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Let us investigate the electrolysis of molten sodium chloride.

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Let us investigate the electrolysis of molten sodium chloride. The electrolyte contains charged particles, such as Na+ and Cl-, which move towards the electrodes during electrolysis. The charged particles are called ions. At the anode: Cl- ions lose an electron to the positive anode and form a neutral chlorine atom. Cl- -> Cl + e- The chlorine atoms then join up in pairs to form molecules of chlorine gas, Cl2. These molecules of chlorine gas now bubble off at the anode. Cl + Cl -> Cl2 The complete half-reaction is: 2Cl- -> Cl2 + 2e- At the cathode: Na+ ions near the cathode combine with electrons on the cathode forming neutral sodium atoms. Na+ + e- -> Na Electron/ion flow: Note the movement of ions in the electrolyte and the movement of electrons in the external circuit. The electric current is being carried through the molten sodium chloride by ions. Na+ ions remove electrons from the cathode and Cl- ions give up electrons at the anode. Investigate the loss in mass of the anode when a copper sulphate solution is electrolysed These are the input variables that I believe could affect this investigation: Surface area of electrode Temperature of electrode Time Current Molarity of copper sulphate solution I have chosen to investigate the variable of current, and its effect on the loss of mass at the anode. Therefore, to maintain a fair test I will have to control the rest of the variables above. ...read more.


Therefore, the mass of the anode will keep changing and the anode will keep changing electrode. Therefore, to avoid this confusion, I will simply use DC. ****It is important to keep the molarity of the copper sulphate solution the same. This is because the greater the molarity of the solution the more copper ions there are available to take the electrons from the cathode. The faster that the electrons are taken from the cathode, the faster the copper at the anode decomposes to give the same number of electrons. This therefore means that the molarity of the copper sulphate solution affects the loss in mass at the anode. To avoid this bias I will keep the molarity of the solution at one each time. Finally, my experiment set up ensures that the electrodes will not touch at any point on the experiment. This will bias the results because there will be no contact with the solution and therefore the electrodes will not gain or lose mass because they will not come into contact with the copper ions in the solution. They will always be clipped at opposite parts of the beaker and if they do come into contact, I will re-do the result. It is also important to maintain an accurate test because this will enable me to see a clearer trend. To ensure that I get accurate results I will measure time to the nearest millisecond. ...read more.


Also, it would have been more sensible if I had not left the electrodes in the Copper Sulphate solution over half-term. It was found that 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. Every copper ion reduced at the cathode, one copper atom must be oxidized at the anode to keep the electrical charge balanced. This made it possible for the loss of copper at the anode to be used as a part of the results. The loss at anode can be more accurately measured since one cannot alter the mass of the anode significantly if it had been dried properly (since the copper does not drop off). The concentration of electrolyte did not have a significant effect on the gain of copper at the cathode. However, when the concentration dropped to less than 0.1 M the conductivity of the solution was changed by such a huge amount that the PSU in use was no longer able to pump through the desired current. This alters the current, thus declaring the results invalid since current was expected to affect, and did affect the mass of copper deposited at cathode. The drop in the mass deposited in very low concentrations was due to the drop in current rather than the concentration itself. If a more powerful PSU was available (say, 100V max), and a more sensitive variable resistor was used, even at low concentrations, it should not have had an effect on the mass of copper deposited at the cathode. ...read more.

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