Readings for this section

Petrucci: Chapters 5-4 to 5-6.

Oxidation Numbers and REDOX Balancing 05/09/2012


Assigning Oxidation Numbers

Oxidation numbers (ON) are not real charges. They are the results of an accounting method whereby we can keep track of electrons during a chemical reaction. In a few cases, the ON is actually a real charge but only in rare monatomic ions like Cl- and Na+.

Oxidation numbers are defined using two premises.

  1. The bonds in compounds for which you wish to assign oxidation numbers are assumed to be 100% ionic in nature.
  2. The electrons in the bonds are thus divided such that the more electronegative element gets both electrons in a bond and the less electronegative element ends up with a net loss of electrons.

Using this method, we need to be up on the relative electronegativities of the elements and it takes time to figure out which way the electrons go and the resulting charges. For example, if we take the bonds in water H2O. There are two O-H bonds in each molecule.

Taking them one at a time, in an O-H bond, there two electrons and we assign them both to the oxygen since it is more electronegative than hydrogen. Doing this to the second bond as well leaves us with an oxygen ion with 8 electrons for a net charge of -2 and two hydrogen ions with no electrons for a net charge of +1. This takes considerable thought processes and can be simplified using a set of rules based on these processes but which gives results directly.

RULES: (taken in order of importance.)

  1. The sum of the oxidation numbers (ON) of the atoms in a species (ion, molecule, atom) is equal to the charge on that species.
    For example,
    He(atom)       ---> ON = 0
    H2(molecule)  ---> both atoms must have ON=0 so sum is 0
    Cl- (ion)         ---> ON = -1
    Na+ (ion)        ---> ON = +1
  2. Alkali metals in a compound with non-metals have ON=+1
    example: NaCl     ---> ON(Na) = +1 \ON(Cl) = -1 [rule 1]
  3. Alkali Earth metals in a compound with non-metals have ON=+2
    example: MgO    ---> ON(Mg) = +2 \ON(O) = -2 [rule 1]
  4. Hydrogen in cpds. has ON = +1
  5. Halogens have ON=-1 as in HF or HCl. (This is always true for fluorine but for the others, except if bonded to other halogens [interhalogen] or to oxygen)
    example ClO-    ---> ON(Cl) is not -1 since O is more electronegative than Cl.
  6. Oxygen in compounds is always ON(O)=-2. Except with Fluroine (rule 5) or in the case of peroxides O22- [ON(O)=-1] or super oxides O2- [ON(O) = -1/2].
  7. For all elements where these rules don't work, use the overriding premise and work out the oxidation numbers, i.e., the more electronegative element gets the electrons (or the negative oxidation number).
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Balancing Redox Equations

Now we can use the oxidation numbers as discussed in the previous section to help us balance REDOX reactions. There are as many different methods of balancing redox equations as there are text books on the subject. Personally, I've used at least 4 different methods and I have my favorites. Herein, I will use the algebraic  method we discussed earlier in the year to balance REDOX equations. Because of the peculiarities of REDOX, I will give you a set of rules to follow to guide you.

A Half-reaction Method for balancing REDOX equations.

  1. Assign ONs and determine which elements change
  2. Divide the reaction into half reactions using the elements which change as a guide.
  3. Add e- and  H2O to both half reactions, and, in acid, add H+, and, in base, add OH-.
  4. use algebraic method to balance the two half-reactions..
  5. At this stage, you have two balanced half-reactions. These are useful if you're doing electrochemistry. If you need to get the overall reaction, you can do so by adding the half-reactions together such that the electrons are canceled out.

Now let's try a few examples.

Sodium nitrate reacts with zinc in sodium hydroxide solution to give ammonia and the tetrahydroxozinc(II) ion, Zn(OH)42? . We have:

Break this up into half reactions according to the arrows drawn here.

Oxidation

            Zn        ?         Zn(OH)42?

Reduction

NO3-      ?         NH3

Now add e-, H2O, and OH- (in base).

Oxidation

Zn  ?  Zn(OH)42? + e? + OH- + H2O

Reduction

      NO3-     ?   NH3 + e? + OH- +  H2O

Use algebraic method to balance.

Oxidation

  

Zn  ? aZn(OH)42? be- cOH- dH2O
Zn 1 a       a=1
O 0 4a   c  d 4a + c + d = 0
H 0 4a   c 2d 4a + c + 2d = 0
Charge 0 ?2a ?b ?c   ?2a ? b ? c = 0
solving:

a = 1, b = 2, c = ?4, d = 0

balanced

Zn + 4OH-? Zn(OH)42? + 2e-


Reduction

  

NO3-   ? aNH3 be-+ cOH-+ bH2O
N 1 a       a=1
O 3   c d c + d = 3
H 0 3a   c 2d 3a + c + 2d = 0
Charge ?b ?c   ? b ? c = ?1
solving

a = 1, b = ?8, c = 9, d = ?6

balanced

NO3- +  6H2O + 8e- ? NH3 + 9OH-

Both these half reactions are balanced. To get the overall reaction, we need to add these together such that the electrons are canceled. That means 4?the first reaction plus the second:

Oxidation

4?  Zn  + 4 OH-     ?  Zn(OH)42? + 2 e-]

Reduction

      NO3- + 6 H2O + 8 e-   ?   NH3 + 9 OH-    


Overall

4 Zn  + NO3- + 6 H2O + 7 OH-  ? 4 Zn(OH)42? + NH3

Now check the balancing to ensure both mass balance and charge balance is correct.


Another example:

Copper reacts with dilute nitric acid to give nitrogen monoxide NO.

Notice that the ON for nitrogen in NO3- is an extremely unlikely +5. This could never be the actual charge on the nitrogen in any real compound (convincing evidence that oxidation numbers don't represent reality very well).

We have the following two half reactions (use the chemicals joined by the arrows):

Oxidation

Cu          ?         Cu2+

Reduction

NO3-      ?         NO

Now add e-, H2O and H+(in acid).

Oxidation

Cu          ?         Cu2+ (See note below)

Reduction

      NO3-      ?         NO + e? + H+ +  H2O

Now, use algebraic method to balance the two.

Oxidation

  

Cu  ?

Cu2+ + 2 e-
In this case, it is very obvious that balancing this half reaction is easily accomplished by adding two electrons to the product side. This done, both charge- and mass-balance are complete.

It's important to realize that although rote methods may always get you the correct answer they are not always the easiest way to solve a problem.

Reduction

  

NO3- ?

 aNO + be? + cH+dH2O
N 1 a a=1
O 3                             c          d c + d = 3
H 0                             c         2d c + 2d = 0
Charge ?1               ?b          c ? b + c = ?1
solving

a = 1, b = ?3, c = ?4, d = 2

balanced

NO3- + 4 H+ + 3 e-   ?  NO + 2 H2

Both these half reactions are balanced. To get the overall reaction, we need to add these together such that the electrons are canceled. That means 3?the oxidation reaction plus 2?the reduction:

Oxidation

             3 Cu  ?         3 Cu2+ + 6 e-

Reduction

2 NO3- + 8 H+ + 6 e-    ?     2 NO + 4 H2O           


Overall

3 Cu  + 2 NO3- + 8 H+   ?  2 NO + 4 H2O + 3 Cu2+

Now check the balancing to ensure both mass balance and charge balance is correct.


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Prof. Michael J. Mombourquette.
Copyright ? 1997
Revised: March 11, 2002