Electrochemistry part 2
The Electrochemical Series
If you can become familiar with the electrochemical series (a few reactions from it are shown in the
table), then it will enable you to answer a lot of questions very quickly.
Depending on the question, you will get different substances in the table so don’t rely on the one that is
in your book, but the principles are the same no matter what they use.
The information that you need to be able to extract is as follows:
1.The more negative the potential then the better a reducing agent it is:
In the table above, the most negative value is for the reduction of Fe 3+:
Fe3+(aq) + e-  Fe2+(aq)
Fe2+ is therefore the best reducing agent. You have to be careful that you don’t say Fe 3+ is the best
reducing agent. Remember the reducing agent is oxidized so it will go from Fe 2+ to Fe3+, which is fine.
If you had said Fe3+ is the best reducing agent, then it would have to be oxidized to Fe 4+, which doesn’t
exist! So remember that the reducing agent has to be oxidised.
2. Using the example above. Fe2+ will reduce everything with a more positive potential. For example, it
will reduce Cl2 to 2Cl-.
3. The best oxidizing agent is that with the highest E o value. Remember that an oxidizing agent will itself
be reduced, so it is the substances on the left hand side that we are interested in this time. In this case
F2O is the best oxidizing agent.
4. F2O will therefore oxidize anything (on the right hand side) with a lower Eo value.
Another example:
Looking at the table above, what is the best reducing agent? We are looking for the most negative value
and something that can be oxidized. In this case it has to be Cr(s).
What is the best oxidizing agent? Look for the most positive value and something that can be reduced. It
has to be Fe3+.
More on Ecell values
Another common question is to compare two different species in the table and whether or not a reaction
will take place between them. All you have to do is a simple subtraction of the two different Eo values to
get the Ecell value.
Using the table above:
(a) Predict the products, if any, when iron metal and aqueous zinc(II) ions are mixed. Use Eo values from
the table to explain your answer.
So we have the reaction between iron metal and zinc(II) ions. You could look at this as iron reducing the
zinc(II) ions or the zinc(II) ions oxidizing the iron. It doesn’t matter.
What is important is that from our theory earlier, we know that iron WON’T reduce anything with a more
negative Eo value than it, so the reaction WILL NOT take place. That is the question answered.
They also want this expressed numerically. Having thought about it chemically, we now know that we
need a negative value to show that the reaction won’t take place.
Earlier we mentioned that:
Ecell = Eo (the more positive value – the more negative value)
This is one of the times we CANNOT use this rule as we would get a positive E cell value, which contradicts
what we have just concluded.
So we need to do the opposite:
Ecell = Eo (-0.76 – (-0.44)) = -0.32 V
(ii) aqueous iron(III) ions with aqueous chromium(II) ions
We do the same thing here as for part i). We have Fe(III) which is the best oxidizing agent in the table and
Cr(II) which has a more negative Eo value. Using the table we know that this reaction will happen (Fe(III)
will oxidise Cr(II)), so it is simply:
Ecell = Eo (+0.77 – (-0.41)) = 1.18 V
Non Standard Conditions
All these values are based upon the standard hydrogen electrode at standard temperature, pressure and
concentration. If any of these conditions are changed then the Eo values change.
Normally they change the concentration. You need to look at each half equation and work out if the value
will become more positive or negative and see how that affects the Ecell value. For example if you have:
Zn + Cu2+ -> Zn2+ + Cu
If the Zn2+ concentration is increased, then the equilibrium would go to the left. This is going against the
Zn reaction so the Eo value would go towards zero.
If they increase the Cu2+ concentration then the equilibrium goes to the right, which is going with the Cu
reaction, therefore the Eo value will go away from zero.