This diagram demonstrates what is said below
Copper (II) Sulphate is an ionic substance. When in solution, the ions are free to move and can carry an electric charge. Ions are also produced from the water:
H2O (l) → H+(aq) + OH- (aq)
CuSO4 (aq) → Cu2+(aq) + SO42- (aq)
The copper ions carry a positive charge and are attracted to the cathode. In the electrolysis of copper sulphate a number of things happen at the anode. If it is made from graphite oxygen will be formed and given off. If the anode is made from copper the copper atoms will lose electrons to form copper ions which will then go into solution. The copper ions are then attracted to the cathode where they gain electrons to form copper metal once more. Only copper will be deposited at the cathode. This means that if an impure piece of copper is used as an anode the copper atoms will ionise and the ions will be deposited on the cathode as pure copper. This is therefore a good way to purify copper.
At the cathode, the copper ions become copper atoms as they gain electrons:
Cu2+ + 2e- → Cu
The copper atoms cling to the cathode and make it heavier, copper is deposited at the cathode.
At the anode, the copper atoms dissolve and become copper ions:
(electricity)
Cu → Cu2+ + 2e-
The anode becomes lighter as it has lost atoms.
I have chosen to investigate the affect that the size of current has on the electrolysis of the solution. I chose this because it is easy to control as you can use a variable resistor and will give reliable and concise results.
Prediction:
If I increase the size of the current I will increase the amount of copper deposited at the cathode. This is because more electrons are passed through the solution. Therefore if I double the current, I will double the amount of electrons passing through the cell. This, if I double the amount of electrons, I will double the amount of copper deposited at the cathode. This is because 2 electrons are needed in order to make a copper ion a copper atom
(Cu2+ +e- →Cu (s)) so if I double the number of electrons, I will double the number of available electrons to combine with more copper ions, forming more copper atoms. I think that the gain/loss in mass of the cathode/anode will be directly proportional to current. I think this means that I will get a graph that has a straight line through the origin if the results are accurate.
For this experiment, I will need:
- 30 copper electrodes (6 for each amplitude)
- 1 beaker
- 2 connecting wires
- electric balance
- copper (II) sulphate solution
- a changeable power supply
- Propanone for cleaning the electrodes prior to use
- Stopwatch
The power supply has to be able to change the current in order for the experiment to work, as the varying factor is the current.
I will set up the apparatus as shown opposite. The current values I will use will be 0.1, 0.2, 0.3, 0.4 and 0.5 amps. I will use these currents as I believe they will give a wide enough range of results giving me a good set of results to analyse fully. I will repeat each reading 3 times and calculate an average to make the results more accurate. I will do this by identifying any anomalous results and retake the reading and then take an average of the three readings. More readings will make more accurate results and a more reliable conclusion. To ensure that this experiment is a fair test, I will keep the temperature the same, the beaker size the same, the amount and concentration of copper (II) sulphate solution the same, the distance between the two electrodes the same and dry the 2 electrodes in the same way. This should keep all the factors the same to eliminate all inaccuracies, apart from the current which is the variable I shall be investigating. This is how I will carry out the investigation:
Firstly, I shall lay all my apparatus out and measure 50ml of copper (II) sulphate solution into a 100ml beaker. I shall then clean two of the electrodes with propanone to remove any dirt or impurities that may be on the electrodes to make the experiment a fair test. The 2 electrodes will then be weighed on a 3-figure electric balance and their masses recorded. The 2 electrodes will be connected to a power supply and placed in the copper (II) sulphate solution 4cm apart and the power supply set to the correct amplitude (0.1 the first time) and turn the power supply on at the same time the stopwatch is started. The circuit will be left on the same amplitude for 4 minutes. After this time is over, the circuit will be disconnected and the two electrodes removed. Both the electrodes will be dried with a paper towel in the same way, and the reweighed. Their new masses are then recorded and tabulated. This will be carried out 3 times for each of the amplitudes 0.1, 0.2, 0.3, 0.4 and 0.5 (electrolysis will be carried out a total of 15 times; 3 times for each amplitude). I am increasing the current each time to prove my prediction, i.e. that increasing the current increases the amount of copper deposited.
Analysis of Results
This graph [1] shows that the average gain in mass is directly proportional to current. It shows that as the current increases, the amount of copper deposited increases. The graph was not a straight line through the origin as expected, so a best-fit line was drawn. A few anomalous results were identified, for example, At 0.4amps; the average gain in mass was 0.0315g, which did not lie very near the line of best fit. If we compare this to the graph showing the average loss in mass at the anode, we can see that the point lies on the line although there is only 0.0005g between them. This may suggest that the results were not very accurate. Although both of these results, 0.0315g gained and 0.032g lost were very close, they both lay on very different places on each of the lines drawn. This suggests either error on those two results, or for the rest of them. On the graph comparing both lines of best fit, we can see a slight difference in the direction of the lines. They both show that the loss/gain in mass is directly proportional to the current. The mass of copper lost at the anode should be exactly the same as the mass of copper gained at the anode, since the concentration of Cu2+ ions in the electrolyte does not change. The graph, however, shows that the gain in mass at the cathode is more than the loss in mass at the anode, and they should be the same. This maybe because it was extremely difficult to measure the gain of copper on the cathode, since by cleaning the electrode some of the copper deposited was knocked off. Nevertheless, this washing process was vital because of the need to eliminate the copper sulphate crystals that may grow on the electrodes. The results might have been better if the experiment was left on for longer to compensate for the error. This would account for the loss in mass of the anode being more than the gain in mass of the cathode, but this was the other way around, so I can only suggest that the measurements were not accurate enough.
In my plan I predicted that if I doubled the current, the amount of copper deposited at the cathode would also double. The results support this to some extent but not as accurately as hoped. At 0.2amps, there were 0.017g of Cu deposited. 2 x 0.017 = 0.034 so I would have expected 0.034g of copper to have been deposited. At 0.4amps, 0.0315g were collected. This has a difference of 0.0025g, therefore is 0.0025g out. Two of the results were the same. This was for 0.2amps, where both the loss and the gain were 0.017g. The difference in most of the results could be because of 2 reasons:
- The loss in mass of the anode was a more accurate reading, as no copper can be removed. The copper atoms deposited on the cathode are very delicate at can be easily removed when cleaned. This may have altered the reading.
- An average of both results was taken, and this could have differed the results
Graph [2] shows the same as graph 1: that the average loss in mass is directly proportional to the current.
Graph [3] shows both graph [1] and graph [2] on the same axis. Although some of the points were very similar, the best-fit lines gradually changed by 0.001g every 0.1amp. For example:
This table shows that as the amplitude increases by 0.1, the difference between the two sets of results also increases by 0.001. This indicates that the error in measuring the loss/gain in mass was a systematic one.
Graph [4] shows the relationship between charge and average gain in mass at the cathode. Like the other graph it is drawn with a line of best fit. It is a straight line and the end of the line occurs at the origin. This means, therefore, that the average gain in mass at the cathode and charge are directly proportional. To explain why the two are directly proportional I need to use Faraday's law of electrolysis. This is explained at the bottom of this page.
According to the ionic equation of the reaction at cathode:
Cu2+(aq) + 2e- → Cu(s)
every copper ion requires two electrons to become copper metal, thus one mole of copper ion plus two moles of electrons gives one mole of copper metal.
1 mole of copper ions + 2 moles of electrons → 1 mole of copper atoms
The current is a measure of how many electrons flow through the circuit in a given time, so if the time is kept constant, the number of electrons passing is only affected by the current. So by increasing the current, the number of passing electrons is also increased, so more electrons can form with more copper ions forming more copper atoms. Therefore, the number of copper atoms, and their combined mass, would only be affected by the current. In fact, if the mass of copper required is doubled, the current must be doubled to allow double the number of electrons to pass through in the same amount of time, this giving twice the number of copper atoms as well as twice the overall mass.
If 63.5g (1 mole) of copper atoms must be deposited at the cathode in one second, a current of 193,000 amps would be needed. This is given by using Faraday’s law and Faraday’s constant.
Faraday's law states that a certain amount of electricity (charge in coulombs) is required to deposit a certain amount of a metal at the cathode. Faradays law also states that 96 500 coulombs of electricity is equal to 1 mole of electrons.
Current (I) = Charge (Q) / Time (T)
Faraday’s Constant:
1 mole of electrons have 96,500 coulombs of charge.
1 mole of electrons = 6x1023
Faraday’s law therefore means the more charge being passed through the solution, the more copper metal is being deposited at the cathode. At the cathode, this is happening
Cu2+ + 2e- → Cu
1 mole + 2 moles → 1 mole
1 mole + 2 x 96500 → 63.5g
Copper has a formula mass of 63.5g and 1 mole of copper is therefore equal to 63.5g. The amount of electricity needed to make 1 mole of copper can be calculated. 2 moles of electrons are needed to make 1 mole of copper. 1 mole of electrons is equal to 96500 coulombs. Therefore if 2 moles are needed then the amount of electricity needed to make 1 mole of copper (63.5g) is
96500 x 2 = 193 000 coulombs. The calculation is shown below:
I = Q/T
Q = 2 x 96500 coulombs
T = 1 second
I = 2 x 96500 /1
I = 193000 amps
From the results obtained, I can calculate the amount of electricity passing through the cell by using the formula:
Q = I (current) x t (time)
I can also calculate the average percentage error. This will show how accurate my results were. I will do this by taking the average gain in mass at the cathode and divide it by the average loss in mass at the anode. I will then multiply this number by 100. Then, I will subtract this number from 100. The results are shown below:
These figures show that some of the results were inaccurate, such as 0∙3amps, which has an average percentage error of 20%, which is very inaccurate.
Conclusion:
From the results obtained, I can conclude that the average gain in mass at the cathode is directly proportional to current. I can also say that average loss in mass at the anode is directly proportional to the current. This experiment’s results were difficult to obtain accurately, which may account for the slight inaccuracy of the results. However, the results gave enough information to support my original prediction. My prediction was correct to a certain extent. The results were not as accurate as I had hoped they would be, but generally supported my prediction, in that they showed that average gain/loss in mass is directly proportional to current. The graph was in the same direction as the way I had thought it would be, but it as accurate as I had envisaged. As I did not obtain the results, I can not tell why the results were a little inaccurate. I have obtained the results I got because of the increase in electrons passing through the cell when the current is increased. They show that I have found out what I predicted, i.e. that average gain in mass at the cathode is directly proportional to current e.g. 0∙2amps there is an average gain of 0∙017g. These two figures are directly proportional, proving my prediction right. This is the same as my prediction as it shows what I thought would happen did actually happen.
