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Investigating the factors that affect the amount of copper deposited during the electrolysis of copper (II) sulphate solution.

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Introduction

Investigating the factors that affect the amount of copper deposited during the electrolysis of copper (II) sulphate solution Introduction: This investigation will examine one of the factors that affect the amount off copper deposited during the electrolysis of copper (II) sulphate solution. Electrolysis is the decomposition of an ionic solution by electricity. The electricity splits ionic chemical compounds into separate components, for example, copper chloride can be split into Cu metal and Cl gas. Electrolysis is used to extract reactive metals from solutions of their ores. It is an extremely powerful way of splitting a compound using electricity, but it is very expensive and is only used for metals higher up in the reactivity series. These metals only way of extraction is by electrolysis as they are more reactive than carbon. These metals are Potassium, Calcium, Magnesium, Aluminium and sodium. This is a simple diagram of an electrolysis cell: When the battery is switched on, it creates and electrical current. The positive and negative terminals create a positive and a negative electrode. The current flows through the electrolyte (the molten ore). The ore has to be melted in order for electricity pass through it. As electricity flows through the molten ore, the compound splits apart and forms ions. These ions are free to move because the ore is a liquid. The ions carry a charge and the negative ions (the non-metal ions) are attracted to the positive anode and the positive ions (the metal ions) are attracted to the negative anode. The metal is always deposited at the cathode and a gas is given off at the anode. If the metal is reactive, it will produce H2 at the cathode (where it is collected) as it reacts with the water present. We had done a previous experiment investigating the electrolysis of copper (II) sulphate as described below. We first got two pieces of copper and labelled them 'C' and 'A' (cathode and anode). ...read more.

Middle

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. ...read more.

Conclusion

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: Current Average gain in mass at cathode/g Average loss in mass at anode/g Average percentage error 0.1 0.0095 0.0115 17.39 0.2 0.017 0.017 0 0.3 0.027 0.0225 20 0.4 0.0315 0.032 1.56 0.5 0.0465 0.0415 12 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. ...read more.

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