How to resolve redox problems?

The below scheme divides the redox reactions in three (green) parts, result of joining direct and indirect redox with weak and strong substances.
This shows spontaneous and non spontaneous redox reactions:

Direct Indirect
strong spontaneous reactions dry and wet batteries
go on
weak nothing effective
happens 		electrolysis and
charging of
batteries

If we want to solve redox reactions, not with oxydation numbers, but with the half reaction equations, than we must follow some rules:
  1. what kind of reaction?
  2. what particles participate?
  3. what are the half reaction equations?
  4. and the total equation?
  5. what conclusions can we draw and what observations can be done?

The following scheme shows 6 steps that always return in solving a redox reaction, now and then dependent on the presence of electrodes and wether it is an indirect or direct reaction.

Without electrodes With electrodes
1. design a scheme of what happens, of what is done.
2a. Make a list of all present substances / particles, including electrodes

2b. Define the reductor and the oxydator (underline them)
3. give the half reaction equations 3. give the half reaction equations
OX reacts at the positive electrode
RED reacts at the negative electrode
4. make all equations certain 4. make all electrode equations certain
5. check possible secondary reactions
6. Write down your observations and conclusions

N.B.: the use of the redox table is needed.

An example:
In the chemical lab we put a small piece of the metal sodium into distilled water (in a kind of aquarium).
The Sodium is a soft metal; we just can cut a piece of it, but be careful:
they conserve Sodium for a reason under petroleum; it reacts spontaneously with air (Oxygen) and with water.

We dry the piece of Sodium (with filter paper we remove the petroleum) and put is carefully on the water surface.
Immediately it begins to react heavily, a gas is formed, the piece of Sodium moves fastly on the water surface, burning with yellow flames.
This goes on until the metal has disappeared completely.

We join a couple of drops of indicator to the water; that shows us that the water became basic.

This is obviously a direct reaction: the Sodium particles collide directly with the water particles, and there are no electrodes.
This reaction is very spontaneous.

Now we obey the six steps:
  1. A sketch of the situation:


  2. the present particles/substances are: Sodium and water, nothing more.
    We look in the table and must conclude that water is an oxydator as well as a reductor (in both cases very weak)
    But Sodium is extremely strong reductor, so it makes sense to think that water in this case will act as an oxydator.
    The reductor has a lower position than the oxydator, so the reaction is spontaneous.
  3. The equations of the half reactions:
    Red: Na Na+ + e- |x2
    Ox: 2H2O + 2e- H2 + 2OH- |x1

    NB: the number of electrons in both half reactions must be the same.

  4. Total equation:
    2Na + 2H2O    2Na+ + H2 + 2OH-
    of:
    2Na(s) + 2H2O(l)2NaOH(aq) + H2(g)        ΔH < 0
  5. Problably there will be one secondary reaction: the Hydrogen gas (a product) is responsible for the fire; it reacts with oxygen from the air and causes the flames.
  6. Observations and conclusions:
  7. The reaction is spontaneous, energy comes free in the form of warmth and light; so it is an exothermic reaction.
  8. A gas is formed (Hydrogen) that easily will start to burn; you can see that.
  9. The product NaOH creates a basic environment (ionen OH-); we can check that with an indicator: the colour becomes violet.