How does the mass of copper plated in the electrolysis of copper sulphate solution vary with the current used?

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Chemistry Sc1                                                 Justin Barauskas No 9

How does the mass of copper plated in the electrolysis of copper sulphate solution vary with the current used?

Aim

The aim of this experiment is to find out relationship between the mass of copper deposited on the cathode and the (variable) current used.

Planning

First of all the factor I am going to vary in this experiment is the current. When the current is varied for example if it is big, the mass of copper deposited will be larger than with a small current.  This is because when there is more current there are more electrons flowing therefore more copper deposited. So really the reason for more copper deposited is that, because there are more electrons flowing at the same time the anode loses copper ions quicker and the cathode gains copper ions quicker and deposits them as copper atoms. Therefore the ionic reactions happen faster leading to more copper deposited in the constant time. This is what happens at the cathode:- Cu2+ + 2e- = Cu. This is what happens at the anode Cu = Cu2+ + 2e-.

Prediction

I predict that in this experiment, when the current is high the amount of copper deposited will be higher that with a small current. This is because the higher the current the higher the flow of electrons which would lead to higher deposition of copper. Therefore also the current should be proportional to charge if the time is kept constant. This is because to work out the charge you have to know the time because Charge = Current multiplied by Time.

 Also I have done the same preliminary experiment and from which I can found out that the mass of copper lost at the anode is equal to the mass gained on the cathode. This is because for every copper ion that is lost from the anode a copper atom is gained at the cathode. I also found out that the bigger the current used the more mass deposited on the cathode. Furthermore, I found out that that the amount of substance formed is directly proportional to the size of the current and for how long it flows.

        From this experiment I got a good idea of what the values I am going to use and how I should control my other factors and how important they are. Also I am going to use my preliminary experiment to compare it to the real one and the repeat. I am going use the same values as I used in the preliminary and the same type of method in doing the experiment so that when I compare it is fair to compare.

Another point is that how much copper will be deposited can be worked out if you know what current you are using and for how long you are using it. The amount of substance formed is directly proportional to the size of the current and for how long it flows. So from that I can work out the amount of substance I am going to get and then measure it to see if it is true. So now I am going to predict the results that I should be getting. This is the calculation I am going to use:

 

 Charge (coulombs) =time (sec) x current (amps)

                       = 600   x 0.25 = 150 coulombs

95000 coulombs = 1/96500 moles of electrons

150 coulombs = 1/96500 x 150 = 0.00155 moles of electrons

2 moles of electrons give 1 mole of copper atoms, 2 mol = 64g, 1 mol = 32g (0.00155x 32) = 0.0496

From this one point I am going to plot a prediction graph because we already know its going to be proportional so one point is going to be twice the other one on my scales. The graph is going to be Current against the mass lost at the anode and at the cathode. This is going to show me the relationship of current and mass deposited. I expect the current to be directly proportional to mass gained at cathode and lost at the anode.

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The Values

For the time allowance I am going to use 10 min. I think that is enough time for electrons be turning copper ions from the anode into copper atoms and letting them cling onto copper at the cathode. This would lead to enough copper deposition on the cathode so that an accurate measurement could be taken down.

 I am going to try and use 5 different current values, 0.25, 0.50, 0.75, 1.00, 1.25 amps. I am going to be using these current values because I think these are the most suitable because. First of all, ...

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