Chemguide: Core Chemistry 14 - 16 ``` ``` Electrolysis calculations ``` ``` This page deals with simple calculations from equations involving electrolysis. You will need to be familiar with the routine calculations from equations covered previously in this section. ``` ``` The Faraday constant The Faraday constant is the single most important bit of information in electrolysis calculations. Make sure you really understand the next bit. Coulombs The coulomb is a measure of the quantity of electricity. If a current of 1 amp flows for 1 second, then 1 coulomb of electricity has passed. That means that you can work out how much electricity has passed in a given time by multiplying the current in amps by the time in seconds. Number of coulombs = current in amps x time in seconds If you are given a time in minutes or hours or days, then you must convert that into seconds before you do anything else. For example, if a current of 2 amps flows for an hour, then: Number of coulombs = 2 x 60 x 60 = 7200 (60 minutes in each hour; 60 seconds in each minute.) That's easy! ``` ``` The Faraday Electricity is a flow of electrons. For calculation purposes, we need to know how to relate the number of moles of electrons which flow to the measured quantity of electricity. Suppose you have a simple electrode equation for the formation of copper on the cathode during the electrolysis of copper((I) sulfate solution. Cu2+(aq) + 2e-     Cu(s) You can read that as 1 mole of copper(II) ions picks up 2 moles of electrons to form 1 mole of copper atoms. It turns out that 1 mole of electrons flowing around a circuit is equivalent to 96500 coulombs. This is known as the Faraday constant, F. F = 96500 C mol-1 (read as coulombs per mole) So 96500 coulombs is called 1 faraday. This has a small "f" when it is used as a unit. Whenever you have an equation in which you have 1 mole of electrons, that is represented in an electrical circuit by 1 faraday of electricity - in other words, by 96500 coulombs. ``` ``` Using the Faraday constant in calculations Example 1 Calculate the mass of silver deposited at the cathode during the electrolysis of silver nitrate solution if you use a current of 0.10 amps for 10 minutes. (F = 96500 C mol-1; RAM: Ag = 108) The first thing to do is to work out how many coulombs of electricity flowed during the electrolysis. Number of coulombs = current in amps x time in seconds Number of coulombs = 0.10 x 10 x 60 = 60 Now look at the equation for the reaction at the cathode: Ag+(aq) + e-     Ag(s) Just as with any other calculation from an equation, write down the essential bits in words: 1 mol of electrons gives 1 mol of silver, Ag. Now put the numbers in. 1 mol of electrons is 1 faraday. 96500 coulombs give 108 g of silver. So, if 96500 coulombs give 108 g of silver, all you have to do is to work out what mass of silver would be produced by 60 coulombs. Mass of silver = 60/96500 x 108 g = 0.067 g Note:  If you aren't happy about the last step go via 1 coulomb: If 96500 coulombs give 108 g, then 1 coulomb would give 108 divided by 96500 g. 60 coulombs would produce 60 times this amount. Example 2 This example shows you how to do the calculation if the product you are interested in is a gas. Calculate the volume of hydrogen produced (measured at room temperature and pressure - rtp) during the electrolysis of dilute sulfuric acid if you use a current of 1.0 amp for 15 minutes. F = 96500 C mol-1. The molar volume of a gas at rtp = 24 dm3 mol-1. Start by working out how many coulombs of electricity flowed during the electrolysis. Number of coulombs = current in amps x time in seconds Number of coulombs = 1.0 x 15 x 60 = 900 Now look at the equation for the reaction at the cathode: 2H+(aq) + 2e-     H2(g) Write down the essential bits in words: 2 mol of electrons give 1 mol of hydrogen, H2. Now put the numbers in. Two moles of electrons is 2 faradays. 2 x 96500 coulombs give 24 dm3 H2 at rtp. So, if 2 x 96500 coulombs give 24 dm3 H2, work out what volume of hydrogen would be produced by 900 coulombs. Volume of hydrogen = 900/(2 x 96500) x 24 dm3 = 0.11 dm3 Don't quote your answer beyond 2 decimal places. The current and the molar volume are only quoted to that degree of accuracy. Note:  If you can't follow the last bit of the calculation: If 2 x 96500 coulombs give 24 dm3 H2, then 1 coulomb would give 24 divided by 2 x 96500 dm3. 900 coulombs would produce 900 times this amount. In other words, you are working out 24/(2 x 96500) and then multiplying by 900. It is a different order from the way it is shown in the simple proportion sum shown above, but the answer is still exactly the same. In addition, if you feel happier working out the value of 2 x 96500 before you do anything else, that's what you should do. As long as you get the answer right, nobody is interested in the exact way you handle the sums. Example 3 This example shows you what to do if the question is reversed. How long would it take to deposit 0.635 g of copper at the cathode during the electrolysis of copper(II) sulfate solution if you use a current of 0.200 amp. F = 96500 C mol-1; RAM: Cu = 63.5. This time you can't start by working out the number of coulombs, because you don't know the time. As with any other calculation, just start from what you know most about. In this case, that's the copper, so start with the electrode equation. Cu2+(aq) + 2e-     Cu(s) 2 mol of electrons give 1 mol of copper, Cu. 2 x 96500 coulombs give 63.5 g of copper. You need to work out how many coulombs give 0.635 g of copper. Number of coulombs = 0.635/ 63.5 x 2 x 96500 = 1930 Note:  And again, if you still don't like simple proportion sums: If 2 x 96500 coulombs give 63.5g of copper, then you would get 1g of copper if you divided the 2 x 96500 coulombs by 63.5. 0.635 g would be produced by multiplying this by 0.635. Now what? You know how many coulombs you need, and you know what the current was in amps. You have got all the information you need to work out the time. Number of coulombs = current in amps x time in seconds 1930 = 0.200 x t t = 1930/0.200 = 9650 seconds. Don't waste time trying to convert that into minutes or hours (unless the exam question specifically asks you to). ``` ``` Example 4 Another gas example: Calculate the volume of oxygen produced (measured at room temperature and pressure - rtp) during the electrolysis of sodium sulfate solution if you use a current of 0.50 amp for 30 minutes. F = 96500 C mol-1. The molar volume of a gas at rtp = 24 dm3 mol-1. Start by working out how many coulombs of electricity flowed during the electrolysis. Number of coulombs = current in amps x time in seconds Number of coulombs = 0.50 x 30 x 60 = 900 Now we need to look at the equation for the reaction at the anode. Unfortunately, there are two ways of looking at this, and you may come across either of them. The first one releases oxygen from water molecules: 2H2O(l)     O2(g) + 4H+(aq) + 4e- The alternative way releases oxygen from hydroxide ions from the ionisation of the water: 4OH-(aq)     2H2O(l) + O2(g) + 4e- Write down the essential bits in words. Both ways of looking at it say the same thing: Releasing 1 mol of oxygen, O2, involves 4 mol of electrons. Now put the numbers in. Four moles of electrons is 4 faradays. 4 x 96500 coulombs give 24 dm3 O2 at rtp. So, if 4 x 96500 coulombs give 24 dm3 O2, work out what volume of oxygen would be produced by 900 coulombs. Volume of oxygen = 900/(4 x 96500) x 24 dm3 = 0.056 dm3 Or work out what volume you would get from 1 coulomb and then multiple by 900. Don't quote your answer beyond 2 decimal places. The current and the molar volume are only quoted to that degree of accuracy. ``` ``` Where would you like to go now? To the calculations menu . . . To the Chemistry 14-16 menu . . . To Chemguide Main Menu . . . ``` ``` © Jim Clark 2021