 There are few conditions for an electrolyte to be used in to be used in salt rich, ok. The first reason is what the ionic mobility of these ions must be equal, ionic mobility is more tendency to move in the solution. Now, suppose why these two ions must have equal ionic mobility, suppose if it is not equal K plus has more ionic mobility than CL minus, then this K plus will utilize this negative charge faster than CL minus. Then also what happens, the electrolyte will not be maintained. The condition is what these two ions must utilize this positive and negative charge simultaneously. If one negative charge will get utilized here, at the same time one positive charge must get utilized on the other side, right. That is why the first condition is what the ionic mobility of ions of electrolyte present in the salt rich must be same, right on the first condition. Now, this point to write I will discuss. The ionic mobility of the ions present in the electrolyte, of the ions present in the electrolyte in salt rich must be equal. Second condition, second and last condition. The ions of electrolyte present in salt rich, the ions of electrolyte present in salt rich must not react with, must not react with, the ions present in the half set, both half set must not react with, the ions present in the half set. Now, the second point means what, suppose we have, suppose I am using K plus and CL minus here, if the electrode is silver or AG plus ion is present in one of the half set, right. So, AG plus and CL minus combines and forms what? The precipitate of AG-CF, AG-CF solid. This precipitate, if this ions, ions of salt rich reacts with the ions of present in the half set, that forms precipitate. If this precipitate forms, then what happens in the half set? There will be some accumulation. This initial precipitate will start accumulating on the bottom of the precipitate and that will also stops the working of, affects the working of electrochemical set. So, second condition is what? The ions present in the electrolyte must not react with any ions present in the half set. Basically, if we have silver electrode, we cannot use KCL. We can use of any other, we can use of any other electrolyte, right. So, this is the condition. Unless you have this, can it out? No. Okay. Next we have of EMF of the cell, right down next. EMF of the cell. Have we discussed Nernst equation? No. Okay, after this. EMF of the cell. Next point. You will not get any numerical question on this electrochemical cell transformation, theoretical question you can give. Now, EMF of the cell can be defined as E cell, for it as E cell. E cell is equals to, we write E of the reduction potential we are taking of cathode minus the reduction potential of, this is the formula of EMF. Both are reduction potential of cathode minus. But we also know what, we also know E reduction potential of cathode is equals to what we can write? Minus of E oxidation potential of cathode, right. And E reduction potential of anode, reduction potential of anode is equals to similarly we can write minus of E oxidation potential of anode. So, we can write EMF of the cell in terms of oxidation potential and reduction potential. If both we are using oxidation potential then what happens? This is the one formula. I will ask you to memorize only this. Okay, you can write down in any other option. But write down the formula, if this and reduction potential will convert into oxidation potential, in the formula becomes what? Can I write this E cell is equals to E oxidation potential of anode minus E oxidation potential of cathode, right. Again you can write E cell is equals to minus E oxidation potential of cathode minus E reduction potential of anode. Or we can also write E cell is equals to E oxidation potential of anode plus E reduction potential of cathode. Is it right for the four different forms? Correct. Zn is equal to form but Zn is equal to form precipitates. Zn is equal to form but it will exist in the form I am Zn plus NCl by Zn plus 1 to Cl minus. So, EGCl forms what? EGCl forms precipitates, it will start building over here, right. We will have a solid building here but Zn stays in any component, it will remain in the form of I am Zn plus 2 to Cl minus, right. Understood this? Okay, now this is actually at any condition, right. At a standard condition what we can write? At standard state, the standard EMF of the cell, right. EMF of the cell at any condition but when we take a standard EMF becomes what? Standard EMF. Standard EMF of the cell is what? E0 cell, right. E0 cell what we can write? E0 of reduction potential of cathode minus E0 of reduction potential of anode. So, you just memorize this one. Okay, all these other three you can write since you know this. Only one, this one is the part of the reduction potential of both if you are taking right cathode minus anode. Since cathode is present on the right side, we also write as E right minus E left. Yes, in some way we will see this also. E right minus E left. Why right? Because cathode is present on the right side. Only one thing you remember, cathode minus anode where both are reductions, right. E, reduction potential of cathode minus anode at a standard state. What is the E0 cell for this Daniel cell or electrochemical cell? What is the E0 value for that? 1.1. So, for Daniel cell what we can write? Daniel cell is nothing but this electrochemical cell. Daniel cell we can write E0 of the cell is equals to E reduction potential of cathode. What is cathode copper? What is the reduction potential value? 0.34 for cathode minus 0.76 for zinc that is anode. This becomes what E0 cell is equals to? 1.1. Next write down cell notation of electrochemical cell. Cell notation of electrochemical cell. For cell notation there are some rules we have. According to that rule we represent a cell notation for a given cell reaction. Cell reaction is different. Cell notation is different, right. Cell reaction is what? Zn gives Zn plus 2. Zn plus u2 plus gifts. Zn plus 2 plus u sorry. That is cell reaction. But cell notation is different. So you should know the conversion of cell reaction to cell notation and cell notation into cell reaction. Both are important. Right on the first point into this cell notation. The first point is the anode is written on the left side and the cathode is on the right. Anode is on the left side and cathode is on the right. Next second rule. The phase boundary are represented by, the phase boundary are represented by a single vertical line. Single vertical line, single vertical line means the phase boundary are represented by this single vertical line. The concentration of electrolytes, the concentration of electrolytes in anode and cathode half cell, in anode and cathode half cell must be written in parenthesis, must be written in parenthesis. In case of the gas, in case of the gas, in case of the gas, the partial pressure is to be mentioned, the partial pressure is to be mentioned in ADM atmospheric, in ADM or MMR. A comma is to be used to separate, to separate two chemical species present in the same solution, in the same solution. Double vertical line is used, double vertical line is used to represent, to represent salt rich, to represent salt rich. EMF of the cell, EMF of the cell, must be written, EMF of the cell, must be written on the extreme right, on the extreme right. Now you see, the first very simple example we are discussing, we have Daniel's cell. And what is the reaction of Daniel's cell? ZL from it converts plus Cu2 plus H1. Suppose this concentration is C1 mole and it converts into, that was C2 mole plus Cu4. This is the reaction. Now the reaction of anode is what, reaction of anode is what, this is the oscillation right. And this is the reduction. Anode is zinc electrode and cathode is copper. Anode must be written on the left side, left side of salt rich. Salt rich is represented by this, double vertical line is salt rich. ZN is converting into ZN plus Cu2. So we will write what ZN, solid and the phase boundary because this is solid, this is aqua solution right. The phase boundary is represented by the vertical line okay. And here we have ZN plus 2 C1, sorry C2 mole. Just after this we will write down the concentration right. After this this is anode, whatever is written on the left side of the salt rich that is anode, that represents anode right. Now cathode is what Cu plus 2, concentration is C1 mole right, converting into Cu4 right. This phase boundary, this is the metal we have and in the last we will write E. So this is cell rotation and this is cell reaction. Understood?