Redox Cells: Revision Checklist

When I teach about redox cells, I like starting with the Daniell cell. It is an example of the world’s first battery (cell) and if you remember the details of it, you can apply to same rules to any unfamiliar redox cells.

First, a diagram of the cell itself:

Note that the zinc half cell is on the left and the copper half-cell on the right. This is important to remember when it comes to calculating the cell potential (voltage). More on this later.

I like this example as you can use some fairly basic, pre-IB chemistry to work out what is going on. The Knowledge you need to draw upon is that of the reactivity series. Zinc is more reactive than copper. Once you have remembered this, everything thing else should logically fall into place.

1. As zinc is more reactive than copper, it reacts. When it reacts it will lose electrons:

Zn(s) –> Zn²⁺(aq) + 2e-

2. Zinc is oxidized or acts as the reducing agent.

3. As it is losing electrons it will have a more negative electrode potential (than copper) – you can think about as being electron deficient – it is missing negative charge so it is more negative (than copper).

The rule here is the more negative half-cell goes on the left-hand side.

4. The more negative half-cell will be the anode.

The electrons released in the oxidation reaction will move along the wire to the copper half-cell (OK, maybe the physicists will not like my wording here but we can get away with it in chemistry).

As the electrons move to the copper half-cell they will attract the positive Cu2+ ions in solution:

Cu²⁺(aq) + 2e- –> Cu(s)

5. The copper ions gain electrons so Cu²⁺ is reduced (or Cu²⁺ is acting as the oxidizing agent).

6. As the half-cell has gained extra electrons you can think that it has a great number of electrons than it would ordinarily have – it has a positive number of electrons so.

7. This means it is the more positive half-cell.

8. The more positive half-cell is the cathode.

When it comes to working out the electrode potential of the cell, think of it as being:

E_cell = Reduction – Oxidation

Some books teach this as right – left, but this can lead to problems, Reduction – oxidation will give you the correct answer.

The Zn / Zn²⁺ half-cell’s potential is -0.76v and the Cu / Cu²⁺ half-cell is +0.34v so in this example:

E_cell = +0.34 – – 0.76 = +1.1v

Once you have this idea in your head, you can apply it to any unfamiliar set of half-cell pairs to deduce the reactions and cell potential.

I hope this helps with your revision!