The investigation I am conducting is to find out how much copper is attracted to the cathode from passing electricity through copper sulphate solution through an experiment called Electrolysis. When the power pack is turned on electrons pass through the circuit and through the electrodes at the cathode, because the copper sulphate solution becomes ionized; that is, their molecules become dissociated into positively and negatively charged components, which have the property of conducting an electric current, the positively ions in the solution move toward the negative electrode and the negative ions towards the positive. When reaching the electrodes, the ions may gain or lose electrons turning them into neutral atoms or molecules. The positive copper ions and Hydrogen ions will be attracted toward the negative electrode (cathode), the copper ions discharge to form metallic copper whereas the Hydrogen ions will stay in the solution as the copper ions release their electrons more readily than the Hydrogen ions. The Sulphate and Hydroxyl ions are attracted to the positive electrode (anode) but as the Hydroxyl ions release their electrons more readily than the Sulphate ions the Hydroxyl will get there first. The half-equations for the electrolysing of copper sulphate solution between two carbon electrodes are: -
             At the anode (+): 4OH (aq) 2H20(l) + 02(g) + 4e
             At the cathode (-): CU (aq) + 2e CU(s)
             In class prior to this experiment we have been studying Electrolysis on Copper Chloride solution, the results showed that the higher the voltage applied to the circuit the more copper had formed at the cathode. In the investigation we are looking at the cathode for our results, the Copper Chloride solution will have the same outcome as the Copper sulphate solution because the copper will always form on the cathode ahead of Hydrogen in both cases.
             In the investigation on Electrolysis I aim to find out if increasing the amoun...