To know how long my experiment will last I will check the amount of amps at the beginning of the experiment and from that work out how long the experiment should run for. I can do this because I will know how many coulombs I want to pass through the circuit. Below is an example of how this can be accomplished:
charge = current X time
Therefore:
time = charge / current
For Example if there was a current of 2 amps and I wanted 100 coulombs:
time = 100 / 2
time = 50 seconds
I will make my experiment as fair as possible. I will always keep my electrodes completely submerged. I will keep all variables constant apart from charge. I will keep the same concentration of solution. I will use the same quantity of solution. I will measure the weight of my electrodes using the metric system and an electronic scale which will allow my results to be more accurate. These measurements will accurate to 2 decimal places. The amount of solution used in the beaker will be measured from the reading at the bottom of the meniscus. I will repeat each test three times and work out an average, as this will make it possible to recognise any anomalies. I will dry off my electrodes (as stated earlier) in such a way that I will only weigh the electrodes and not any excess materials.
I will adhere to basic laboratory safety rules such as keeping bags under the table, tying hair back and not running as well as other basic rules. I will also set up my experiment away from any hazards to stop it from getting knocked over. I will keep my working area as clear as possible so there is nothing on my desk except the apparatus needed. I will wear goggles in order to protect my eyes. I will be careful to dry my hands thoroughly after washing my electrodes to avoid any chance of shock.
I am unsure of what quantities would be best to use for my experiment. If my quantities are too low then a great deal of accurate equipment will be required or if my quantities are too high then I will waste a lot of time. Therefore I will do preliminary work to see what kind of quantities would be worth looking at. I set up my experiment as shown in the diagram above. I tested for 5 coulombs, then 20 coulombs, then 100 coulombs, then 200 coulombs and finally 500 coulombs. These experiments gave me a better understanding of what I was dealing with and what quantities would be appropriate.
From my preliminary work I decided that I will use 100 coulombs, 150 coulombs, 200 coulombs, 250 coulombs, 300 coulombs, 350 coulombs and 400 coulombs.
To understand my experiment and to give a prediction it is important to understand electrolysis
Electrolysis is the process of passing an electronic current through a liquid solution so that compounds joined by ionic bonding can be separated into pure elements. Electrolysis splits a compound into both negative ions and positive ions. Negative ions are attracted to the anode and positive ions are attracted to the cathode. In my experiment there will not be negative ions that will go to the anode. Instead the anode will be active and discharge into the solution the same amount of copper that the cathode gains leaving the solution at a constant morality.
In electrolysis the ions allow electrons to flow around the circuit. An ion is an atom or molecule which has acquired an electrical charge. An ion which carries a positive charge is called a cation and an ion which carries a negative charge is called an anion. In my experiment the copper will become a cation before it gains 2 electrons to become a naturally charged atom.
To permit the flow of electricity electrons must move from negative to positive. This can be done bone electrons themselves simply moving or through ions giving off and gaining electrons. Electrons are not normally found in aqueous solutions because they react with water. Ions, however, are quite stable in aqueous solutions and can carry charge as they move through it
The ions must be in a solution or in molten form so that particles can move. In my experiment I will use a copper sulphate solution. A solution which contains ions is called an electrolyte solution (sometimes simply an electrolyte). Electrolyte solutions conduct electricity because the charged ions can move through them. Electrolyte solutions are ionic conductors.
Electrodes are needed to allow for the electrons to pass through the solution and through the external circuit but also so that the copper can be collected. The copper ions will be attracted to the negative electrode where they can be weighed.
I predict that copper will migrate from the anode to the cathode. The copper sulphate will be ionised which will result in the copper and sulphate separating:
CuSO4(aq) ==> Cu2+(aq) + SO42+(aq)
The copper will then be attracted to the cathode because it has a positive charge. At the cathode it will gain two electrons and form neutrally charged copper atoms (this is a reduction reaction) which will remain at the cathode:
Cu2+(aq) + 2e- ==> Cu(s)
At the anode the copper loses two electrons and becomes copper ions (this is called a reduction reaction) which go into the solution:
Cu(s) - 2e- ==> Cu2+(aq)
The sulphate does not take part in the reaction and the concentration of the solution should not change. The anode will lose the same amount of copper that the cathode gains.
It is possible to take this prediction further and calculate what mass of copper should appear on the cathode. This is possible if I use Faraday’s first law and his calculations. Faradays first law states that the mass of a given element liberated during electrolysis is directly proportional to the quantity of electricity consumed during the electrolysis.
This effectively means that the higher the amount of electricity passed through the circuit the more mass that will be ‘liberated’. The higher the electricity the higher the mass given off.
I can use Faraday’s calculations I can calculate the mass that should be given off. I need to work out how many Faradays I will use in my experiments.
1 Faraday = 96 500 (approximately)
Therefore
Quantity of electricity passed (F) = current (A) X times (s)
96500
Since current multiplied by time is equal to charge this equation can be re-written:
Quantity of electricity passed (F) = charge (coulombs)
96500
In my first experiment I used 100 coulombs, which is equivalent to:
Quantity of electricity passed (F) = 100
96500
= 0.001036 F
Copper ions have a 2+ charge so therefore need 2 Faradays to liberate one mole of copper atoms (64g).
Therefore I need to divide the number of faradays in my experiment and multiply that by 64. For example 100 coulombs should produce
0.001036 / 2 X 64 = 0.033161g
Because the results can be calculated in an equation like the one above I predict that when I draw my graph I will get a straight line of best fit. I also predict that 100 coulombs will result in half the weight change of 200 coulombs. I believe that my graph will look something like this:
I have drawn in what I think the line of best fir will look like.
Using Faraday’s equations I can predict results for all my planned experiments and later compare these to my actual results. My predicted results are as follows:
I will also set up a basic control experiment to see if anything else might affect the weight of electrodes. I will leave a copper electrode which is not connected to anything in a copper sulphate solution for 5 minute, dry it and see if there are any changes in its weight.
There was no weight change so there are no other factors that I have not thought of which may contribute to weight change in the electrode.
My results for my experiment were as follows:
. *This is an average of the amount of weight change for all three repeats of each experiment. This includes both weight gained and weight lost. I worked out the change in the 3 anodes and the 3 cathodes and divided by 6
** This is an anomaly in my experiment as it only changed its weight by 0.05 whereas the average (for which I have excluded this anomaly) is 0.9, almost double the weight change in the anomaly.
My results show that the higher the amount of coulombs the more mass that will be produced
It is useful, to test accuracy, for me to compare my average weight changes with the quantitative predictions I made earlier. This is shown in my chart below.
I can compare these 2 results on one graph
NB Quantitative predictions may appear in a slightly different position than stated in my table, as they are accurate to more decimal places. For example the first point actually lies at 0.033161 which is rounded off in my table to 0.3
This graph shows that my experiment yielded results which show that my average weight change is lower than they should have been. My first results were more accurate than the results later in the experiment. At 250 coulombs there is a noticeable dip in the average weight change. This shows that there may have been some inaccuracies in my work.
In my prediction I stated that I would expect a straight line in my graph. This did not happen but would have if my results were perfect.
Because the quantitative predictions form a straight line, the equation of this line can be used to make any further calculations in a much simpler way than before. This may however work only for copper ions. Further investigation would be needed to see if this would work elsewhere. The equation of the line is:
Y=0.0003316062x
Or
Weight change = 0.0003316062 multiplied the number of coulombs
I can test this by using 300 coulombs as an example
Weight change = 0.0003316062 X 300
=0.099482
To see if this is correct I will use faradays method of calculation:
F = 300 / 96500
= 0.0031088
Weight change = 0.003188 / 2 * 64
= 0.099482
Therefore, because both yielded the same results, both methods of calculations work
I can conclude that my results do back up my prediction. The cathode gained weight and the anode lost weight. The copper cathode gained approximately the same weight as the anode lost. This is because the copper migrated from the anode to the cathode. At the anode the copper atoms lost two electrons and become copper ions:
Cu(s) - 2e- ==> Cu2+(aq)
The copper atoms were then attracted to the cathode because they had gained a positive charge. At the cathode the copper gained two electrons and forms a neutrally charge copper atom which was deposited at the cathode:
Cu2+(aq) + 2e- ==> Cu(s)
I believe that my experiment was fairly accurate. My method was good enough to get good results. In particular my way of removing excess materials from the copper electrodes was very good. I also believe I planned my experiment out to be quite fair as stated when I gave reasons why this is a fair test. I collected a sufficient number of results in order to get enough evidence for my conclusion. My first 3 results are particularly accurate although after that there were some inaccuracies. There are several possible reasons for this:
All my experiments were not carried out on one day. Therefore the room temperature may have changed as I did not keep any record of it. Heat would have sped up the movement of the molecules which might increase the rate of reaction but it would also increase the amount of resistance in the experiment. Because current = voltage / resistance the higher the resistance the lower the current and therefore the lower the amount of coulombs present in the circuit. To correct this in any further experiments I could use a water bath so that temperature is kept constant.
Also during my experiment I noticed that the current as measured by my ammeter would decrease. This made it slightly difficult and unfair, as the amount of coulombs would change if the current change. I also noticed that after a while my beaker which I used for my experiment heated up whilst I carried out my experiments. This would have increased resistance and therefore lowered the current. This may be why my results were slightly lower than I would have expected. To compensate for this in any further work I would probably use a water bath to keep temperatures constant.
At the end of each lesson I noticed that there were bits of copper floating around in my beaker. These must have somehow dropped off during the course of the experiment. These bits are another factor as to why my results were lower than anticipated.
My cathode and anode curved in towards each other during my experiment. Although the overall trend in the current was decreasing the anode and the cathode, being slightly closer, would increase the current, as there would be less distance for the electrons to travel. They curved in towards because they have opposite charges. As they were close together in the beaker they exerted an attraction on each other that would have resulted in them getting closer together. To stop this from happening in further experiments I would keep my electrodes further apart so that they would not exert any attraction on each other.
I do not believe that I used an accurate enough scale in my experiment. Even though the electronic scale was accurate to two decimal places to get a better picture of how close my results were to my quantitative prediction I would need a scale that is perhaps accurate to 4 or 5 decimal places although this would leave no room for error.
I could make these changes and then see if I could not get results, which would, when plotted on a graph, match up, completely with my quantitative predictions.
To further this investigation I could investigate a different variable although I still believe charge is the best variable to investigate. I could also take a wider range of results as a range of 100-400 coulombs is effectively insignificant compared to a faraday (96500 coulombs). I would also be interested to see if 193000 coulombs (2 faradays) does actually produce 64g of copper. For this though I would obviously need to scale up my investigation to accommodate for the huge increase in mass change. To back up my current results I could get more evidence from different amounts of coulombs. More evidence would further back up my conclusion. I could also see how different elements with different types of ions would be affected by changing amounts of charge. I could use an electrode which is not active and I could see how my results would differ.
