+ -
As stated in the ionic equation for the reaction at the cathode, it shows us that for every Copper (II) ion, 2 electrons are needed to become copper metal. Therefore two moles of electrons are needed to liberate 1 mole of copper ion, giving one mole of copper metal. The current measures how many electrons flow through the circuit in a set time, so if the time is constant the only factor which will affect the mass of copper collected is the current as the concentration of the solution does not have much of an affect.
The copper ions will be attracted to the cathode where it will combine with the electrons form the cathode and those which are being lost from the anode, forming a coating of copper on the graphite cathode. If more of the hydroxide and copper ions are available, more copper will be formed at the cathode because more electrons are being lost from the hydroxide ions.
I think that the mass of copper will be directly proportionate to the current. So if the current is doubled, then the mass of copper collected will also double and if the current is tripled then the mass will also triple. We can work out the approximate mass of copper collected at the end of the reaction with Faraday’s Law. So to work out the mass of copper collected after 5 minutes (300 seconds) with a current of 0.1A I would be able to do so using Faraday’s Law and Constant ( 1 mol of electrons have approximately 96500coulombs of charge).
0.1A=
Time * Current= Charge
300s * 0.1C/s = 30.0 C
Charge/Constant = Mol of Electrons
30.0 C/ 96 500 C/ mol= 0.00031088 mol
At the Cathode:
Cu2+ (aq) + 2e- Cu (s)
2 mol of electrons are needed to liberate 1 mol of Copper, so 0.00031088 mol of electrons will be needed to liberate 0.00015544 mol of Copper. So the mass of 0.00015544 mol of copper is:
Mol of Copper * Molecular Weight = Mass of Copper
0.00015544 mol * 64g/ mol = 0.00994819g
This tells us that 0.00994819g of copper will be formed at the cathode.
We can use the same formula to work out how much copper will be formed with a current of 0.2A, 0.3A, 0.4A and 0.6A after the same time- 5 minutes.
0.2A=
Time * Current= Charge
300s * 0.2C/s = 60.0 C
Charge/Constant = Mol of Electrons
60.0 C/ 96 500 C/ mol= 0.000621761 mol
At the Cathode: Cu2+ (aq) + 2e- Cu (s)
2 mol of electrons are needed to liberate 1 mol of Copper, so 0.000621761 mol of electrons will be needed to liberate 0.00031088 mol of Copper. So the mass of 0.00031088 mol of copper is:
Mol of Copper * Molecular Weight = Mass of Copper
0.00031088 mol * 64g/ mol = 0.01989632g
This tells us that 0.01989632g of copper will be formed at the cathode.
0.3A=
Time * Current= Charge
300s * 0.3C/s = 90.0 C
Charge/Constant = Mol of Electrons
90.0 C/ 96 500 C/ mol= 0.000932642 mol
At the Cathode: Cu2+ (aq) + 2e- Cu (s)
2 mol of electrons are needed to liberate 1 mol of Copper, so 0.000932642 mol of electrons will be needed to liberate 0.000466321 mol of Copper. So the mass of 0.00031088 mol of copper is:
Mol of Copper * Molecular Weight = Mass of Copper
0.000466321 mol * 64g/ mol = 0.029844559g
This tells us that 0.029844559g of copper will be formed at the cathode.
0.4A=
Time * Current= Charge
300s * 0.4C/s = 120.0 C
Charge/Constant = Mol of Electrons
120.0 C/ 96 500 C/ mol= 0.001243523 mol
At the Cathode: Cu2+ (aq) + 2e- Cu (s)
2 mol of electrons are needed to liberate 1 mol of Copper, so 0.001243523 mol of electrons will be needed to liberate 0.000621761 mol of Copper. So the mass of 0.000621761 mol of copper is:
Mol of Copper * Molecular Weight = Mass of Copper
0.000621761 mol * 64g/ mol = 0.039792746g
This tells us that 0.039792746g of copper will be formed at the cathode.
0.6A=
Time * Current= Charge
300s * 0.6C/s = 180.0 C
Charge/Constant = Mol of Electrons
180.0 C/ 96 500 C/ mol= 0.001865284 mol
At the Cathode: Cu2+ (aq) + 2e- Cu (s)
2 mol of electrons are needed to liberate 1 mol of Copper, so 0.001865284 mol of electrons will be needed to liberate 0.000932642 mol of Copper. So the mass of 0.000932642 mol of copper is:
Mol of Copper * Molecular Weight = Mass of Copper
0.000932642 mol * 64g/ mol = 0.059689119g
This tells us that 0.059689119g of copper will be formed at the cathode.
From these figures, I can predict the shape of the graph. I predict it will go up in a positive correlation and will be directly proportionate- as the current increases, the mass does. So if the current is doubled, tripled, etc, so will the mass of copper collected. The graph below will show a similar graph as to the final graph:
Apparatus:
- 100ml glass beaker
- 6 wires
- 2 crocodile clips
- 1 light bulb
- 1 stopwatch- hh mm ss.ss
- 2 graphite rods (electrodes)
- 50ml copper sulphate solution (electrolyte)
- 2 elastic bands
- 1 small piece cardboard to hold the electrodes
- 1 2d.p. scale
- 1 variable resister
- 1 multimeter
- Sand paper
- Acetone
- Power pack.
Fair Testing:
- Use the same amount of copper sulphate solution
- Time the reaction for the same amount of time
- Use the same apparatus each time
- Weigh the cathode before and after every experiment so you know how much copper has been deposited
- Sand off the copper after each experiment
- Keep the Digital ammeter on either 10Amps or on Milliamps.
Safety:
- Make sure the electrodes do not touch while in solution, because they can cause a short circuit if the electricity supply is on.
- Wash your hands after touching the electrodes or electrolyte, as you may start to have a small rash or irritation on the hands.
- Always wear goggles when doing any chemical experiment.
- Do not keep the voltage too high, because it may cause a short circuit.
- Turn off power supply after use, and any other equipment needed to be switched off.
Method:
- Set up equipment as shown below in the diagram:
- Weigh the cathode and record its mass.
- Measure out 50 ml3 of Copper sulphate solution into a beaker and place the electrodes into the solution, making sure they do not touch as they can create a short circuit.
- Make sure that the stopwatch is clear. When all the equipment is checked and in working order, and the current you wish to record is displayed on the ammeter, switch on the power supply at the same time as you start the stopwatch. Time this for 5 minutes.
- After 5 minutes, switch off the power supply and remove the cathode.
- Weigh the cathode again. There should be an increase in the mass of the cathode. Deduct the mass before from the mass after. This is the amount of copper formed during the reaction.
- Next, using the sand paper, sand away the copper so that only graphite is visible.
- Repeat steps 2-7 another 2 times to gain the average results for that current.
- Repeat steps 2-8 again for the other 4 currents.
Analysis:
From the graph and results, I can tell that as the current increased, so did the mass of copper formed at the cathode. The graph shows a positive correlation and somewhat matches the pattern that I had predicted it to have, also seen in the predicted graph drawn previously. I had predicted that the graph would be directly proportionate- that is as the current increased, the mass would also. If it doubled or tripled so would the mass. This did work for some of the currents I used. When 0.3A was doubled to 0.6A, the results went from 0.03g to 0.06g. When 0.2A was doubled to 0.4A, the results went from 0.3g to 0.6g. This shows that there were some direct proportionate results.
My predicted results were not exactly the same as my final results, but they did show some similarities that as the current increased, so did the mass collected. In my predicted graph, the current and mass were exactly the same but this did not happen in my final, because I had conducted the experiments on different days, so the conditions under which I conducted the experiments were different, giving me the change in results. If I had done my experiment all on the same day, my results could have possibly matched those that I had predicted.
From this experiment, I have found out that as the current increases, so does the mass of copper formed on the cathode during electrolysis. This is because when current is increased, it means that the flow of electrons has also increased. This obviously means that more ions are present in the electrolyte. If more ions are present, it means that more copper (II) ions will attract to the cathode and more hydroxide ions will be attracted to the anode. The hydroxide ions will loose one electron to make a complete outer shell, and so it will be happy as it will be neutral, because it will be easier to loose one electron on its outer shell, rather than having to try and gain 7 electrons to complete its outer shell. The electrons that are lost will go and combine with the copper ions and give it a full outer shell, to form the copper atom. It is easier for copper to gain 2 electrons than loose 6 to complete its outer shell. The ions and electrons combine together to form copper metal. If more ions are present then the speed will increase and so will the mass of copper collected.
Evaluation:
I used a very easy method to follow, because there was not much to change. It was effective and gave good results. In the investigation, I only had 2 anomalous results, which I re- tested and they were correct the next time. The accuracy of results was very good, there were no big gaps between the results I got each time I repeated the test. They were almost the same, so that ensured that I was working correctly. I think I may have got the anomalous results because the multi- meter switched off during both those test, so that may have made a difference. Also because the experiment was not carried out all on the same day and equipment may have been different, therefore making a difference.
The results were reliable, even though I did have a few errors which were corrected. The results did however show a clear pattern that as the current increased, so did the mass of copper collected.
I could improve my method by carrying out the experiment under the same conditions every time I did it, using the same equipment, measuring more accurately and taking a bit more time, so that things would not be rushed. I could also have the room temperature constant, so that if temperature affects the reaction, then it wouldn’t have much of a difference.
If I were to extend the investigation to make it better, I would keep the
Table of results
