 Hello everyone, welcome back to the series of lecture on actinide chemistry and I would like to have some recap of whatever we have learned in the previous lectures. So, in the previous lecture we have started with the actinide hydrolysis and this is the journal equation that I have given for the actinide hydrolysis and we have talked about this tendency of hydrolysis is dependent on the oxidation state and we have seen that oxidation state you can say plus 3, plus 4 and plus 6 as I have said that when we talk about plus 6 it is basically plus 3.3 and when we say plus 5 it is basically plus 2.2. So, we have seen that that this oxidation state which are higher if they are having hydrolysis in the range of near neutral pH whereas, for the lower that is the plus 2.2 and their hydrolysis will start at pH greater than 8. So, we have also compared that hydrolysis constant of different actinide ions compared to different oxidation state and the trend is basically the tendency is very high the tetravalent followed by the hexavalent followed by trivalent and then pentavalent which is very much according to their chart that is 4 plus 3.3 plus 3 plus and 2.2 plus. So, this is very much according to their chart is and if you just compare the first hydrolysis constant of different oxygen state I have given you for the different matter man then the thorium and uranum which is compared the thorium k value is 10 to the power minus 2.5 and for uranum this is 10 to the power minus 5.5. So, we can say that it is having tendency to get hydrolyzed compared to the uranum and when you compare uranum in amyretium again it is almost 100 times more tendency of hydrolysis of uranum because of higher charge that is 3.3 compared to 3 in the amyretium. The lowest possible is for nepunium which is having the lowest possible charge of 2 units and this is very weakly hydrolyzed and pH will vary for this kind of hydrolysis generally more than 8 or so for the hydrolysis and then we have seen that how the trend of hydrolysis vary in lengthenite series we have seen that the lengthenite series that hydrolysis will start at a lower pH when we are moving that is because of the Zerbaya ratio because of the higher anic potential of the lengtheners when we are moving from the left to right. We have also seen that the hydrolysis of the tetravalent ion they are forming some sort of polymers or we can say the polynuclear species in which you are having more than one metal ion and when you have tetravalent ion and they just precipitate into the solution which they just get hydrolyzed into the solution these are called amorphous or freshly prepared precipitates they are having mainly hydroxyl bridging but the moment you leave it for some time they are making kinds of compound which are known as oxo bridge compounds or you can say oxo bridge polymers and this kind of polymers are very difficult to dissolve and with aging their stability is going to be increased and the relative stability or you can say the inertness of these polymers were the acid. When I say acid as you have seen from this equation that suppose you just prepare a hydroxide and after the formation they are just prepared you add acid and from this equation you can get back your metal ion of interest but if you leave it for certain amount of time or maybe for the kinetics of going from here to here is very slow so if we leave it for a good amount of time they are forming these kinds of compounds and they are very difficult to dissolve and the difficulty will depend on the percentage of oxo bridging higher the oxo bridging more difficult is the polymer to dissolve or you can say more inert is the hydroxide polymer then we have also seen that how to get the composition or you can say the percentage of the hydroxide species at a given pH if the log k values are known and with that we have seen that we have taken an example also with log k value having assuming that it is the only mono hydroxy complex and we have seen that if your pH is almost one unit lower than the log k value in the positive sense because it is 10th so the 95 percent as present as free but when your pH is close to the log k value when I say close I am just assuming the positive pattern not with the negative 11.3 when you are in that range then we are assuming that for the mono hydroxy if you are having this kind of range more than 50 percent of the total metal ion is free what it means that 50 percent is hydrolyzed and we have also seen that this kind of speciation diagram and I requested you to try this diagram because the two important things that you require for this diagram is the material concentration and the log k value that has already given here and you can just try to draw this diagram using this value and I have given you example of neptunum here that I have discussed with the value of log a minus 11.3 that if you are having a pH of 11.3 then 50 percent is free and 50 percent is hydrolyzed what it means that before this pH that is pH 11.3 the system is dominated by the free metal ion and after this pH the system is dominated by the hydroxy complex we have also seen with the speciation diagram or the formation of hydroxy species for Andericium and the diagram is shown here here also you can see the tendency of formation of hydroxy complex is more for Andericium compared to the European then this is for a protonome and uranium so we have also discussed some concept that in assuming that that we have a concept of PE that is very similar to the pH or analogous to the pH you can say just as a pH is the major of availability of proton but if we are having a concept or we are having some quantity which represents the availability of electron or is the major of electron then we can just change the quantity and we can get idea about the redox behavior of the metal ion in the solution so as I have shown you that as pH is increasing the proton availability is getting lower and lower and when the pH is decreasing the proton availability is getting higher and higher in a very similar sense because you have a variable called PE and when the PE is increasing we are having lower the probability what it means that you are having a more and more oxidizing condition and when the PE is getting lower then you are having more and more electrons what it means that you are having a reducing condition so with this assumption that yes there is some variable called PE that will tell about the availability of electron here I just want to mention that the free electrons are not available we are not directly measuring any concentration or any number of free electron that are present so this is not about the free electron but just let us assume that there is some concept there is some variable that you can measure and you can correlate with the PE that will give you some indirect information about the condition whether the system is reducing or oxidizing so we are having two kind of concept one is called PE and one is called pH so as I have shown you that for the pH you can always plot pH versus the fraction of hydrolyzed species or you can say and you are getting some kind of curve in which you say the fraction is decreasing and hydrolyzed fraction is increasing in a very similar case as I have shown you that suppose you have something called PE this is that given you some reduction in which you assume that iron 3 plus is taking an electron giving you iron 2 plus and you have PE you are somehow able to measure the PE value of your solution and you are able to vary that PE of your solution what will happen when your PE value is on the higher side but I mentioned that when your PE value is on the higher side when I say on the higher side this condition is oxidizing in nature when your PE values are on the lower side the conditions are reducing in nature it is just like pH when your pH value is on the higher side means less proton availability when your pH value is on the lower side more proton availability similarly when your PE is on the lower side more electron availability and when your PE is on the higher side less electron availability so based on this very simple concept what we can say that if we can have PE so if we just try to see that how the redox potential is varying with respect to PE you can see that when the conditions are oxidizing for the PE values on the higher scale you are getting iron 3 plus and as we are moving in this line the iron 3 plus is getting decreased and the formation of iron who takes place so simply by having a variable that is similar to the pH which is giving me information about the hydrolysis I can get information about the different kind of redox species that is printed with the system so now the question arises okay we are having two sets of information then what is called a pH that is you have a pH and then you can vary the pH and get the information about the hydrolysis in PC and in a second set of experiment I can get the value of PE by varying the PE I can get information about the redox species so we need two plots right but it is not true generally these two plots that is pH and EH they combine with each other to give only one plot and the plot looks like this this is one of the plot that I have taken from a review paper this is for an actinium and here you can find certain line certain reasons in certain lines which are vertical horizontal or with some slope we will discuss about these lines and how we can arrive on this kind of line but in short I can say that this pH or PE curve or pH PE curve these are basically the stability field diagram they show how in a very comprehensive way your pH of the system and electrons simultaneously shift the equilibrium of reaction under various conditions it will also show you that under a given set of pH and PE which species will predominant so we will discuss this diagram which is also known as Paubeck's diagram in bit detail and after those discussions I'll come back to this slide and then we'll try to understand that how we define this kind of line why the lines are straight why they are bent on where the different area what the different area mean and why we are in this line this boundary that we will try to understand and then we'll come back to this slide to understand what all these lines mean to us as I've seen that you cannot measure PE directly but electrons are not available as such into the media so you cannot measure them directly but from our basic chemistry point of view we all know certain relations certain very basic relation that is delta G is equal to minus and that is the free energy relation with the reduction point here you also know the Noist equation is equal to E naught plus 0.059 divided by N log of oxidized species upon the reduced species and from this very well known equation we'll try to derive this equation and we'll try to understand what I mean by PE and we'll also see one term that is EH but we'll mainly try to understand what do I mean by PE so these are very known equation and then we'll try to deduce this equation of the PE from the non-concept these we already know so let us assume that an equation that you have an oxidized species that is a it takes N number phylethrone to give some kind of reduced species that is B and from the general chemistry point of view we can always write there's some equivalent constant for this reaction and the equivalent constant for the reaction is simply this k into the reduced species divided by the oxidized and the electron and we just do some juggling mathematical juggling here and there and then we'll find this term here what that this term is let us see what do we mean by ph, ph is nothing but minus log of proton concentration or I can write ph is equal to log of 1 plus h plus so we are having 1 upon h plus notation kind of thing since I want to write PE which is I want to put a similar concept like ph I am having this 1 upon e so this term I am already having like 1 upon h 1 upon e I am already having but if I take a log of this equation which I have done so when you take log of this this equation what we are ending is log of this quantity this is very much similar to log of 1 prime proton which is nothing but ph so this quantity is nothing but our PE and then we have other quantities that is already there and we have taken log and we can obviously arrange them this quantity we have defined as PE now the relation is like this that your PE is nothing but 1 upon n log k plus 1 upon n log of oxidized therefore reduced species here I just want to mention that this log k and this ratio we are talking about the half cell reactions because we have started with this reaction so this is the particular half cell reaction we are talking about that it takes a electron and getting reduced to some form so now let us assume that this is a constant because k is a constant and for a given reaction the number of electron let us assume 1 or 2 so if we assume that for a given reaction this as a constant which is given by PE naught so what we can write is nothing but PE is equal to PE naught plus 1 upon n log of oxidized species upon the reduced species and this we can write for any general equation having this kind of equation where then when your oxidation is there when your oxidized species takes a electron and gives some reduced species for this kind of half cell reaction we can write this kind of equation so now the Nernst equation we know is this one e is equal to e naught plus 0.0592 divided by n log of oxidized species upon the concentration of the reduced species and a very similar thing you can write for PE also you can see PE is nothing but PE naught plus 1 upon n log of oxidized upon reduced species here the term is given that is called EH if you compare these two equations if you compare these two equations and you try to find out then you find that there is a relation between PE and EH and the relation is nothing but f upon 2.3 RT and I have also found you that what we have assumed that log e divided by n which is a constant I presume it to be equal to PE naught and this log is nothing but log k value of a reaction of a half cell reaction and this n is nothing but the number of electron that are involved in that particular half cell reaction with this understanding let us try to get this equation for different kind of system but different kind of half cell reactions so these are several reactions I have given you and you can see this is a half cell reaction you have a standard electron potentials and from there you can get the log k values I hope this is not a very tedious term because you already know that relationship of delta g with the electron potential you also know that relation of delta g with RT and k and then you can easily work around this and you can get relation between the log k and the standard reaction potential and from here you can easily derive these values so for all these couples we can derive the log k values that already evaluated with respect to their standard electron potentials these values are basically taken from the book that is the equity chemistry from one instrument and more than so now suppose you have a system with this kind of equilibrium that is going taking some electron some half cell reaction and making some species that is iron 2 plus and I would like to ask that what is the PE of this system suppose I know somehow the concentration of iron 2 and iron 3 in my system and I would like to ask what is the PE of this system the equation is very simple as this one is that P is nothing but PE0 plus 1 by n log of oxidized species upon reduced species we have to find out PE value as shown that PE0 is nothing but log k upon n and from the previous table you can see that and for this reaction this reaction we are talking mainly so n is 1 oxidized species and reduced species concentration suppose you know you just simply put this value in this equation that PE is nothing but PE0 which is coming 13 log of this concentration you get an PE value of 11.0 or from there you can compare the electron activity so you see that from this simple equation having some redox potential we have tried to find out the concentration you can see the activity of the electron in the media so this is not the direct electron that are involved in the media but there are certain redox reactions are there that are always happening in the media and because of those redox reaction there is always some some redox or oxidizing condition that are maintained into the media and we can always try to get PE corresponding to those equations if we know the concentration of those oxidized or reduced species we can do the reverse also suppose you know the PE in that case you can always calculate at least the ratio of the oxidized species to the reduced species now as we are talking about a media which is generally across media in our case the water itself has tendency to get either oxidized to O2 or reduced to H2 and here we have given the relation of oxidation of water this is basically a redox reaction but this is basically a redox reaction and we have calculated to the log k value and this PE0 which is obviously log k divided by n and since one electron is involved so it is the log k value so for this kind of reactions also what we can calculate is this kind of equilibria so when we talk about the water as a media what is the problem then when water is a media and you are having some redox couple whose potentials are on a very higher side or on a very lower side in both the cases if the potentials of the couples are so that then they can oxidize water or you can say the water itself gets oxidized because of electrons that are present in the media or from any other couple then you cannot work with the media so for water as a media we have certain limitation or rather we should say certain boundary condition under which we have to work and these boundary conditions are basically decided by two things one the oxidizing limit of the water that can be the reducing limit of the water so when we talk about the oxidizing limit what we mean that what is getting oxidized to O2 when we talk about the reducing limit we say that what is getting reduced to the hydrogen so this is the equation that is in the reduced form we have written for the oxidizing limit of this water media and we know this equation that pe is nothing but pe0 plus 1 upon n log of oxidized species upon reduced species and from this equation we can write all these components here and the pe value again we can get from the table that is 20.75 and the value of these other quantity except oxygen is assumed to be 1, proton you have to assume is like this and then we take log and this proton will be doing nothing but pH and activity of H2 is also assumed to be 1 and with this assumptions what we are getting this this is a linear equation pe is equal to 20.75 minus pH if you compare it with a linear equation that is y is equal to mx plus t in which your y is nothing but your pe scale and x is nothing but your pe scale but you are getting a negative slope right so you are getting a negative slope like this similarly when we talk about the reducing limit you can write this equilibria and from this equilibria if we had another equilibria that is your hydroxyane combining with the proton to give you water molecule and when you club you are getting this equilibria and again for this particular equilibria what you have to do you have to just write the pe equation that is pe is equal to pe0 plus 1 upon n log of proton upon hydrogen gas to the power 1 by 2 again this if you see from the table is nothing but 0 so this value is 0 this you can always convert into the pH scale that is proton that is you can put always minus here and then here and then you can always convert into the pH scale so here what we are getting is nothing but pe is equal to minus of pH so here again if you compare y is equal to mx plus c t is not there so if you compare this y is equal to mx plus c where x is nothing but pH y is your t you are again getting a negative slope this is there so now this is my boundary why the boundary because if we are having a pe value above this range it can oxidize my water and if we are having a pe value below this range it can reduce my water so i need to work for those system which stable into the water medium or i can say the system which does not react with my solvent which does not make my solvent unstable to water as a medium and we do water as a medium we have to think of this boundary line so we are having two boundary lines one is because of the oxidation of water and one is because of the reduction for the oxidation we are having this boundary line for the reduction we are having this boundary line both are having a slope of minus 1 that we have seen this and they said working limit into the equity media so now as i shown you in the previous slide that there is some boundary so now we have some idea that why the boundary is and what is the concept of boundary now we will try to learn about lines that why we are getting certain straight lines certain band lines or certain horizontal lines into the into our pH pe diagram then we will try to see okay so here i have taken a redox reaction in which you have substrate ion reacting with proton some electron n giving you h is minus n 4 h 2 for this equation the log k value is 34.0 you have to set up the equation of whichever pe pe not equation and from this equation we should try to add all these values that you are having what you are getting pe is nothing but monopon and log of k since there are eight electrons this is eight this log k value is 34 we are having 34 and this you are having eight then if you plot this you for h you are having pH what you are left with these two you are getting this equation with these two what we assume that at a particular line that is there in the pH pe diagram the concentration of the activity of these two ions are equal so when i assume the concentration or activity of these two species are equal what will happen they will be equal and assuming that they are equal we will get this kind of equation and which can easily be plotted onto the graph so this line is shown here here this is the line which is shown what is it means that the activity of h s minus and s 2 full 2 minus is theme of this line right so you can see this is a slopey line so the slope means the equation contains both pe term and ph term and pe term is basically electron activity ph term is basically proton activity so when you are having some sort of slope some sort of slopey lines it means that the reaction equilibria that you are talking about is controlled by both pH and pe so your proton as well as electron both are influencing that particular equilibria what about the straight lines let us see let us take a very simple example of sulphuric acid or sulphate plus proton which is giving h s before minus with a low k value of minus 2.0 so here since we are not having any electrons we need not to set up the p equation but what we can easily write is the k you can always write the k for this equation which is the sulphate proton and then s is over minus and since we are having low k value and we can just put this value and as I told you that but we assume that that along this line the activities are same so the moment you get this and you put the activity of this equal to activity of this but we are getting pH is equal to 2 when we say pH is equal to 2 and I have to plot on pH p diagram where pH is in the x axis and pe is on the y axis but we get a straight line that is falling on the pH scale so this kind of redox equilibria which are only controlled by the proton they are giving you a straight vertical line what about the redox system when I say the redox system suppose you have a certain redox system as I have shown in the previous right that is iron 3 plus electron giving rise to iron 2 plus suppose you have this kind of equilibria in which only electron activity is to be mattered what kind of lines we are getting in this kind of system I hope you can work easily for this kind of system and you will get a horizontal line like this please try to work on this and if any problem is there we can discuss so we can we know how this equation or how the factors that are controlling this equation will give different kind of line whether they are vertical whether they are horizontal some kind of line with this understanding let us this is one more example I think you can easily work on that here I have shown you that if the system also incorporates some solid species how to work with that so I think you can easily work on that so now with this understanding that we are having three kind of line that is the more horizontal lines or more vertical line some are slopey lines but information we can get so let us try to see the simple graph of net union that is this one and we say net union this is always np2 plus we start with not this because we are in the convex diagrams so oxygen states will keep on changing so I should not say like it is this it is net union only because I have not specified the in state so in this simple diagram the first thing you can see the slant lines two lines that they have already seen that this is the limits because of the medium out of this it will be having some oxidizing nature it will be having a reduce so what it will get water itself will get oxidized above this and get reduced below this so we have to work in this and you can see this limit is in the all diagrams because they all are quoted into the water media and if you see the equation here that water is going to o2 and water is getting to and I have shown you the equations also for this is nothing but it is equal to minus pp and for this for the top one is like p is equal to 20.75 minus ph so you have equations for these lines so now we know about these lines now we are having certain vertical sign what the vertical lines mean as I showing in the example where I am talking about the dissociation or you can say the combination of circuit and the protons where we are getting hs before minus we have seen we are getting certain straight line so we have seen there this equilibria is totally dependent on the so when the equilibria is totally dependent on proton or you can say the two species are mainly dependent on each other with respect to the protons and there is no change in the oxygen state then we are getting this kind of straight line so here we see napkinam is plus 5 here you see napkinam is also plus 5 but the difference in the species is because of the involvement of the proton so it is carbonated which is not carbonated because we are having different ph what about these flat lines flat lines are mainly controlled by the electron activity or you can say e so here if you see the species are almost same but the oxygen states are very different here we are talking about the pentavalent or you can say npo2 plus and here it is becoming the pentavalent and slot line as I have said that they are dependent on both ph as well as the electron activity into the medium so with this basic idea that how the ph and electron activity will control my equilibria we can do this kind of diagram and in this kind of diagram we can get all the information in one diagram that how my ph and p of the system is changing my equilibrium both my relative equilibrium as well as my chemical equilibrium and here also I just want to mention one thing that if you see there are certain areas which are having different colors again if you take the example of napkinium you can see this area is almost everywhere this is one color area then this the second one is this one and third one is this one what it says that the stability of a particular oxygen state in this working range so you can say that in this particular working range the stability of napkinium 5 is very high so if you are maintaining somehow this kind of condition of ph and pe you will always bound to get napkinium 5 whereas if you are changing the pe value you may reach to this which is the green one where napkinium 4 is predominant so just by playing pe and pe of the media and by seeing this diagram you can always tell under what condition my napkinium 4 will be predominant and under what condition my napkinium 5 will be predominant and the higher the area obviously higher is the stability of that particular species so we have discussed about the 4 batch diagram and what we mean by this horizontal vertical or a floppy lines and we have also seen that how one can draw this kind of diagram using very normal chemistry or very simple equations that I have shown you and I would like to request you that you will try these equations to draw by yourself and you can take certain other equations also certain pure equilibrium in which you can have the reaction dependent on either proton or electron or certain reactions in which both are playing a role and you will try to draw this ph pe diagram that will give you an idea about the stability fields of that particular species that you are interested in so with that thank you thank you very much