 we have seen the setup of an electrolytic cell before and one of the applications of electrolysis is in purification of metals. So let's say if one of the electrodes is made up of copper but it also has some impurities and after electrolysis copper gets deposited on the other electrode and the impurities fall into the solution and you get pure copper. But the question is when this sort of setup is used? How much of copper will be deposited on the electrode and in 1830s Michael Faraday had a similar question and he figured out a way to calculate how much of the deposition takes place on the electrode which is called Faraday's first law of electrolysis. So let's see how we can describe this relation. Now the first thing to note is that the weight deposited will be proportional to the charge through the electrode and instead of thinking of this charge if you look at this circuit you can also think in terms of the current through this circuit. So the weight or the amount of deposition will be proportional to the current i times the time t and we are writing this in this form because we know that current is nothing but the charge per unit time. So the weight or the amount of the deposition will be proportional to the current times the time for which the current passes through the electrode. So we can get rid of this proportionality sign and write this as w is equal to some z times the current times the time and this z is called the electrochemical equivalent and if you look at the units of this z this z is actually equal to w by it and we know from before that instead of writing it in terms of current we can write this in terms of charge as w by q. So if you look at the units of z it will be in the form of kg per coulombs so mass per charge. So one way to think about this z is that it is telling us how much of the material is deposited when some q charge passes through it. So let's say that the charge that is passed is one coulomb so then we can think of this electrochemical equivalent or z as the amount of substance deposited when a charge of one coulomb is passing through this electrode. So if we know the value of z and we know the current i passing through this circuit and we know that for how much time we allow the current to pass through this circuit we can use these three values to find out the amount of substance that is deposited. But the thing is this z is a value that is calculated by experiments. So the question is can we rewrite this z in terms of some quantities that we already know for a given electrode material and if we do that it makes our life easier because we can then just plug in those values and if we know the value of current and if we know the value of the time for which the current is passed we can easily calculate the amount of substance deposited. So let's see if there is a way in which we can rewrite the z. Before we look at the electrochemical equivalent let's look at the electrode where the deposition is happening. Let's take an example let's say we have silver ions that gain electrons and silver is deposited. So we can think of this as one mole of silver ions reacting with one mole of electrons to give one mole of silver and we know that the molar mass of silver is approximately 108 grams per mole. So that means for every mole of electrons one mole of silver is deposited or we can say that 108 grams of silver is deposited for every one mole of electrons. Now what if the metal that is deposited is copper. So in case of copper we have this copper ion which gains two electrons to get deposited at the electrode. Now because there is a factor of two here for every two moles of electrons we get one mole of copper and again because the molar mass of copper is 64 grams per mole I can write this as two moles of electrons give me 64 grams of copper that is deposited at the electrode and let's take one last example which is of aluminum. So we have this aluminum ion which is gaining three electrons and getting deposited as aluminum at the electrodes and as you can see in this case we now have three moles of electrons which is giving you one mole of aluminum that is deposited and since we know that the molar mass of aluminum is around 27 grams per mole we can say that for three moles of electrons we get one mole or 27 grams of aluminum that is deposited. So in all of these cases if I want to think how much of the deposition will happen for one mole of electrons I can calculate that by dividing the weight deposited by the number of moles of electrons. So in this case the weight deposited per mole will be 108 divided by one. Now in this case since 64 grams are deposited for two moles of electrons the weight deposited per mole of electrons will be 64 divided by 2 and you can already see the trend in this case the weight deposited per mole of electrons will be 27 divided by 3. So looking at this pattern we can generalize this to write that the weight deposited in grams for one mole of electrons will be equal to the molar mass which I am going to denote by this m in grams per mole divided by n where n is the number of moles of electrons that are taking part in this reaction. Now let's look at this one mole of electrons. So let's say I want to find out what is going to be the charge of one mole of electrons. So I can write the charge of one mole of electrons in coulombs as equal to the charge of one electron times Avogadro's number. Since we know these values let's plug them in. So the charge of an electron is 1.602 into 10 to the power minus 19 coulombs times Avogadro's number which is 6.022 times 10 to the power 23. So if you solve this this comes out to be around 96472 coulombs and we can approximate this a bit further and we can write this as 96500 coulombs. So we have two pieces of information now we have the weight deposited in grams for one mole of electrons and we know the charge of one mole of electrons and now if we go back to the electrochemical equivalent we know that it was of the form of w by q. So let's use both of these pieces of information and try to come up with a simpler way of expressing z. So we saw before how the weight of substance that is deposited at the electrode is equal to z that is the electrochemical equivalent times the current times the time for which the current flows and we also saw that the electrochemical equivalent will be of the form of weight per charge or kilograms or grams per coulomb. So we can simplify this equation by plugging in the value of z from here. So we can write this weight of substance deposited is equal to this z will become w by q or m divided by n divided by q that is this 96500. So instead of dividing it here I can write it as multiplied by 1 over 96500 times current times t. So this will be equal to the weight of the substance deposited and because to plug in z we use grams here this will be in grams. So this right here is the relationship that represents Faraday's first law using which we can calculate the weight of the substance deposited in grams if we know the molar mass the number of moles of electrons taking part in the reaction the current through the electrodes and the time for which the current is passed and usually this current is given in amps or amperes and this time is given in seconds. So if we know all of these values we can use them to calculate the weight of the substance deposited in grams and this relationship is the quantitative form of Faraday's first law. And one more thing sometimes in problems you will see situations where instead of a deposition happening at the electrode gases will be produced. So in that case can you still use this relationship to find out how much of the gas is produced? Well yes but you have to make some simple rearrangements here. So if we just take this m to the left hand side we have w by m on the left side and everything else that was known on the right side. So if we calculate this value this w by m that is the weight divided by molar mass is also equal to the number of moles. So if we know these values what we are calculating is the number of moles of gas that are produced at the electrode and we also know that one mole of gas occupies a volume of 22.4 litres at standard temperature and pressure conditions. So for a mu moles of gas that we calculated using these numbers let's say that it will occupy a volume of v litres. So we know the value of mu from this calculation so to find v we can simply just cross multiply and we can get the volume of gas that is released at one of the electrodes using the same expression that we saw before.