Redox (II): Standard Electrode Potential E⦵(3)

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Standard Electrode Potential E⦵(3)

Here are three further learning objectives:

14/7 To be able to calculate a standard emf, Eocell, by combining two standard electrode potentials.

14/8 To be able to write cell diagrams using the conventional representation of half-cells.

14/10 To be able to predict the thermodynamic feasibility of a reaction using standard electrode potentials.

Writing Cell Diagrams

Having to draw diagrams of cells composed of two half-cells that have been constructed from beakers and glass and metal electrodes etc. is tedious at best.

There has to be a simpler notation for the representation of a cell.

Chemists of course being what they are masters of abbreviation have massively simplified the drawing of a cell from two half cells

So the hydrogen copper cell now looks like this:

Pt [H₂(g)] | 2H⁺(aq) || Cu²⁺ (aq) | Cu(s)

Isn’t that a lot simpler?!! Yes!!

Let’s explain how it’s done:

Pt [H₂(g)] | 2H⁺(aq) || Cu²⁺ (aq) | Cu(s)

As electrons flow from the left hand to the right hand electrode then hydrogen is being oxidised and copper ions are being reduced.

A positive emf will be recorded on the high resistance voltmeter or galvanometer which in this example is +0.34v.

But if copper was replaced with lithium and lithium ions then the reading would be –3.04v but the cell diagram would still be:

Pt [H₂(g)] | 2H⁺(aq) || Li⁺ (aq) | Li(s)

The negative sign would mean that in this example lithium metal is losing electrons and being oxidised and hydrogen ions are being reduced to hydrogen gas.

Calculation of Standard Cell Emfs: E^⦵_cell

Suppose we want to know the emf of a cell composed of the copper half–cell and the lithium half cell, what do we do?

First, we list the two half–cells according to their emfs: the most negative first:

So

                  Half cell                        E^⦵/v

                  Li⁺(aq) | Li(s)             –3.04

                  Cu²⁺(aq) | Cu(s)         +0.34

Second, from the list we apply a fail-safe rule of thumb “top right reduces bottom left”.

The electrode potentials tell us, in other words, that lithium will reduce copper (II) ions to copper.

Lithium will release electrons in the cell and copper ions will take them up.

So we put the lithium half-cell on the left of our cell diagram and the copper half-cell on the right.

The cell diagram looks like this:

Li(s) | Li⁺(aq) || Cu²⁺ (aq) | Cu(s)

Electrons flow from left to right in the diagram.

Third, to calculate the cell emf we use: 

                                    E _cell  =  E _{right hand side} – E _{left hand side}

                                    E _cell  =  +0.34 – -3.04

                                    E _cell  =  +3.38v

If we were to write the cell diagram in the opposite way with the right hand electrode being the lithium electrode on the assumption that copper releases electrons in its reaction with lithium ions then we would find that the cell emf would be –3.34v and the negative sign would tell us that the reaction was not feasible nor spontaneous and would not go.

As it is however, the positive cell emf means the reaction as written is feasible, lithium reduces copper ions to copper in aqueous solution.

Now the reaction between lithium and copper ions is not the only reaction that will go on in the reaction flask. 

Why? Well, we know lithium reacts with water to produce hydrogen and lithium hydroxide and this reaction will also take place.

In other words, we cannot assume in measuring cell emfs that just because one reaction takes place between two species that reaction excludes all others, it does not.

What’s more the two species in the cell reaction may not only react in a redox reaction but one species may produce a precipitate with the other and remove it from the reaction mixture all together.

Neither can we assume anything about the rate at which the two species in the cell reaction will combine.

It can be the case that the rate of reaction between the two species in the cell reaction is so slow that nothing effectively happens even though a positive value of the cell emf was measured. 

Cell emf merely tells us that the cell reaction is thermodynamically feasible not whether the activation energy of the cell reaction is so large (or small) that the reaction cannot proceed.

Cell emf gives us no information about the rate at which the two species in the cell reaction combine together. 

The positive sign of the cell emf tells us that the cell reaction is feasible and spontaneous in aqueous solution. 

We should also look for the final cell emf to be greater than 0.6v to ensure that not merely an equilibrium has been established between the two species but that the reaction has gone to completion. 

If the cell emf lies between 0 and 0.6v then we can assume that the reaction has established an equilibrium.