mod12lec58

[Music] [Music] [Applause] [Music] [Applause] [Music] hello again so welcome back to our latest lecture session so first again a couple of minutes to understand what we have been up to in the last session right so the last session we looked at what we say a couple of minor what do we say derivations where we try to understand the system right say and relate PE in terms of let's say molar Gibbs energy of the electron for a particular redox couple and again to be able to relate that to the redox potential right so we were able to do that and then we went ahead and looked at I guess two half reactions at the standard states and non standard states and try to understand that let's say as an in terms of both the p-values are looking at the molar Gibbs energy of the electron in each of those redox couples to be able to understand let's say in which particular direction would the what do we say reaction proceed as in which half reaction would proceed as reduction and which one would have to be oxidation for the redox process - both right so in that context obviously we looked at the case that the electron obviously would want to travel from a higher energy state to a lower energy State right so obviously I guess once this electron transfer or during this l-tron transfer from higher energy state to lower energy state so what are you going to observe you are going to obviously observe a release in energy right so obviously what what we are driving it here is that in this ordering this what we say transfer of electrons right and because the electron is or was at higher energy state compared to its destination where it's at a lower energy state you are going to have a release energy meaning when this redox process goes through let's say right you are going to have a release energy and we can calculate that - so let's again look at the example that we had earlier so this was the example at non-standard conditions right non-standard conditions the non-standard conditions were at pH 7 and that the activities of H was CL + CL - were respectively 10 power minus 4 and minus 3 but for the couple of s4 for two months and H 2 s it's their 10 power minus 3 and then we understood the system and saw that because the energy of the electron would share molar Gibbs energy of the electron and the couple or with respect to the couple s four four - - and h2s is higher than the one with H was CL + CL - or the one in equilibrium with HCl and Cl - we know that the reaction or the electron would like to move from this couple to here right or this couple meaning this particular one will go through as reduction right because it's going to accept the electron obviously and this is going to go through as oxidation because it's going to release or donate the electrons right so here as you see again there is going to be an energy release so we can calculate that so delta g of that particular reaction is going to be equal to what now right it's going to be equal to minus 120 point four eight which is the final state right - twenty point nine six right and that I guess is approximately equal to 141 point four and the units as written here were kilo joules per mole right per mole right so whenever there is one electron transfer of electron transfer or one mole of what you see electrons transferred here from the couple s po 4 - - and H 2 s 2 HOCl and Cl minus 141 point 4 kilo joules will be released during that particular redox process right so I guess that helps understand the system in a slightly better manner - right so we will now move on to the eh approach right so until now we have discussed the aspects related to PE and so on right so we are now going to be cut the EH approach I guess right so this is relevant to your altro chemical cell eh approach or the redox potential let's see right altro chemical cell and let's look at what some of these components are right so you'll have the electrodes right and the electrodes are cathode where you have the reduction taking place and a node at which you have the oxidation occurring areas right and then you will also have your applied solutions right and then more importantly you will also have this salt bridge I guess let's see the relevance of this and within a short time right short salt bridge partner right shelter chemical cell let's just try to visualize what we are up to here and so let's say here we have cell and our outward here cathode and anode let's say we have a semi permeable membrane or if it's not semi permeable we need a salt bridge and we'll discuss again why we need that and let's say this is the cathode here at cell one and this is the anode here right cathode is where the reduction is occurring right so the electron transfer will be from the anode where the oxidation is occurring as in the electrons are being released right and they through the conductor right and we also need to have the external circuit that's something I missed writing up here we are going to have the external circuit right and why do we need this external circuit because you need a conductor for the electrons from one electrode to the other right so here this is the external circuit and obviously without this external circuit you will not have what do we say the electrons being connected from one cell to the other and the redox reaction will not go through right so now let's just try to understand the relevance of all these what do we say aspects that we looked at electrodes the solution salt bridge and external circuit I guess right Celt roads I guess there are two kinds as in reactive and nonreactive reactive as in these electrodes themselves take part in the half reactions let us say or are involved in the half reactions so what time you'll have loss of these electrodes let's see the non reactive electrodes similar to let's say the platinum electrodes let's say our non reactive and I guess they just act as let's say the media not media I guess the surface at which the relevant half reactions occur now right so again we have reactive and nonreactive what do we see outwards now so again then we have the care third and anode right this at the surface of which are which they themselves are involved in the half reactions yes so at the anode you have the oxidation the electrons are released at the cathode you have reduction electrons are gained the anode and cathode but how do the electrons go from here to there now so that is the reason why you need the what is it now external circuit I guess right so you need the external circuit and obviously without the external circuit you will not have a redox process go through right so that's the key one aspect and again just because the electrons are being transferred now does that mean that the relaxed process is going through or are there any other issues we need to consider now the issue is that you know now there's going to be a charge imbalance now right so the electrons are being transferred right so there's going to be net what is increase in the need to charge in your let's say the solution on the cathode I guess right and then what's going to happen now again that will what we say this u8 the redox process from going through so for that I guess your obviously for the charge balance to occur you are going to have to look at the salt bridge or have a semi permeable membrane let's see right when let's see where that would allow the what we say transport all the relevant salts are the cations and a neon or a neon said say such that there is a balance with respect to the charge right again so there are different cells as in different kinds of cells as in both the electrodes in the same solution are outwards in different what we say cells or solutions pardon me and so on different configurations but the major aspects are the need for the two L Truths the external circuit and the salt bridge I guess right so these are the aspects again one aspect here is you know what's the relevance of these what do we say electrodes are these cells are such let's see so let's look at one particular example right we are going to consider a case let's say where we have a water in equilibrium with water not water pardon me the atmosphere so we have oxygen and hydrogen in the gaseous phase and let's say they are going to be now at equilibrium are in equilibrium with your water so you are going to have o to in the aqueous phase and what do in h2 in the gas aqueous phase part right we have o to in equilibrium with the aqueous phase again h2 in the gaseous phase in equilibrium with the h2 in the aqueous phase gaseous and aqueous phase equilibrium so again now as we know we have relevant half reactions right and if we look at the relevant overall reaction I think we know that o2 plus 2h2 can lead to formation of your h2o right so this is one reaction that can occur right let's look at what are the conditions required for that to occur now as we know this reaction occurs only when there is an electron transferred not right and for that to occur what needs to happen your o2 molecule right needs to collide with the h2 right the actress forms of both odd go to an h2 need to collide for the electron transfer to take place again let's say o2 and h2 right so this can obviously sometimes what does it create issues not sometimes I guess what we say cleared issues with the kinetics let's say so instead of that let's say if I have an Altoid let's say a platinum electrode let's write how old US assistant furthering this particular reaction now as in the relevant half reaction with the oxygen let's say can occur here and the relevant electrons let's say can be transported to the oxygen molecule from here while the relevant half reaction right with respect to the h2 + H+ can occur out here and the electron can transfer through the conductor let's say or obviously the electrode will act as a conductor for the electron transfer as in now you do not need the o2 and h2 molecules let's say oh two axis forms of water and h2 let's say to collide or there need not be any physical collision now so their reactions or the half reactions can occur at different places on the surface of the platinum electrode and the electron transfer can take place from the site of what we say the h2 related half reaction to the site of the what related half reaction now right this one obviously now promote the relevant what do we see overall reaction this is just by one example so obviously though if we get you know look at such particle cases do we cannot let's say constructively utilize this flow of electrons from one what do we say a redox couple to the other let's say if you have your battery such what are we trying to do we are trying to construct you constructively utilize D what do we say current are let's say the flow of the electrons let's see right so to be able to do that obviously you are going to have let's say two electrodes let's say right two electrodes right leg again cell one and cell two let's say and you have electrode one here an electron to here and now we have separation of the two what do we say half reactions instead of the half reactions occurring at just de on the same electrode right we are now going to have only one half reaction at one electrode another half reaction at another what we see electrode and you are going to have the transfer of these particular electrons here from one cell to the other which depending upon the relevant redox process and the relevant energies that would be released let's say right you can utilize accordingly constructively now right so here again the aspect that we need to look at is what drives this reaction now right the redox process or this transfer of salt on again keep in mind that everything you know in nature out there or the reactions involve out there see to it that the energy of the system decreases now right so in that case again what is it about now so the reaction would only go through when there is redox dis equilibrium right only when there is redox disequilibrium right well let's say the electron transfer take place from one couple to the other now right so but when the equilibrium is reached right what does this mean now so for example let's say you have redox disequilibrium and they try to approach each other right and now when there is equilibrium now there is no more potential now right so there is no more potential now yes there is no more potential and Thurs no more what is a transfer of electrons right so obviously let's say for your electron transfer to occur obviously the system needs to be at redox disequilibrium once the two cells reach equilibrium let's say right you are not obviously going to have any for the transfer of electrons let's look at the previous slide to help illustrate this so here let's say you have what we say eh here right the docs potential for what do we say one cell and let's say redox potential for cell to now so how is it going to go through now that both the potentials are they'll travel in such a direction that both the potentials are going to be equal now right so one the h1 is going to decrease and the h2 is going to increase and once this h1 is equal to h2 what we have now we have redox equilibrium right so once again when we have that h1 is equal to h2 right we have equilibrium and there are no more potential and no more transfer of the relevant electrons right so again this is the whole aspect that we need to keep in mind again as in one example we can consider is your battery straight oh so what do you mean by when the battery has exhausted or is useless now right it means that you know the redox equilibrium has been reached right and you have and there is no further potential driving the you know transfer of the electrons I guess so now we are going to move on to the relevant aspects now okay so let's again moving on I guess right so one particular aspect it's worth discussing here is to be able to calculate or measure eh I guess right ultra redox potential how do I do that people look at the standard hydrogen electrode let's say right you know I guess hydrogen gas at 1 atm pressure right and the relevant half reaction let's say and in reference to this standard hydrogen electrode you try to measure the potential of your particular or redox potential of your particular cell or the half reaction so based on that you can calculate the relevant pH of your particular cell now usually you know carrying this or you know standard hydrogen electrode making that might be relatively different so even if you do not have the reference of strain hydrogen electrode which we said to be let's say the H to be 0 that's it right that's the reference here let's say similar to your mean scene sea level now right the standard hydrogen electrode let's say but let's say let's say if you most people might have use your pH meter right what's that measuring now right it's measuring again the potential there right again if you are measuring the potential you need two electrodes again right so one helped road would be this ibis the standard calomel electrode right where you have I believe H do hedge ECL to the solid rate you know don't take my word on that you have the standard calomel electrode now so you know you have within the pH meter your standard calomel electrode and let's say the other electrode would be let's say in your solution let's say where you have let's say semi permeable membrane where H+ would tend to what we say be adsorbed onto the surface of the membrane let's say or this semi permian you are going to again measure what is it now the potential difference between the standard calomel electrode right and the half reaction at the other platinum electrode so when you dip your pH meter in your solution what are you measuring again you are measuring the potential difference there right so again you are measuring the potential difference and calibrating that at page 3 7 and 11 and so on right and you are telling the system that or the pH meter that at this potential difference it's equivalent to 3 at such a potential difference it is equal to 7 and so on so that's how you end up measuring your what do we say your pH I guess right so again and just an illustration I guess of where we come across this particular concept so again when we dip your electrode what we say your pH probe into your solution in effect you are measuring the potential difference between two half-reactions there right anyway standard calomel electrode but our standard hydrogen electrode so reference electrodes to be able to measure your redox potential of your particular half-reactions I guess so moving on though we come to a standard equation which is the Nernst equation for each cell ok or for each half reaction I guess and this is something that people might understand once it's written down eh equal to H naught minus RT by n F natural logarithm of Q - right so I guess once we write this down you should immediately be able to see the analogy with respect to PE is equal to P naught right - 1 by M and being the number of electrons being transferred natural out them off q - right and again earlier we looked at the relationships between PE eh and molar Gibbs energies of the electrons in the last session again right so again looking at the same principle or we can even now plate this again here I guess or if I'm a plate the first equation this equation has eh and - EF by this value will be 2 point 3 0 3 RT let's say will be equal to again eh not into F by 2.303 RT minus 1 by M log it should not be natural autumn here its lot of them here right block you - that's the reason why we have the 2.303 value here the P one with with respect to the logarithm and not the log them to the base 10 not didn't log them to the base e that was a minor error there so here again we looking at this what can you understand here I guess right you can understand that the PE value is nothing but equal to e H into F by 2.30 3rt right again this is something similar to what we you know looked at or derived in our previous sessions here right so again you see that you see the relevant and all these right and from that we can understand the system again and this particular equation is called the Nernst equation for each cell here right and here these are the redox potentials that people use or look at here right so again let's moving on so you have two cells here let's see H 1 and H 2 let's see so how would this particular reaction proceed or how do I understand this particular system let's here right so let me assume that H 1 precedes as reduction right uh-1 proceeds as reduction and he has to proceeds as oxidation let's see so the H would be equal to e h1 minus h2 alright so let's try to plug that in here so that would be equal to e H naught of 1 minus RT by n1 F into natural logarithm of q1 - right - eh - naught minus RT by NF and - F and being the number of electrons being transferred in the half reaction let's say so in this case if it's the same number of electrons being transferred let's say n1f right into natural or damoff q2 - right so what do we end up with you it's going to be equal to H 1 not minus H 2 naught minus RT by NF and National logarithm of q1 - by q2 - right q2 - yes would that be the right way to express that yes I believe that is true and how would that transform into that's going to be equal to H 1 naught - eh - naught minus RT by NF natural out the mauve Q of the overall reaction right Q of the overall reaction right so when is this particular value going to be past you now right so this head cell would be equal to Delta of H naught minus RT by NF RT by NF national anthem of Q - war all right Q - of the overall half reaction right so here let's say if it's not only n what we see the same number of electrons being transferred as in in one half reaction if we had N 1 and then n 2 so the N would have to be replaced by N 1 and 2 right the number of electrons being transferred and Q - 1 and Q - 2 would have to be the ones when you multiply let us say Q 1 - by n 2 and Q 2 - by N 1 right so we are looking at the case as in let's say for the Q 1 - 2 element half reaction we had N 1 electron transfer and for the Q 2 - relevant reaction we had n 2 electron transfer but we have the same number of electrons being transferred we need to multiply Q 1 - by n 2 and Q 2 - by N 1 right so then here the end would be N 1 and 2 and the relevant Q - Q 1 dash and Q 2 - should be modified such that you multiply Q 1 - by n 2 and Q 2 - by N 1 and the relevant stoichiometric coefficients I guess right anyway so that's one particular option here not option pardon me the way to go about it so obviously when the F II cell is greater than zero done then what does that mean now then the out road 1 let's say is cathode right or we're reduction occurs right and electrode 2 is the anode where the oxidation occurs right Azen eh is higher each one is higher than eh - right so again we can understand this in terms of what we had the half reactions we had earlier the relevant setup here so here eh 1 is higher than eh - right so again what do you see here that would mean a reduction with respect to this couple and oxidation with respect to this couple right so again the relevant cathode and anode out here so again that's one particular case to go about it so obviously if each cell right is less than zero what does that mean now then the reverse needs to be applied right then what we see now right is going to have to be oxidation right or one the ultra with cell one would have to proceed as or it would be the anode and two would be the reduction let's say right and that would have to be the cathode if a cell is negative I guess right so this is one particular case do so but major aspect is let's say how unfavorable reaction are the e cell turns out to be negative now let's see right so here let's say we are talking about redox potentials right and electro chemical cell same cells and we looked at the electrochemical cell calculation and saw that where say 1 H 1 greater than eh 2 we looked at reduction in oxidation but let's say if it's less than zero let's say we see that it's unfavorable or that oxidation should take place at anode and reduction at cathode but let's say if I want to force a particular reaction to proceed as reduction even though the thermodynamics or sanch let's say say that you know it's not going to proceed as the relevant what do we see a redox process a yes so what can I do now so in that context obviously I can impose an external potential now right so I'm going to say I can impose right a potential or a external potential all right impose let's say an external potential by imposing this external potential I can turn this each cell which is less than zero right now right so that one proceeds as oxidation and two proceeds as reduction right by imposing this particular what do we say external potential I can see to it that each cell plus this external potential that I'm applying will turn out to be zero right and in that case again you are going to have reduction at your particular electrochemical cell 1 right which will then be the act as the cathode right an oxidation here and again what are the applications obviously if there is a compound let's say that you want to reduce and under thermal dynamic conditions it is not going to be reduced now so you can have these external potential supplied to be able to force these particular what do we say reactions or half reactions are the redox reactions to go through and let's say practical applications what do we look at we look at chrome plating I guess right chrome plating right as in what are we trying to do we are trying to plate the metal onto the electrode right and as on CR 3 plus we want to make this favorable which is usually not right so for that let's say we apply the external potential so how do we go about chrome plating or someone refers to chrome plating right I want to be able to reduce CR 3 plus 2 CR not out with the neutral charge let's say right and let's say this reaction is not favorable though so how do I go about that or how do I make that happen I make that happen by imposing a external potential now right second there are again applications with respect to these particular aspects but here with respect to understanding the system let's say with respect to e cell right and the external approach people I guess need to look at the terminology or the relevant science involved now so here as in to which but what would you count the positive terminal and which part well what would you can't be negative terminal and so on so again we are going to look at these aspects and then look at also warp II measurements right as in warping measurements as in to be able to get what we say generic or you of the reduction potential rate oxidation reduction potential of a particular solution you have these warp emitters that you can use I mean there are some applications but not a great lot so we are going to look at or in greater detail with respect to these terms respect to or with respect to the science with respect to the external potential and also look at with respect to the warp II measurements and then move on to let's say understanding how corrosion happens and how you can limit corrosion in the next session right so I guess with that I learned today's session and thank you [Music] [Applause] [Music] [Applause] [Music] [Music]