Redox (II): Standard Electrode Potential Eo(4): Disproportionation and electrode potentials

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Standard Electrode Potential Eo(4)

Here is a further learning objectives:

14/14 To understand how disproportionation reactions relate to standard electrode potentials.

Disproportionation and electrode potentials

Disproportionation you will remember occurs when in the same reaction a species (atom, ion or molecule) is both oxidized and reduced.

A classic example of this chemistry found in almost all chemistry courses in school or college is the reaction between sodium thiosulphate (Na₂S₂O₃) and hydrochloric acid (HCl):

Na₂S₂O₃   +   2HCl   ⟶   2NaCl   +   S   + SO₂  +  H₂O

I want you to notice how the oxidation numbers of the sulfur atoms change in the reaction.

Look first at the sulfur atom in sodium thiosulfate. It carries oxidation no. +2.

Now look at the elemental sulfur atom (on the right).  It carries oxidation number 0

And finally look at the sulfur atom in the sulfur dioxide (SO₂) molecule (also on the right).  It carries oxidation number +4.

One of the sulfur atoms in sodium thiosulfate increases in oxidation number by 2 units because it becomes sulfur dioxide (SO₂)

You can also see now that the other sulfur atom in sodium thiosulfate decreases in oxidation number by 2 units because it becomes elemental sulfur (S).

Here is the equation again with the oxidation numbers applied:

Na₂S₂O₃   +   2HCl   ⟶   2NaCl   +    S   +   SO₂  +  H₂

2(+2)                                    0     +4               

How come this change in oxidation number can happen?

Standard electrode potentials (E^⦵_cell) exist for the two half cells involved.

Those half cells have two half equations and they are:

2H₂SO_3(aq)  + 2H⁺_(aq)  ⇌  S₂O₃^2–_(aq)  +   3H₂O_(l)      E^⦵ = +0.4v

S₂O₃^2–_(aq)  + 6H⁺_(aq)  ⇌   2S_(s) + 3H₂O_(l)     E^⦵ = +0.47v

If we apply our rule of thumb to these two half equations then top right will reduce bottom left.(Highlighted in blue)

You will notice that in this example thiosulfate on the top right reduces thiosulfate of the bottom left!!

That's what we mean by disproportionation!!

Now we can also calculate what the cell emf (E_cell) will be if we use this relationship:

                             E_cell  =  E_{right hand side} – E_{left hand side}

                             E_cell  =  +0.47 — +0.40

                             E_cell  =  +0.07v

The value of +0.07v you have probably realized is the gap in volts between the two half–cells.

Now more importantly in this example, you should look at this value for the E _cell and realise that 0.07v suggests that this reaction is in almost perfect equilibrium.

Yet we know that when you carry out this reaction it goes, albeit slowly, to completion.

The sulphur forms a yellow precipitate.

Now I think that the reaction goes to completion because the sulphur produced no longer remains in solution.

This precipitation effect pulls the equilibrium to the right and the reaction to completion.

Once the sulphur is removed from the solution it can no longer play a part in the reaction.  So the reaction cannot go into reverse.

Here is another example of a disproportionation reaction.

The reaction between red copper(I)oxide and dilute sulphuric acid is a good example to think about.

You’ll probably remember that if you warm black copper(II)oxide with sulphuric acid you get a blue solution of copper(II)sulphate.

So you might think that if you warmed red copper(I)oxide with dilute sulphuric acid you’d end up with colourless copper(I)sulphate.

But in fact you end up with a red precipitate of copper and a blue solution of copper(II)sulphate!!!!

The explanation is that the copper(I) ions disproportionate.

Let’s look first at the two half cells that are involved:

Cu²⁺_(aq) | Cu⁺_(aq) | Pt       E^⦵    +0.15v

Cu⁺_(aq)  | Cu_(s)                 E^⦵   +0.52v

We can see that the gap between the two half cells is

+0.52  – +0.15  =   +0.37v

So the reaction between top right and bottom left should be feasible.

Therefore, looking at the half cells they should suggest to you that copper(I) ions react with copper(I) ions to form copper and copper (II) ions.

In other words, you should see that these half cells suggest that copper (I) ions disproportionate as we know in fact that they do from the reaction between copper(I)oxide and dilute sulphuric acid.

Cu⁺_(aq)     +       Cu⁺_(aq)      ⟶      Cu²⁺_(aq)   +   Cu_(s)

Cu₂O_(s)   +   H₂SO_4(aq)      ⟶    CuSO_4(aq)     +   H₂O_(aq)    +   Cu_(s)

      red                                       blue solution                         brown ppt

Summary:

So finally notice the conditions you need to look for in the half–cells to realise that the feasible reaction is going to be a disproportionation:

First, you list the half–cells with their E^⦵ values, the most negative first:

Cu²⁺_(aq) | Cu⁺_(aq) | Pt       E^⦵    +0.15v     

Cu⁺_(aq)  | Cu_(s)                 E^⦵    +0.52v

Second, you should see that the species that disproportionates is in both half-cells.

Third, if the reaction is going to be feasible the species that disproportionates is at the top right and bottom left of the two half–cells (Highlighted in colour in the half cells above). 

Fourth, the gap between the two half–cells will be greater than 0.6v to be sure that the goes to completion.