Redox - Oxidation and Reduction.





Original Definition:



Oxidation is the reaction with oxygen:



S + O2 Ž SO2



CH4 + 2 O2 Ž CO2 + 2 H2O



4 Fe + 3 O2 Ž 2 Fe2O3



In this latter case a metal is oxidized to its oxide. It was noted that this was exactly the opposite to the production of the metal from its ore, a process already termed "reduction". Thus reduction became generally accepted as the opposite of oxidation.



Summarizing these definitions:



Oxidation is the addition of oxygen, or the loss of hydrogen.

Reduction is the addition of hydrogen, or the loss of oxygen.








Definition in terms of electron transfer.


If sodium metal is oxidized:

4 Na + O2 Ž 2 Na2O

the product is ionized into Na+ and O2- ions.



If sodium metal reacts with chlorine gas:

2 Na + Cl2 Ž 2 NaCl

the product is ionized into Na+ and Cl- ions.





In both these cases, as far as the sodium is concerned, the sodium metal has lost an electron. In the first case this is obviously oxidation.



Definition:



Oxidation is the loss of electrons, reduction is the gain of electrons.






Two Definitions?


Neither of these two definitions, one in terms of atom transfer, the other in terms of electron transfer, can stand alone.





Atom transfer:

The oxidation of sulphur or the burning of methane produce covalent compounds and it is difficult to see how the electron transfer definition could apply. In these cases the atom transfer definition is appropriate.





Electron transfer:

On the other hand the reaction of sodium metal with chlorine gas does not fit the atom transfer definition but does fit the electron transfer definition.








The question arose:

Is there a way in which both definitions can be rolled into one and the process generalized to cover atom transfers other than hydrogen and oxygen?








Oxidation Numbers.


The way in which both types of oxidation and reduction were put under one definition was to invent a number which indicated how the polarization of an atom changed during a reaction. This number would indicate whether the electron density at a particular atom had increased or decreased and oxidation could be defined as a decrease in electron density; reduction as an increase.





Defining oxidation numbers.



The oxidation number for an atom in a compound is an arbitrary number assigned according to the charge (full or partial) on that atom.



In working with oxidation numbers it is important to note that since they represent either full or partial charges, they cannot be compared between atoms. Comparisons can be made only with the same atom in different molecular environments.



Thus Na+ has an oxidation number of 1

Sulphur in SO2 has an oxidation number of 4.



It cannot be said that the sulphur is more charged, or has lost more electron density, than has sodium. In one case, Na+, the oxidation number is indicating a full electron transfer. In the other, SO2, only partial electron transfer is occurring in this covalent molecule.








Rules for Assigning Oxidation Numbers.


1. The sum of all oxidation numbers of all atoms in a species must equal the charge on that species.



From this: all elements in their elemental form have an oxidation number of 0.



2. Fluorine has an oxidation number of -1.



3. Groups 1, 2, and 3 (except boron, B) have an oxidation equal to their group number.



4. Hydrogen has an oxidation number of +1.



5. Oxygen has an oxidation number of -2.



6. The oxidation numbers of any elements not yet determined by applying the above rules, as appropriate, in numeric order, are obtained by giving the most electronegative unassigned element its most negative oxidation number. This most negative oxidation number is the element's group number less 18.





The above rules are applied in numerical order until all atoms in a species have been assigned an oxidation number.



Once a number has been assigned by one rule, it is not changed by a later rule.








Application of the Oxidation Number Rules.


SO3



Rule 1: ON(S) + 3 x ON(O) = 0

Rule 5: ON(O) = -2.

So ON(S) = +6







KMnO4



Rule 1: ON(K) + ON(Mn) + 4 x ON(O) = 0

Rule 2: ON(K) = 1

Rule 5: ON(O) = -2

So ON(Mn) = +7









S8



Rule 1: ON(S) = 0





NaBH4



Rule 1: ON(Na) + ON(B) + 4 x ON(H) = 0

Rule 2: ON(Na) = 1

Rule 4: ON(H) = 1

So ON(B) = -5









H2O2



Rule 1: ON(H) + ON(O) = 0

Rule 4: ON(H) = 1

So ON(O) = -1











Fe3O4



Rule 1: 3 x ON(Fe) + 4 x ON(O) = 0

Rule 5: ON(O) = -2

So ON(Fe) = +8/3











PCl5



Rule 1: ON(P) + 5 x ON(Cl) = 0

Rule 6: ON(Cl) = 17 - 18 = -1

So ON(P) = +5













NH4+



Rule 1: ON(N) + 4 x ON(H) = +1

Rule 4: ON(H) = +1

So ON(N) = -3








Use of Oxidation Numbers.


Comparing oxidation numbers of a particular element in the reactants and the products will determine whether a redox process has occured or not:



If the oxidation number of an atom has increased during a reaction, that atom has been oxidized.

If the oxidation number of an atom has decreased during a reaction, that atom has been reduced.








1. Is a redox reagent required for the following transformations, or some other type of reagent?


Example 1

To convert POCl3 into H3PO4?



POCl3 H3PO4
ON(P) = +5 ON(H) = +1
ON(O) = -2 ON(P) = +5
ON(Cl) = -1 ON(O) = -2


In this case no oxidation number changes, so no redox reagent is required.









Example 2

To convert KCl into ICl3



KI ICl3
ON(K) = +1 ON(I) = +3
ON(I) = -1 ON(Cl) = -1


In this case a redox reagent is required as the iodine is oxidized from its -1 state to its +3 state.













2. Are the following processes redox or not. If they are redox, state which atom is oxidized and which reduced.

(The equations are not balanced)





Example 1

H2O + SO2 Ž H2SO3



Reactants

H2O SO2
ON(H) = +1 ON(S) = +4
ON(O) = -2 ON(O) = -2


Products

H2SO3
ON(H) = +1
ON(S) = +4
ON(O) = -2

In this case no redox is occurring.







Example 2

FeCl2 + KMnO4 + HCl Ž FeCl3 + KCl + MnO2 + H2O



Reactants
FeCl2 KMnO4 HCl
Fe: +2 K: +1 H: +1
Cl: -1 Mn: +7 Cl: -1
O: -2



Products
FeCl3 KCl MnO2 H2O
Fe: +3 K: +1 Mn: +4 H: +1
Cl: -1 Cl: -1 O: -2 O: -2

In this case, the process is a redox process. Fe is being oxidized, and Mn is being reduced.









3. Oxidation numbers can also be used to balance redox equations such as the one immediately above. If you have learned this way in High School, and are comfortable with it, you may go on using it in this course. Otherwise I will not show it to you.








Balancing Redox Equations.


Because it is very easy to make a mistake in assigning oxidation numbers, I prefer to use the half-reaction method for balancing redox equations.



The half reaction method involves splitting up the reaction into the oxidation half and the reduction half. Each of these is balanced separately and then they are added to give the full equation.







Example.

Balance by half reactions:

Ag+ + Mg Ž Ag + Mg2+



For the two half reactions,

1. pick out the related reactants and products,

2. balance for atoms,

3. balance for charge, adding electrons to the appropriate side.

4. multiply the half reactions by factors which will give the same number of electrons for each.



Ag+ Ž Ag Mg Ž Mg2+
Ag+ Ž Ag Mg Ž Mg2+
Ag+ + e- Ž Ag Mg Ž Mg2+ + 2 e-
2Ag+ + 2 e- Ž 2 Ag Mg Ž Mg2+ + 2 e-


Finally, add the two half reactions together (the electrons should cancel each other out):



2 Ag+ + Mg Ž 2 Ag + Mg2+










Example 2. Balance the following equation by the half reaction method. (in acid)


Cu, HNO3 Ž Cu(NO3)2, NO



By writing the equation this way, it is implied that it is incomplete. Water, H+, or OH- may have to be added.



For the two half reactions,

1. pick out the related reactants and products,

2. balance for atoms, adding H2O, H+ or OH- where required.

3. balance for charge, adding electrons to the appropriate side.

4. multiply the half reactions by factors which will give the same number of electrons for each.



Cu Ž Cu2+ HNO3 Ž NO
Cu Ž Cu2+ 3 H+ + HNO3 Ž NO + 2 H2O
Cu Ž Cu2+ + 2 e- 3 H+ + HNO3 + 3 e- Ž NO + 2 H2O
3 Cu Ž 3 Cu2+ + 6 e- 6H++ 2HNO3 + 6e- Ž 2NO + 4H2O


Add the two half reactions together to give the full, balanced equation:

3 Cu + 6 H+ + 2 HNO3 Ž 3 Cu2+ + 2 NO + 4 H2O



Finally, put back any common ions:

3 Cu + 8 HNO3 Ž 3 Cu(NO3)2 + 2 NO + 4 H2O










Example 3. Balance the following equation by the half reaction method. (in base)

CH3OH, KMnO4 Ž HCOOH, MnO2

By writing the equation this way, it is implied that it is incomplete. Water, H+, or OH- may have to be added.



For the two half reactions,

1. pick out the related reactants and products,

2. balance for atoms, adding H2O, H+ or OH- where required.

3. balance for charge, adding electrons to the appropriate side.

4. multiply the half reactions by factors which will give the same number of electrons for each.

CH3OH Ž HCOOH

MnO4- Ž MnO2

CH3OH + 4 OH- Ž HCOOH + 3 H2O
MnO4- + 2 H2O Ž MnO2 + 4 OH-
CH3OH + 4 OH- Ž HCOOH + 3 H2O + 4 e-
MnO4- + 2 H2O + 3 e- Ž MnO2 + 4 OH-
3 CH3OH + 12 OH- Ž 3 HCOOH + 9 H2O + 12 e-
4 MnO4- + 8 H2O + 12 e- Ž 4 MnO2 + 16 OH-

Adding these two half reactions:

3 CH3OH + 4 MnO4- Ž 3 HCOOH + 4 MnO2 + H2O + 4 OH-

And putting in the common ions:

3 CH3OH + 4 KMnO4 Ž 3 HCOOH + 4 MnO2 + H2O + 4 KOH

Balancing the matter in a half reaction.








Gain or loss of Hydrogen and/or Oxygen (summary):


1. In Acid



a. Loss of oxygen: (O) + 2H+ + 2e- Ž H2O

b. Gain of oxygen: H2O Ž (O) + 2H+ + 2e-

c. Loss of hydrogen: (H) Ž H+ + e-

d. Gain of hydrogen: H+ + e- Ž (H)





2. In Base



a. Loss of oxygen: (O) + H2O + 2e- Ž 2 OH-

b. Gain of oxygen: 2 OH- Ž (O) + H2O + 2e-

c. Loss of hydrogen: (H) + OH- Ž H2O + e-

d. Gain of hydrogen: H2O + e- Ž (H) + OH-










Example 3. Balance the following reaction, in base.

NH3, KOCl Ž N2H4, KCl





NH3 Ž N2H4
KOCl Ž KCl
2 NH3 + 2 OH- Ž N2H4 + 2 H2O
KOCl + H2O Ž KCl + 2 OH-
2 NH3 + 2 OH- Ž N2H4 + 2 H2O + 2e-
KOCl + H2O + 2e- Ž KCl + 2 OH-
2 NH3 + 2 OH- Ž N2H4 + 2 H2O + 2e-
KOCl + H2O + 2e- Ž KCl + 2 OH-


Adding the two half reactions:



2 NH3 + KOCl Ž N2H4 + KCl + H2O