so today we're going to be doing an experiment into electrolysis electrolysis is used to break down compounds using electricity now today we're going to look at copper chloride and sodium sulfate the first thing you need to do is to fill your beaker about 50 cm cubed using the chemical you want so we're going to start with copper chloride now once that's in you need to make sure that the electrodes are submerged inside the liquid but not too close to the crocodile clips as you can see there's plenty of room between the crocodile clips and the top of the liquid next thing we're going to do is connect up the power pack now we want to go with about 4 volts for this practical for each sake we're going to use the red wire for our positive terminal that will be our anode and the black wire into the negative terminal that will be our cathode now when we turn this on we're expecting a reaction to take place we're expecting to see some chemicals being formed or elements being formed at the different electrodes so I'm going to turn this on and leave it to run when you leave it to run you need to leave it to go for about 5 minutes just to make sure you've got enough of the element to test it properly and what we're going to do is look over that time period what happens in the electrolyte our electrolyte here is the copper chloride the copper chloride contains both copper ions which have a two plus charge and chloride ions which are Cl minus now what we would expect to see is that at the positive electrode we're going to have a gas being formed and we can see that with bubbles now in order to test for the chlorine ions we're going to hold a piece of blue LMUS paper just above the anode and what we should see where the chloride ions have formed chlorine the lmus paper is bleached the bleached LMUS paper shows a positive test for chlorine at the cathode i'll turn this off so we can see this we should see that copper started to form around the bottom of it so where it was submerged in the copper chloride copper metal has formed from the copper ions at the anode we will get chloride ions being attracted to that side chlorines have a one minus charge and they will give up an electron chlorine travels around diatomically in that case we will need two chloride ions being attracted to the anode giving off two electrons to become a stable diatomic molecule at the cathode copper 2 plus ions are drawn towards the negative charge here they gain two electrons this in turn forms copper metal as a solid the diatomic chlorine forms as a gas at the anode where electrons are lost the chloride ions are oxidized to chlorine gas at the cathode where they gain electrons we have a reduction copper ions are reduced to copper metal these are called half equations they show what happens at each electrode we're now going to do a test with sodium sulfate now sodium sulfate contains a metal that is more reactive than hydrogen sodium you can see is far higher in the reactivity series than hydrogen is once again making sure that the electrodes aren't touching I'm going to put the electrodes into the sodium sulfate making sure the bottom of the electrode is covered again I'm going to turn the circuit on so the electrons can flow and that should allow us to see a reaction starting now this time we've got bubbles forming at both electrodes at the negative electrode we've got hydrogen forming now the hydrogen formed in this experiment won't be enough to actually test it via our usual pop test however this time we've got oxygen formed where something is less reactive than the chlorine or the halide ions we get oxygen formed instead of the sulfate ions which will stay in solution they'll react and bond with the sodium ions and stay in solution at the anode we said we had oxygen being formed now these oxygen molecules are formed from the hydroxide ions in water now once again because it's at the anode we have an oxidation reaction taking place where we have a loss of electrons we know the product is oxygen as a gas and we also get electrons given off however the question is what has happened to this hydrogen within the hydroxide ion and with oxidation reactions involving hydroxide ions we get water formed as an extra product and that goes back into the solution within the electrolysis cell we need to balance this at present we don't have enough oxygens from the hydroxide ions or hydrogens's to fulfill the equation we have so we have four hydroxide ions forming our oxygen and our water now because we've got four negative charges on this side we need to have four negative charges on this side we now can balance the hydrogens so we should have four hydrogen's from the hydroxideion going to form two lots of two hydrogen form the water at the cathode we said we had hydrogen being formed h+ ions from the water gain electrons to form hydrogen gas now once again hydrogen is a diatomic molecule so we need to balance this with equal numbers of hydrogen's but also equal numbers of charges so with half equations it's not important only to make sure that all the elements balance but also that all of the charges balance within the equation