Chemguide: Support for CIE A level Chemistry ``` ``` Learning outcome 6.2(d) This statement is about finding a value for the Avogadro constant by doing simple electrolysis reactions. Before you go on, you should find and read the statement in your copy of the syllabus. ``` ``` What follows assumes that you have read and understood the page about learning outcome 6.2(c). ``` ``` The calculations You know that 1 faraday is a mole of electrons. You also know that 1 faraday is 96500 coulombs. The quantity of charge carried by a single electron is known from physics experiments beyond the scope of this A level chemistry course. The Data Booklet tells you that this charge is 1.60 x 10-19 coulombs. If you find out how many of these charges are needed to make up the number of coulombs in 1 faraday, then you will know how many electrons there in a mole of electrons. That is the Avogadro constant. So how many times does 1.60 x 10-19 go into 96500? Use your calculator to find out. If you have entered everything properly, you should get an answer of 6.03 x 1023. Given the rounded-off nature of the data we are working with, that is what you would expect. ``` ``` But what if you didn't know how many coulombs there were in 1 faraday? What this topic is essentially about is how you would find that out. Let's think about the electrolysis of silver nitrate solution using silver electrodes. The cathode reaction is: Suppose you electrolysed silver nitrate solution using a current of 0.100 amp for exactly 30 minutes. You then found that the cathode had gained 0.201 g. How many coulombs were involved in the experiment? Number of coulombs = current in amps x time in seconds Number of coulombs = 0.100 x 30 x 60 = 180 Now look at the equation for the reaction at the cathode: For every 1 mol of electrons involved, you would get 1 mol of silver. We had 0.201 g. How many moles of silver is that? Ar of Ag = 108. Number of moles of Ag = 0.201/108 = 0.00186 That means that there must have been 0.00186 moles of electrons in that 180 coulombs we have already calculated. We can find out how many coulombs there are in 1 mole of electrons by dividing 180 by 0.00186. Note:  If your maths isn't very good, you might not like this! What I would suggest you do in cases like this is to decide what you would do with much easier numbers, and then do the same thing with the more complicated ones. Suppose 2 moles of electrons were 180 coulombs. How many coulombs would there have been in 1 mole? You would obviously just divide by 2. Do the same thing here with the more awkward 0.00186. If you work this out, it comes to 96800 coulombs to 3 significant figures. That is to within about 0.3% of the value we normally use. Having found that number experimentally, we could then divide by the charge on the electron to get a value for the Avogadro constant, as above. The answer comes to 6.05 x 1023. The accepted value is 6.02 x 1023. If you actually did this experiment, and got a value that close, you would be really, really pleased with it! ``` ``` You will find that the CIE Chemistry Coursebook use the electrolysis of copper(II) sulphate solution with copper electrodes as an example for this statement. That is very slightly more complicated because the charge on a copper(II) ion is 2+, and so you would have 2 moles of electrons involved rather than the 1 in this case. It would be worthwhile looking at their example if possible. ``` ``` The experiment There are two possible measurements you could make. You could measure the mass gained by the cathode, or you could measure the mass lost by the anode. At the anode, silver goes into solution. We discussed this in statement 6.2(b). For every electron that travels around the external circuit, one atom of silver is added to the cathode, and one is removed from the anode. That means that (provided the silver anode is pure - see below) the mass gained by the cathode is equal to the mass lost by the anode. So it doesn't matter which you measure. It seems more obvious to measure the mass gained by the cathode, but there are practical problems in doing this. Unless the current density is very low, the silver doesn't plate on to the cathode very firmly. There is a danger of it falling off, or being rubbed off when you clean and weigh the electrode. Note:  Current density is just a measure of the current relative to the surface area of the electrodes. Provided the silver is pure, it is safer to find the mass lost from the anode. Obviously, if it isn't pure, the impurities will also be lost from the anode, either as ions going into solution or as an anode sludge. Note:  If you read the piece about the purification of copper previously, you will have come across this as a sludge which forms under the ande containing various impurities. If you didn't, it doesn't really matter. If there are impurities being lost as well as silver, then your weighings are fairly meaningless! So, in outline: You need clean silver electrodes and some silver nitrate solution in a beaker. You need to weigh the silver anode. Set the electrolysis up, and carefully control your measured current for a measured amount of time. At the end of that time, switch off the current, and remove the anode from the silver nitrate solution. Wash it with pure water to get rid of any silver nitrate solution, and then with propanone to get rid of the water. The excess propanone will evaporate quickly in the air. Finally, reweigh the electrode, and do the sum. ``` ``` Go to the Section 6 Menu . . . To return to the list of learning outcomes in Section 6 Go to the CIE Main Menu . . . To return to the list of all the CIE sections Go to Chemguide Main Menu . . . This will take you to the main part of Chemguide. © Jim Clark 2011 (last modified April 2014)