Cu Cu + 2e
So the anode wears away, making its mass decrease, and the cathode grows thicker, making its mass increase. When they are weighed it should be found that the mass of copper lost at the anode equals the mass of copper gained at the cathode. This is because there will be the same number of electrons being lost from the copper at the anode as there are gained at the cathode because the current is the same at both electrodes. Also all the copper ions released at the anode will float through the solution to the cathode where it becomes copper. For every copper ion reduced at the cathode, there must be one copper atom oxidised at the anode in order to keep the electrical charge balanced.
The power pack acts as an electron pusher pushing the electrons to the cathode. If the current of the solution is electrolysed is increased it means that there are more electrons being pushed around the circuit. This means that more copper atoms will be formed at the cathode because there are more electrons to be given out to the copper ions. Every ion only needs two electrons to form an atom which means the electrons can be given out to more ions therefore forming more copper and therefore increasing the mass of the cathode. The anode will lose more mass if the current is increased because there are more electrons being given out from the copper atoms to the anode so that there are more copper atoms becoming ions at the anode, therefore decreasing its mass even more.
There are two laws made up by a man called Faraday concerning electrolysis. His first law states that:
“The mass of an element deposited during electrolysis is directly proportional to the number of coulombs of electricity passed”.
Faraday’s second law states that:
“The mass of an element deposited by one Faraday of electricity is equal to the atomic mass in grams of the element divided by the number of electrons required to discharge one ion of the element”.
These statements can be used to predict the mass of copper deposited at the cathode and will also tell us a little more about how the results will turn out.
Each electron that goes round the circuit has a tiny charge. A current of 1 ampere flowing for 1 second carries a charge of 1 coulomb. To find the charge of any current we can use the formula: charge = current x time. 1 mole of electrons always has a charge of 96 500 coulombs. This, along with Faraday’s laws will also help as to predict the results because once the charge has been found the moles of copper can also be found.
Equipment List
copper sulphate solution copper electrodes
stopwatch emery paper
ammeter propanone
wires ammonium hydroxide solution
power pack water
variable resistor heatproof mat
beaker electric scales
measuring cylinder
Diagram:
Method:
- Collect equipment together
- Put on safety glasses and overall
- Clean the copper foil pieces with emery paper on the heatproof mat, so as not to scratch the desk, until all the loose particles on the surface have gone
- Wipe the electrodes with a tissue soaked in ammonium hydroxide solution which will get rid of any grease that is on the surface which may affect the results
- Weigh each electrode using the electrical scales being careful not to touch them too much as this will put the grease back onto them which you have just wiped off. Record their weights in a table
- Decide which foil will be used for the anode and which for the cathode and fix them in the holder
- Connect the circuit together as shown in the diagram above making sure the cathode is connected to the negative side of the power pack and the anode to the positive side
- Measure out 75ml of copper sulphate solution using the measuring cylinder, pour into the beaker and place the electrodes in so the holder is resting on the top of the beaker
- Turn on the power pack and quickly adjust the variable resistor so that 0.2 is showing on the ammeter then switch the power pack off
- Start the stopwatch and at the same moment switch the power pack back on. Leave for 10 minutes
- Once the 10 minutes is up turn off the power pack and take the electrodes out of the copper sulphate, disconnecting them from the wires
- Being careful not to muddle the electrodes up wash them with water which will get rid of the copper sulphate resting on the surface because you do not want to weigh this. Then drop a few drops of propanone onto each electrode using a pipette; this will help the electrodes dry quicker. Finally, dry them thoroughly using a hair drier.
- Weigh both electrodes and record their weights in the results table
- Start the whole process again only this time using the currents 0.3A, 0.4A, 0.5A and 0.6A. Once these have all been completed repeat all the currents again to get more accurate and reliable results
- It is a good idea to swap the anode and cathode round each time as this will mean that there is not a huge build up of copper on one of the electrodes.
Safety
Copper sulphate is not harmful unless swallowed, however at 0.5M it can be irritant to the eyes and skin so be sure to wear safety glasses and an apron and wash any spillages immediately. Propanone is also highly flammable so do not let it get near to any hot objects such as the hair drier.
Prediction
Using my background knowledge I can determine that there will be an increase of mass at the cathode due to the copper being deposited, and there will be a decrease in mass at the anode because it is losing its copper atoms to the cathode. This agrees with the results from my pilot study also.
I also predict that the current will be directly proportional to copper deposited because in my pilot study when the current increased the change in mass also increased. Both of Faraday’s laws support this prediction.
It is also possible to predict that there will be the same mass lost at the anode as the mass gained at the cathode. This is because, using my background knowledge I found that all the copper which is released from the anode floats into the solution and over to the cathode, which it is attracted to. The half equations at both electrodes also support the prediction that the mass lost is the same as the mass gained because the product at the anode is the same as the ingredients at the cathode. For example the anode produces Cu and 2e from Cu and the cathode uses Cu and 2e to make Cu.
Using Faraday’s laws we can also predict the mass of copper which will be deposited at the cathode for the different currents.
0.2A:
charge = current x time
0.2 x 10 x 60
120 C
we then have to convert the charge to moles of electrons:
moles of electrons = 120 / 96 500 (it is known that 1mole of electrons has a
charge of 96 500 coulombs)
= 1.244 x 10
this can now be converted into moles of copper:
moles of copper = 1.244 x 10 / 2 (using the equation at the cathode we
know that 2moles of electrons gives 1mole
of copper)
= 6.218 x 10
to find the mass:
mass = moles x RAM (relative atomic mass found on periodic table)
6.218 x 10 x 64
= 0.04g
Using these equations I am able to form a “theoretical” table of values that can be plotted against my actual results obtained. I will then be able to spot any anomalous results.
Below there is a table showing these “theoretical” values:
These masses should be the same as the mass lost at the anode.
I have drawn a graph showing these results; it supports the prediction of direct proportion, as it is a straight line that goes through the origin.
Results
Repeat:
Analysis
My results support my prediction that as the more the current is increased the more copper there is deposited at the cathode, hence increasing the mass, and the more copper there is lost from the anode, hence decreasing the mass. This is shown on the graph by the gradient being a positive slope for the cathode and a negative slope for the anode. The reason for this happening is that when the current is increased it means that there are more electrons being pushed around the circuit. So at the cathode there are more electrons to be gained by the copper ions so more copper is produced which increases the mass. At the anode there are more electrons to be lost from the copper atoms, therefore more copper ions are produced, therefore dissolving the anode more because these ions are now attracted to the cathode because of their positive charge. The reason for the copper ions becoming atoms at the cathode is because the two negative electrons balance out their positive charge:
Cu + 2e Cu
This is exactly what happens at the anode only the reverse so that the electrons of the copper atoms, which are attracted to the anode, leave the atoms, therefore unbalancing the charge so that they are positively charged.
Cu + 2e Cu
You can see from these two equations that there are only ever two electrons needed for each copper ion so this is why when there are more electrons in supply more ions can be balanced out to become atoms.
I also predicted that the mass of copper deposited would be directly proportional to the current. My graph supports this prediction as it shows a straight line through the origin. This means that as the current is increased so does the mass of copper deposited.
I decided to draw a line of best fit through the points plotted on the graph as this will help us analyse the results more easily as the line is not affected by the anomalous results. It is true to say that if the current is doubled the copper deposited is also doubled because this can be seen from the graph. Using the line of best fit we can see that at 0.2A the change in mass is 0.35g, when the current is doubled to 0.4A the change in mass is shown to be 0.7g. 0.7 is exactly double 0.35.
In my prediction I used Faraday’s second law to predict the results and these were proven to be very similar to my results, however, my results had some error in them due to the difficulties of gaining complete accuracy throughout the experiment. Comparing the lines of best fit on the graph of my actual results with those on the graph of the results using Faraday’s law it is noticeable that they are very similar.
My final prediction which was also proved to be correct was that the mass gained at the cathode would be the same as the mass lost at the anode. My results do support this to some extent, however, there was not enough accuracy in the experiment to be able to support the prediction fully. The reason for the mass gained being similar to the mass lost is that the copper which is lost from the anode drifts across the solution to the cathode to increase its mass in the same way it decreased the mass of the anode. All the ions that are lost from the anode will be attracted to the cathode because of the opposite charges. For every copper ion reduced at the cathode, there must be one copper atom oxidised at the anode in order to keep the electrical charge balanced.
Evaluation
This was a very successful investigation in that all of my predictions were proved correct, however, not to the accuracy I would have liked. There were some factors that could have been improved upon to make the experiment even more successful. One of these factors were the electrodes which, even after a very good clean with the emery paper, were still a little dirty and could have possibly had some irremovable substances attached to them possibly from being used before or maybe from being subjected to the air for too long. If this experiment were to be repeated for a second time the electrodes would have to be made from very new copper having never been used before and it would also increase the accuracy to have a new pair for each current.
Another factor that could have affected the overall outcome of the investigation may have been the fact that the practical work was carried out over two lessons meaning that variables such as the temperature may have varied between the lessons. Also when weighing the electrodes there were two different sets of scales used, these scales may have had slight differences between their balances. To overcome this only one of the scales should be used.
My results obtained did not match perfectly what the results should have been which I had found from using Faraday’s law showing that there was some error in the experiment. Looking at the graph we can see that the major anomalous results were found on the cathode where for 0.3A the reading was much lower than that of 0.2A which should not happened. The reason for the cathode being less precise than the anode is that whilst washing the cathode after it had been in the solution some of the copper may have fallen off which would have altered the mass. As for the anode there is nothing to be washed away by only the water. This is a very hard factor to overcome as it is essential to wash the electrodes in order to remove the copper sulphate, a lot more care could be taken though, perhaps running the water over the electrode much more mildly.
Despite the anomalous ones the results did follow the basic pattern that I had predicted. However, on very few of the currents did the mass gained at the cathode prove to be equal to the mass lost at the anode. This may have been because the solution was not left to electrolyse for long enough. If the experiment were to be repeated the solutions should be left for longer, perhaps 20 minutes, however, it was impossible to leave them for this long this time because our lessons were not long enough.
Although my results do support my predictions made they are not accurate enough to support a firm conclusion. The experiment would have to be done using more technical equipment, perhaps in a proper laboratory where the temperature is always constant, in order to produce reliable findings. It would also have been nice to do some more repeats in order to make the results more reliable, however, this was once again impossible due to the time factor.
I found this investigation very interesting yet I would have liked to taken it to further levels of investigation. Perhaps investigating how different factors affect the experiment or maybe seeing how electrolysing copper sulphate with copper electrodes could be used in business such as the copper plating industry and what current would have to be used there, and so on.
