Redox (Reduction-Oxidation)
Reactions involve the oxidation of a reducing agent. The reactions
involve the transfer of electrons from the reducing agent to the
oxidizing agent. The oxidizing agent gains electrons and is reduced
while the reducing agent loses electrons and is oxidized. Two common
oxiding agents are potassium dichromate
(K2Cr2O7) or its congeners potassium
chromate (K2CrO4) and chromium trioxide. All
three of these oxidants are chromium VI (+6) species and the
oxidations are normally conducted in aqueous acid. The final form of
chromium is Cr +3. Potassium permanganate (KMnO4; Mn +7)
is usually used as an oxidant in aqueous alkali. Its reduction
product is MnO2 (Mn +4).
Consider the oxidation of 3-pentanol with
K2Cr2O7 in aqueous sulfuric acid.
Under these conditions the oxidant is effectively
H2Cr2O7 but we can work with
dichromate ion Cr2O7=. The first
order of business is to write a half reaction for the reduction.
Cr2O7=
----> Cr+++
Clearly, there must be an additional
chromium atom on the right side of the equation. Thus,
Cr2O7=
----> 2Cr+++
Now deal with oxygen. Seven oxygen
atoms are required on the right side of the equation. The only other
source of oxygen is water. Add one water molecule for each oxygen
atom to be balanced and two protons on the opposite side of the
equation for each water added to balance the two hydrogens introduced
via water. That is H2O - 2H+ = O. [If you
need to balance only hydrogens, use protons.] This operation
leads to:
14H+ +
Cr2O7= ----> 2Cr+++ +
7H2O. (equation
1)
Now deal with the oxidation of 3-pentanol.
CH3CH2CH(OH)CH2CH3
---->
CH3CH2C(O)CH2CH3
In this reaction two hydrogens have been
lost from 3-pentanol. Adding two protons to the right side of the
equation gives:
CH3CH2CH(OH)CH2CH3
---->
CH3CH2C(O)CH2CH3 +
2H+ (equation
2)
Now that the two equations have been
balanced in respect to atoms, the issue of electrons must be
addressed. In equation 1 the left side of the equation has a total
charge of +12 while the right side is +6. This change is a gain of 6
electrons (+6). That is, the oxidant is reduced. On the other hand,
equation 2 has zero charge on the left and +2 on the right. This is a
loss of two electrons (-2). Therefore, equation 2 must be multiplied
by three [3 x (-2) = -6] to balance the gain of electrons in
equation 1. Equation 2 becomes,
3
CH3CH2CH(OH)CH2CH3
----> 3
CH3CH2C(O)CH2CH3 +
6H+
(equation
2-1)
Adding equation 1 and equation 2-1 together gives,
14H+ +
Cr2O7= + 3
CH3CH2CH(OH)CH2CH3
----> 2Cr+++ + 7H2O + 3
CH3CH2C(O)CH2CH3 +
6H+
Subtracting 6H+ from each side
leaves,
8H+ +
Cr2O7= + 3
CH3CH2CH(OH)CH2CH3
----> 2Cr+++ + 7H2O + 3
CH3CH2C(O)CH2CH3
(equation 3)
This equation is balanced in both atoms and
electrons.
Now we shall address the oxidation of 3-pentanol with potassium
permanganate in aqueous alkali. Permanganate is reduced to manganese
dioxide.
MnO4-
----> MnO2
To add oxygen in aqueous alkali, each
oxygen equals 2OH- - H2O = O and H2O
- OH- = H. Therefore, two water molecules are added on the
left side and four hydroxyls on the right. The electron change is +2
electrons.
2H2O +
MnO4-----> MnO2 + 4OH-
+3 electron change (equation
4)
The oxidation of 3-pentanol requires the
equivalent of two hydrogens added to the right side of the
equation.
CH3CH2CH(OH)CH2CH3
---->
CH3CH2C(O)CH2CH3
Accordingly, the above equation
becomes
2OH- +
CH3CH2CH(OH)CH2CH3
---->
CH3CH2C(O)CH2CH3
+ 2H2O;
-2 electron change(equation 5)
Equation 4 must be multiplied by 2 and equation 5 multiplied by 3 to balance electrons.
Equation 4 becomes
4H2O + 2MnO4-----> 2MnO2 + 8OH- (equation 6)
and equation 5 becomes
6OH- +
3CH3CH2CH(OH)CH2CH3 ---->
3CH3CH2C(O)CH2CH3 + 6H2O (equation 7)
Adding equation 6 and equation 7 gives,
4H2O +
2MnO4- + 6OH- +
3CH3CH2CH(OH)CH2CH3
----> 2MnO2 + 8OH- +
3CH3CH2C(O)CH2CH3
+ 6H2O
Simplifying the hydroxyl groups and cancelling
the water one obtains,
2MnO 4- +
3CH3CH2CH(OH)CH2CH3
----> 2MnO2 + 2OH- +
3CH3CH2C(O)CH2CH3 + 2H2O
Two moles of permanganate accommodates
three moles of 3-pentanol.