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.
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