apart as:
Na
Na+ + e- oxidation
2e- + Cl2
2 Cl-
reduction
The term redox comes from the two abbreviations
indicated in blue.
1563
Balancing Redox Chemical reactions
1564
Balancing Redox Chemical reactions
We treat the two principal cases:
1565
Balancing Redox Chemical reactions
We treat the two principal cases:
(1) Reactions in acidic conditions.
1566
Balancing Redox Chemical reactions
We treat the two principal cases:
(1) Reactions in acidic conditions.
(2) Reactions in basic conditions.
1567
Balancing Redox Chemical reactions
We treat the two principal cases:
(1) Reactions in acidic conditions.
(2) Reactions in basic conditions.
Reactions that are not in acidic or basic conditions
are generally straightforward to balance.
1568
The basic idea is to balance the two half-equations
separately.
1569
The basic idea is to balance the two half-equations
separately. This will be much easier than working
directly with the complete reaction.
1570
The basic idea is to balance the two half-equations
separately. This will be much easier than working
directly with the complete reaction.
The two balanced half-equations are then
combined so that the electrons on the right and left
of the equation cancel out.
1571
We make use of two symbolic reactions. Note these
symbolic reactions are not real reactions, and they
are not balanced in the conventional sense.
1572
We make use of two symbolic reactions. Note these
symbolic reactions are not real reactions, and they
are not balanced in the conventional sense. They
are simply useful ways to think about removal of an
“O” atom from a species in different conditions.
1573
We make use of two symbolic reactions. Note these
symbolic reactions are not real reactions, and they
are not balanced in the conventional sense. They
are simply useful ways to think about removal of an
“O” atom from a species in different conditions.
2 H+(aq) + “O”
H2O
(use this for acidic conditions)
1574
We make use of two symbolic reactions. Note these
symbolic reactions are not real reactions, and they
are not balanced in the conventional sense. They
are simply useful ways to think about removal of an
“O” atom from a species in different conditions.
2 H+(aq) + “O”
H2O
(use this for acidic conditions)
H2O + “O”
2 OH-(aq)
(use this for basic conditions)
1575
We make use of two symbolic reactions. Note these
symbolic reactions are not real reactions, and they
are not balanced in the conventional sense. They
are simply useful ways to think about removal of an
“O” atom from a species in different conditions.
2 H+(aq) + “O”
H2O
(use this for acidic conditions)
H2O + “O”
2 OH-(aq)
(use this for basic conditions)
“O” means an oxygen atom part of some species, e.g.
an O atom in the ion MnO4-.
1576
Some cleanup might be necessary at the final stage
of putting the two balanced half-equations
together.
1577
Some cleanup might be necessary at the final stage
of putting the two balanced half-equations
together. This tends to be more important for
reactions in basic solution.
1578
Some cleanup might be necessary at the final stage
of putting the two balanced half-equations
together. This tends to be more important for
reactions in basic solution.
Stepwise process:
1579
Some cleanup might be necessary at the final stage
of putting the two balanced half-equations
together. This tends to be more important for
reactions in basic solution.
Stepwise process:
1. Write the skeleton equation containing the
oxidizing agent and reducing agent in ionic form,
and separate it into two half-equations.
1580
Some cleanup might be necessary at the final stage
of putting the two balanced half-equations
together. This tends to be more important for
reactions in basic solution.
Stepwise process:
1. Write the skeleton equation containing the
oxidizing agent and reducing agent in ionic form,
and separate it into two half-equations.
2. Balance each half reaction separately. For
reactions in acidic conditions, add H2O to balance
oxygen atoms and H+ to balance hydrogen atoms.
1581
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
1582
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
3. Add electrons to one side of each half-reaction to
balance the charges. Do this only after you have
atom balance.
1583
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
3. Add electrons to one side of each half-reaction to
balance the charges. Do this only after you have
atom balance.
4. Equalize the number of electrons by multiplying
one or both half-reactions by appropriate
coefficients.
1584
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
3. Add electrons to one side of each half-reaction to
balance the charges. Do this only after you have
atom balance.
4. Equalize the number of electrons by multiplying
one or both half-reactions by appropriate
coefficients.
5. Add the two half-reactions together and cancel the
electrons.
1585
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
3. Add electrons to one side of each half-reaction to
balance the charges. Do this only after you have
atom balance.
4. Equalize the number of electrons by multiplying
one or both half-reactions by appropriate
coefficients.
5. Add the two half-reactions together and cancel the
electrons.
6. Do any cleanup as necessary.
1586
In basic solution, add H2O to balance hydrogen atoms
and OH- to balance oxygen atoms.
3. Add electrons to one side of each half-reaction to
balance the charges. Do this only after you have
atom balance.
4. Equalize the number of electrons by multiplying
one or both half-reactions by appropriate
coefficients.
5. Add the two half-reactions together and cancel the
electrons.
6. Do any cleanup as necessary.
7. Do a check on the final balanced reaction.
1587
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
1588
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
This example can be balanced by eyeball inspection,
but we will use the half-reaction approach.
1589
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
This example can be balanced by eyeball inspection,
but we will use the half-reaction approach.
Clearly
Sn2+(aq)
Sn4+(aq)
Fe3+(aq)
Fe2+(aq)
1590
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
This example can be balanced by eyeball inspection,
but we will use the half-reaction approach.
Clearly
Sn2+(aq)
Sn4+(aq) + 2e- (1)
e- + Fe3+(aq)
Fe2+(aq)
(2)
1591
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
This example can be balanced by eyeball inspection,
but we will use the half-reaction approach.
Clearly
Sn2+(aq)
Sn4+(aq) + 2e- (1)
e- + Fe3+(aq)
Fe2+(aq)
(2)
Now in order to combine the two half-reactions so
that the electrons cancel out, we need to multiply
the second equation by 2.
1592
Example 1: The following is a simple example (not
in acid or base conditions). Balance the reaction:
Sn2+(aq) + Fe3+(aq)
Sn4+(aq) + Fe2+(aq)
This example can be balanced by eyeball inspection,
but we will use the half-reaction approach.
Clearly
Sn2+(aq)
Sn4+(aq) + 2e- (1)
2 e- + 2 Fe3+(aq)
2 Fe2+(aq)
(2)
Now in order to combine the two half-reactions so
that the electrons cancel out, we need to multiply
the second equation by 2.
1593
Now add the two balanced half-reactions to obtain:
Sn2+(aq) + 2 Fe3+(aq)
Sn4+(aq) + 2 Fe2+(aq)
Note the final balanced equation is in net-ionic form.
1594
Now add the two balanced half-reactions to obtain:
Sn2+(aq) + 2 Fe3+(aq)
Sn4+(aq) + 2 Fe2+(aq)
Note the final balanced equation is in net-ionic form.
Check:
species L R
Fe
2 2
Sn
1 1
charge +8 +8
1595
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
1596
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
1597
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
1598
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
Fe2+(aq)
Fe3+(aq)
(1)
1599
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
Fe2+(aq)
Fe3+(aq)
(1)
Cr2O72-(aq)
Cr3+(aq)
(2)
1600
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
Fe2+(aq)
Fe3+(aq)
(1)
Cr2O72-(aq)
Cr3+(aq)
(2)
The first half-reaction can be balanced by adding
the appropriate charge.
1601
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
Fe2+(aq)
Fe3+(aq) + e(1)
Cr2O72-(aq)
Cr3+(aq)
(2)
1602
Example 2: In acidic solution potassium dichromate
can convert Fe2+ to Fe3+, and Cr3+ is also formed.
Give the balanced net ionic equation.
The skeleton equation is:
Cr2O72-(aq) + Fe2+
Cr3+(aq) + Fe3+(aq)
The start of the two half equations are:
Fe2+(aq)
Fe3+(aq) + e(1)
Cr2O72-(aq)
Cr3+(aq)
(2)
To balance the second half-equation, we need to
remove all seven “O” atoms from the dichromate
ion.
1603
Cr2O72-(aq)
Cr3+(aq)
(2)
1604
Cr2O72-(aq)
Cr3+(aq)
Symbolic equation: 2 H+(aq) + “O”
(2)
H2O
1605
Cr2O72-(aq)
Cr3+(aq)
Symbolic equation: 2 H+(aq) + “O”
(use this for acidic conditions)
(2)
H2O
1606
Cr2O72-(aq)
Cr3+(aq)
(2)
Symbolic equation: 2 H+(aq) + “O”
(use this for acidic conditions)
14 H+(aq) + Cr2O72-(aq)
H2O
2 Cr3+(aq)
(2)
1607
Cr2O72-(aq)
Cr3+(aq)
(2)
Symbolic equation: 2 H+(aq) + “O”
(use this for acidic conditions)
14 H+(aq) + Cr2O72-(aq)
H2O
2 Cr3+(aq) + 7 H2O
(2)
1608
Cr2O72-(aq)
Cr3+(aq)
(2)
Symbolic equation: 2 H+(aq) + “O”
(use this for acidic conditions)
14 H+(aq) + Cr2O72-(aq)
H2O
2 Cr3+(aq) + 7 H2O
(2)
Now an eyeball inspection indicates that this halfequation is balanced for atoms. Now balance it for
charge.
1609
Cr2O72-(aq)
Cr3+(aq)
(2)
Symbolic equation: 2 H+(aq) + “O”
(use this for acidic conditions)
6 e- + 14 H+(aq) + Cr2O72-(aq)
H2O
2 Cr3+(aq) + 7 H2O
(2)
Now an eyeball inspection indicates that this halfequation is balanced for atoms. Now balance it for
charge.
1610
So the two balanced half-equations are:
Fe2+(aq)
Fe3+(aq) + e(1)
6 e- + 14 H+(aq) + Cr2O72-(aq)
2 Cr3+(aq) + 7 H2O (2)
1611
So the two balanced half-equations are:
Fe2+(aq)
Fe3+(aq) + e(1)
6 e- + 14 H+(aq) + Cr2O72-(aq)
2 Cr3+(aq) + 7 H2O (2)
Now multiply the first half-equation by 6 then add
the two half-equations.
6 Fe2+(aq) + 14 H+(aq) + Cr2O72-(aq)
2 Cr3+(aq) + 7 H2O
+ 6 Fe3+(aq)
1612
So the two balanced half-equations are:
Fe2+(aq)
Fe3+(aq) + e(1)
6 e- + 14 H+(aq) + Cr2O72-(aq)
2 Cr3+(aq) + 7 H2O (2)
Now multiply the first half-equation by 6 then add
the two half-equations.
6 Fe2+(aq) + 14 H+(aq) + Cr2O72-(aq)
2 Cr3+(aq) + 7 H2O
+ 6 Fe3+(aq)
Check:
species
L R
Fe
6 6
H
14 14
O
7 7
Cr
2 2
charge +24 +24
1613
Example 3: Sodium sulfide is oxidized in an alkaline
solution of potassium permanganate to yield sulfur
and manganese (IV) oxide. Give the balanced
equation.
1614
Example 3: Sodium sulfide is oxidized in an alkaline
solution of potassium permanganate to yield sulfur
and manganese (IV) oxide. Give the balanced
equation.
We will look at this example using two different
approaches.
1615
Example 3: Sodium sulfide is oxidized in an alkaline
solution of potassium permanganate to yield sulfur
and manganese (IV) oxide. Give the balanced
equation.
We will look at this example using two different
approaches.
(1) The first approach is to treat the problem
directly in alkaline solution.
1616
Example 3: Sodium sulfide is oxidized in an alkaline
solution of potassium permanganate to yield sulfur
and manganese (IV) oxide. Give the balanced
equation.
We will look at this example using two different
approaches.
(1) The first approach is to treat the problem
directly in alkaline solution.
(2) The second approach is to treat the problem as
though it were in acidic solution, and then clean up
hydrogen ions at the end.
1617
Example 3: Sodium sulfide is oxidized in an alkaline
solution of potassium permanganate to yield sulfur
and manganese (IV) oxide. Give the balanced
equation.
We will look at this example using two different
approaches.
(1) The first approach is to treat the problem
directly in alkaline solution.
(2) The second approach is to treat the problem as
though it were in acidic solution, and then clean up
hydrogen ions at the end. Many students prefer
this second approach.
1618
Approach 1: The skeleton equation is:
1619
Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
1620