Chemguide: Support for CIE A level Chemistry ``` ``` Learning outcome 6.2(c) This statement is about simple electrolysis calculations. Before you go on, you should find and read the statement in your copy of the syllabus. ``` ``` Electrolysis calculations are no more difficult than any other calculation from an equation. In fact, you may well have done them as a part of whatever course you did before you started doing A level. We will just look at four examples. ``` ``` 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. (Use the Data Booklet to find the value of the Faraday constant and the relative atomic mass of silver.) You will find a copy of the Data Booklet towards the end of the syllabus. The Data Booklet gives the value F = 9.65 x 104 C mol-1 (or 96500 C mol-1 if you prefer). Ar of 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: 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 your maths is really bad, so that you aren't happy about simple proportion sums, then think of it like this: If 96500 coulombs give 108 g, then 1 coulomb would give 108 divided by 96500 g. 60 coulombs would produce 60 times this amount. It doesn't matter in the least how you work this out - all that matters for your chemistry is that you get the answer right! ``` ``` 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 sulphuric acid if you use a current of 1.0 amp for 15 minutes. (Use the Data Booklet to find the value of the Faraday constant and the molar volume of a gas at room temperature and pressure.) The Data Booklet gives the value F = 9.65 x 104 C mol-1 (or 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: 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. ``` ``` 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) sulphate solution if you use a current of 0.200 amp. (Use the Data Booklet to find the value of the Faraday constant and relative atomic mass of copper.) The Data Booklet gives the value F = 9.65 x 104 C mol-1 (or 96500 C mol-1). Ar of 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. Write down the important bits of this in words: 2 mol of electrons give 1 mol of copper, Cu. Now put the numbers in. 1 mol of electrons is 1 faraday. 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 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, included because the syllabus mentions the electrolysis of sodium sulphate solution as well as sulphuric acid. Calculate the volume of oxygen produced (measured at room temperature and pressure - rtp) during the electrolysis of sodium sulphate solution if you use a current of 0.50 amp for 30 minutes. (Use the Data Booklet to find the value of the Faraday constant and the molar volume of a gas at room temperature and pressure.) The Data Booklet gives the value F = 9.65 x 104 C mol-1 (or 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: The alternative way releases oxygen from hydroxide ions from the ionisation of the water: 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 H2, 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 Don't quote your answer beyond 2 decimal places. The current and the molar volume are only quoted to that degree of accuracy. ``` ``` 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)