 E01 is given that is 0.15 mole and this E01 is minus 0.15 mole, this is what we asked. Now this Cu2 plus the second reaction gives Cu and E02 is 0.34 mole. So we calculate what minus N1F even though and for this it will be minus N2F E02. In the net reaction plus one electron gives Cu. So delta G3 will be minus N3 0. Since we have added these two equations for this we can write delta G0 3 should be equals to delta G1 0 plus delta minus N3. F E3 0 is equals to minus N1F even 0 plus minus N2F G2 0. F gets cancelled, the negative sign will get also cancelled and E3 0 is equals to what? N1 value is what? N1 is 1 right? So 1 into even 0 is minus 0.15 plus N2 value is 2 and into 0 is 0.34 divided by N3 value is what? 0.5, always remember delta G you have to add a subtract, E cannot add a subtract. Next question we write on calculate the potential of the cell reaction and the cell notation is given that is CR to CR3 plus 0.1 mole that F E2 plus 0.011 mole to F E. You have to find out E cell for this. The data given E0 for CR3 plus to CR is minus 0.74 mole and E0 of F E2 plus to F E minus 0.44 EF of the cell reaction. Write down the reaction of anode, then write down the reaction of anode, then you find out the complete cell reaction and apply another situation. Ok, that is all this one. Ok, if the reaction reaction is what? The first reaction is CR3 plus 3 electron gives CR. And the second reaction is F E2 plus 2 electron gives, IL is what? IL is cathode, IL is cathode so it will get reduced, right? This is reduction, this is also reduction, both are reduction potential. To write down the reaction, one you have to convert in oxidation. Oxidation takes place at chromium. This we have to write down the oxidation reaction. Oxidation reaction is what? Chromium gives CR3 plus 3 electron. So its E0 value is what? E0 of anode it is. That will be 0.74 mole minus becomes plus. Why IL? I write some oxidation reaction here because this is anode, right? This must be oxidation, this must be reduction. This is reduction reaction and this is oxidation reaction. And this is given E0 of cathode is what? Minus 0.444. When you add these two, you get net reaction and emf of the cell E0 of the cell is what? E cathode minus E anode. E cathode minus E anode. E cathode is minus 0.44 minus E anode that is the reduction potential of anode. Reduction potential is minus 0.774, 1.74. E0 of the cell is 0.3. Both are reduction potential. So I have taken the reduction potential value of this minus 0.7. Now when you add these two, the electron will get cancelled. So we have to multiply this by 2 and this by 3. So total number of electron exchange is what? 6. So n value is? 6. And the reaction is 2CR plus 3FE2 plus, this is solid, aqueous. Gives 2CR3 plus aqueous plus 3FE2. Solid we don't write in concentration term, right? We write only aqueous part. So we write D cell is equal to E0 cell minus 0.059156, concentration of product by reaction. The concentration of product? CR3 plus 1FE2. Right? But here one thing you have to keep in mind, the coefficient is 2 and 3. So it's coefficient you have to write here, 2 and 3. It's like Q or, so in Q we'll write the coefficient in the power of the concentration of pressure term, right? You must take care of this thing, whatever the coefficient we have here, let me go into the power here. Understood this? Yes or no? Yes. But all E0 cell we have that is 0.30 minus, see here in the denominator we have 6. So 0.0591 we can consider this as 0.06 also to make the calculation easy. 0.059 what we can write? 0.0656 approximately. Log of, what is the concentration of CR3 plus? CR3 plus concentration is 0.1. So 0.1 is square divided by 0.01, so this becomes 0.30 minus 0.01 log of, this is 10 to the power minus 2 divided by 10 to the power minus 6. So this becomes what? 0.30 minus 0.01 into 4, right? Now it is 0.1. What? Understood this? Initially when you are solving the question, try to write down the reaction. Initially 5-10 problems you write down the reaction to keep in mind. So I suggest you write down the, then you apply last equation for initial 5-10 questions. After that you can do it easily then. The next question you write down. The solution of C us of 4 in which the solution of C us of 4 in which copper rod is placed in a solution which is diluted to 10 times. Find out the reduction electrode potential. Reduction electrode potential. No, it's not given. It's not given. But how can you know C us of 2 plus C is 0.3? That's not required, the value is not given. You can use this, but this is not required, without this also you can do. Concentration initially. And the final concentration you have. The final is 1 by 10 of the initial. Concentration is diluted to 10 times. This is actually a half cell reaction. Like I said, last equation you can apply in half cell reaction as well as complete cell reaction. So half cell reaction you write down copper, C 2 plus, the sweet electron you see. Apply last equation into that. Write down the expression. You'll get C you concentration in the last equation expression. That is first case when C you concentration is suppose X. In the second case X becomes what? 0.1 into X. 0.1 into X. That's what you have to put and subtract both the equation you will get. That's why E naught value will get cancelled when you subtract. That's why it is not given. You write down first the equation you understand what you have. Sir, what is 0.09? 0.29? Yeah. 0.03 minus 0.03. Minus 0.03. Sir, what should we find? What should we find? What we have to find? I know the negation. E cell you have to find. Sir, we can approximate it to 0.06. Calculation is easy. You can use 0.06. You can use 0.03. You can use 0.06. Sir, it's not a problem. I mean, that is quite funny. We think it's wrong. What is the answer? 0.31 0.3 0.03 Minus of 0.03 Yeah. Thank you.