One thing chemists want to know is whether or not a particular reaction can happen, whether
it is feasible, without going to all the bother and expense of actually trying the reaction out.
A reaction is said to be feasible if it can happen according to the thermodynamics.
Bear in mind this tells us nothing about the rate in which a reaction may happen.
For many redox reactions we can use the potential difference, also known as voltage, of an
electrochemical cell to find out which reactions are thermodynamically feasible. I'm about to
start my experiment so don't forget to put your safety goggles on. In the reaction between zinc
metal and copper(II) sulfate solution the zinc metal transfers electrons to the copper 2+ ions.
If the zinc metal transfers electrons to the copper 2+ ions directly the energy available
to do work, the free energy, is released as heat. Let's look at that reaction more closely.
As we can see our copper 2+ ions have been reduced to form solid copper metal on the surface of the
zinc. The zinc metal has been oxidised to form zinc 2+ ions. If we force the electrons through
an external circuit to get from the zinc metal to the copper ions we can use the change in
free energy to do work. In the electrochemical cell I am about to set up here, we have zinc
in zinc sulfate solution connected to copper in copper sulfate solution via a salt bridge.
The salt bridge contains ions in solution that are free to move. Oxidation occurs at the zinc half
cell making it the negative electrode. Reduction occurs at the copper half cell causing it to be
the positive electrode. Ions flow from the salt bridge to keep the charge of the solution neutral.
Here I am using a voltmeter to measure the potential to do work. Those potentials can
be used to predict which reactions are feasible and which are not. The feasible reaction is the
one where oxidation occurs at the negative half cell and reduction at the positive half cell.
Now we have seen one electrochemical cell we are going to look at multiple cells. Compared to the
large scale apparatus we just saw, this microscale equipment is easy to set up and only uses small
volumes of solution. To start with I'm going to cut a flower petal shape out of my filter paper.
As you can see I have laid our filter paper into a petri dish. I'm now going to add our four metals
to the different petals of the flower. As you can see I've also labelled the petals using pencil.
I'm now going to add a few drops of one mole per decimeter cube solutions containing those metal
ions onto the filter paper underneath the pieces of metal. I'm using sulfate solutions here. The
metal in contact with the solution containing its ions forms a half cell. To connect these
half cells with a salt bridge we simply add a few drops of a saturated solution of potassium
nitrate to the centre of the filter paper. We need to add enough so that it soaks out
and meets the other damp areas of the metal solutions. The salt bridge now connects the
half cells in the electrical sense because ions can flow from and into the salt bridge.
Now we can start to measure the potential difference of our four different half cells using
a voltmeter. If your voltmeter requires setting we need to set it to the values we are expecting
to measure. In this experiment it's up to about 2 volts, so this is the most appropriate setting.
A voltmeter has a positive terminal and a negative terminal. Here the positive
terminal is connected to a red lead and the negative terminal is connected to a black lead.
I'll start with our familiar zinc and copper. I touch the lead connected to
the positive terminal of the voltmeter to the piece of copper and the lead connected
to the negative terminal of the voltmeter to the zinc. Make a note of the potential difference.
The same thing is happening here as with the cell I showed you in the introduction,
just on a smaller scale. Notice the direction of electron flow. Electrons will leave the negative
electrode. Here the zinc atoms are becoming ions, they are being oxidised. The copper
half cell is more positive than the zinc so electrons flow into the copper half cell.
Copper ions are reduced. Ions flow from the salt bridge to keep the charge of the solution neutral.
Look what happens to the reading on the voltmeter if we swap over the leads.
The zinc half cell is more negative than the copper because zinc is more readily
oxidised to its ions. So the zinc metal is losing electrons. The copper is gaining them. Now we can
set about measuring the other combinations of half cells. First, zinc and magnesium.
If we get a negative reading we simply swap over the connections to get the positive value.
Record the potential difference and which metal was connected to the negative terminal
and which to the positive terminal to give a positive value.
Next, iron and magnesium.
And now, copper and iron.
We measure the potential difference between the copper and zinc to be 0.96V.
The potential difference between copper and iron is 0.50V.
Think about the reactivity series. Which metal, iron or zinc, is higher in the series? With this
in mind, can you predict what the potential difference between iron and zinc will be?
Now let's see if you were right: let's measure the iron and zinc cell.
Lastly, we measure the magnesium and copper combination.
In chemistry, we represent cells using a kind of shorthand called a cell diagram which,
despite its name, is not actually a diagram. We put the more negative half cell on the left which,
in the case of the zinc-copper cell, is the zinc half cell. Remember the negative half
cell is where oxidation is occurring so we put the components in order of increasing oxidation state.
The oxidation state of zinc metal is zero, whereas zinc ions are in the +2 oxidation state.
Because they are in different phases from each other, zinc metal being solid and zinc ions in
solution, we put a single solid line between the two called a phase boundary. We represent
the salt bridge using two vertical lines. The copper half cell is the positive one, so
because reduction is occurring we need to put the components in order of decreasing oxidation state.
Again we add a phase boundary between the solid copper and the copper ions.
Keep in mind the flow of electrons when building cell diagrams to help you build them correctly.
Now we have the full cell diagram we can use this to explore the relationship between the
other cells we have tested. In this video, for example, iron-copper. Use the additional
resources attached to this video to practice creating cell diagrams for those other reactions.
In a further experiment, we can investigate how the concentration of solution affects the
potential difference of a half cell. Again, I have cut a piece of filter paper to form four petals.
This time with a piece of copper in each section
and this time I am going to add copper sulfate solutions of different concentrations.
To two petals I have added one mole per decimeter cubed copper sulfate solution.
To another petal I have added 0.1 mole per decimeter cubed copper sulfate
solution. And to our final petal I'm adding 0.01 mole per decimeter cubed copper sulfate solution.
Just like before, I'm connecting the petals with a salt bridge
of potassium nitrate. Two of the concentrations are the same: one mole
per decimeter cubed. What will the potential difference be between these two half cells?
As we can see here our potential difference is 0.00V.
What do you think will happen if I try two different concentrations?
I'm now going to try one mole per decimeter cubed and 0.1 mole per decimeter cubed.
As you can see, our potential difference has changed.
I will now try all the other possible combinations of half cells.
Now we have seen that concentration of solutions is one condition that can affect the potential
difference of a half cell. Can you think of any other variables that can affect
the potential difference of a half cell? How might you investigate them? In this experiment
have seen how we can measure cell potentials and how these relate to the relative reactivity,
or reducing power, of some metals. We have seen the significance of which half cell is negative
and which is positive. Hopefully this has got you thinking about how we can use these to predict
reactions and how changing conditions might affect them. There is more practise than all
of this in the additional resources. Thank you for watching and have fun with your experiments.
Inspiring your teaching and learning.