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What Effects the Reaction in the Electrolysis of Copper Sulphate.

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Introduction

What Effects the Reaction in the Electrolysis of Copper Sulphate I will investigate what affects the reaction in the electrolysis of copper sulphate. In order to investigate what affects the reaction I will electrolyse a copper sulphate solution using two copper electrodes. As these electrodes participate in the reaction they are called active electrodes. I will set up my experiment as shown in the diagram: I will pour 75cm3 of copper sulphate solution into a beaker. This amount is sufficient to fully cover the electrodes. I will weigh both the electrodes before the experiment using an electronic scale. I will connect the electrodes to a power source and an ammeter as shown in the diagram above. I will run the experiment for the duration which I have explained later in this plan. Then when the experiment has ended I will turn off the power and take out the electrodes. I will wash off any excess copper sulphate which could crystallise when heated. I will then, using an electric heater, evaporate off any water which is now on the electrodes. I will measure the weight change in the negative electrode (the cathode) and the positive electrode (the anode) and see how this changes as my chosen variable changes. I will give reasons for why I will use each piece of apparatus: Piece of apparatus Reason for use Beaker A beaker is sufficiently sized for the experiment and will not required me to use excessive amounts of copper sulphate to fill it Copper electrodes 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 Ammeter I need this to measure the current in the experiment and after knowing the current I can work out how long I need to continue my experiment for. (This is explained later) Power source This is needed to provide the experiment with electricity so that the ions can be made and transferred Electronic ...read more.

Middle

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. ...read more.

Conclusion

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. Gareth Knott Science Coursework 5/8/2007 Page 1 ...read more.

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