in this video we're going to talk about how to calculate the cell potential using the nurse equation now there's two equations that you need to be familiar with here's the first one the non-standard cell potential is equal to the standard cell potential minus RT over NF times the natural log of q r is the energy constant which is 8.3145 T is the temperature in Kelvin n is basically the number of electrons and the balance half reactions it's the ratio between the moles of electrons per mole of substance f is fad constant 9648 cols per mole of electrons so typically you need to use this equation if the temperature is not at 25° or 298 Kelvin now let's say if it is you can still use that equation or you can use this version the non-standard cell potential is equal to the standard cell potential minus 0591 / n * log Q now you might be wondering what is q q is the reaction quotient it's equal to the ratio of the products divided by the ratio of the reactants I mean it's one single ratio it's the ratio of the products to the reactants the same way you would calculate K is basically the same as q but Q is associated with the initial concentrations as opposed to the equilibrium concentrations Now what is the difference between e and e e represents the non-standard cell potential whereas e not represents the standard cell potential at e not all the concentration of the ions that participate in the reaction is 1 mole per liter for non-standard E the concentrations do not equal 1 M as in the cas of the problem that we have the concentration of aluminum is .1 and for copper 2 plus is 2.5 so the cell potential is going to change so here's a question for you do you think the non-standard cell potential e how is it going to relate to the standard cell potential eot is it going to be greater than equal to or less than now it's not going to be equal to because the concentrations are not standard it's not 1 mole per liter so is it going to be greater than or less than what's going to happen if we increase the concentration of the reactants we know that the reaction is going to shift to the right so it's going to become more spontaneous in the forward direction for a spontaneous reaction the cell potential is positive so if it's shifting to the right it's becoming more positive which means that the cell potential is going to go up if you decrease the concentration of the reactants it's going to shift to the left making it spontaneous in the reverse direction or non-spontaneous in the forward Direction so the cell potential becomes more negative less positive so it decreases likewise if you increase the concentration of the products it shifts to left the cell potential decreases if you decrease the concentration of the products it's going to shift to the right and the cell potential is going to increase so therefore you need to know that increase in the reactants and decrease in the products will increase the cell potential and decrease in the reactants or increase in the products will decrease the cell potential so the non-standard cell potential all right did not want to erase the whole thing the non-standard cell potential is greater than the standard cell potential whenever you have a large amount of reactant or a small amount of products the non-standard cell potential is less than the standard cell potential if you have a small amount of reactants and a large amount of products so if we look at our problem here we have a huge amount of reactants cu2+ is a reactant and we only have a small amount of product so therefore having a small amount of product and a high amount of reactants means that the non-standard cell potential is going to be greater than the standard cell potential so we should have an answer that's greater than two volts so now let's go ahead and calculate it now the first thing that we need to do is calculate Q so Q is basically the products over the reactants so we have the aluminum 3+ ion which is a product and we have a two in front of it so that's going to be squared and then the copper 2 plus ion that's a reactant so keep in mind you can't include any liquids or solids they should not be included in reaction quotient expression now the coefficient is three so the exponent is gonna be three here so we can't include these two solids and besides we don't have the concentration for it because it's a solid the amount of al3+ that we have is 0.1 and for cu2+ it's 2.5 .12 / 2.5 5 ra 3 power that's equal to 6.4 * 10- 4 so that's Q now the next thing that we need to calculate is n so let's take this uh net reaction and separate into half reactions so we have two aluminum atoms on the left side turn into two aluminum cations how many electrons do we need to balance it the total charge on the left is zero because there's no charge on the right it's 2 * three which is six so six electrons are required now for the other half reaction the reduction half reaction we have three cu2+ on the left and on the right 3 CU so 3 * 2 is 6 we need six electrons on the left so therefore n is equal to six in this case now that we have the value of n we can calculate the non-standard cell potential so let's write the equation first it's equal to the standard cell potential minus 0591 / n * log Q since the temperature is at 298 K or 25° we don't need to use the other equation only the temperature is different should we use it so this is going to be 2us 0591 / 6 time log of Q where Q is this number so that's 6.4 * 10us 4 and just go ahead and type it in exactly the way you see it so I got 2.03 volts as you can see the non excuse me the nonstandard cell potential is greater than the standard cell potential because we have a large amount of products and I mean a large amount of reactants and a very small amount of products number two a galonic cell is made using 25 molar zinc sulfate and an unknown amount of copper two sulfine the cell potential is 1.05 volts at 50° C what is the concentration of copper 2 sulfate in the cathode compartment and also how many grams of cuso4 are dissolved if the cathal compartment contains 250 M of solution so let's begin by finding n so let's write this in terms of half reactions to balance the oxidation half reaction two electrons are needed and to balance the reduction half reaction two electrons are needed as well so clearly we can see that n is equal to two now the next thing that we need to do is convert the Celsius temperature into Kelvin so that's going to be 50° C plus 273 so the temperature is 3 123 Kelvin now what equation should we use because the temperature is not at its standard value of 298 Kelvin we need to use this one e is equal to eus 0591 divid by actually not that one minus RT over NF times the natural log of Q this is the one that we need to use now we have the concentration of the zinc 2 plus ion as we can see is 0.25 what we need to do is calculate the concentration of the copper 2+ ion that's going to be equal to the concentration of copper sulfate because the copper 2 plus ion and copper sulfate they're 1 to one ratio in order to find the missing concentration we need to calculate the value of Q the reaction quotient so how can we go ahead and do that how can we go about doing it how can we find it well first let's rearrange the equation let's subtract both sides by eot so we're going to have rt over NF time the natural log of Q next let's multiply both sides by NF so that those will cancel and on the left side it's going to be NF * e minus E so that's equal to RT Ln Q at this point let's divide both sides by RT so Ln Q is equal to everything on the left side now to get Q we need to convert the logarithmic expression into its exponential form e raised to everything on the left side is going to equal to Q so the reaction quoti Q is e raised to the NF time e minus E over RT that's the equation that we need to use to calculate Q once we have q we can determine the missing concentration so we know n n is a negative well n is two but we have a negative sign in front of that f is 96,00 485 the non-standard cell potential is 1.05 the standard cell potential is 1.1 R is the energy constant 8.3145 and T is the temperature in Kelvin which is 323 now let's go ahead and plug everything in so you should get 3633 now keep in mind Q is equal to the ratio of the products divided by the reactants so the zinc 2+ ion is the product and copper 2+ is the reactant we can't include any solids the coefficient for both of these ions is one so it's raised to the first power so the concentration for zinc we have is25 so now we got to calculate the concentration of copper so Q which is 3633 that's equal to .25 times the concentration of cu2+ so you can cross multiply if you want but the concentration of cu2+ when you uh do the algebra it's going to be 0.25 / 3633 so you should get 0. 688 so that is the concentration of the copper 2 plus ion which is the same as the concentration of copper sulfine so notice that the amount of product which is represented by zinc is significantly greater than the amount of reactant so we have a very a relatively high amount of product and a relatively low amount of reactant and that's why e is less than e any time the non-standard cell potential is less than the standard cell potential it means you have relatively low amount of reactant in this case copper or a relatively high amount of product in this case zinc now let's clear away a few things and let's focus on Part B how many grams of copper sulfate are dissolved in the cath compartment if it contains 250 M of solution so keep in mind the concentration of copper sulfate is 0. 688 so let's just keep that there now in order to get the grams of copper sulfate we need to find the M Mass the mol mass of copper is 6355 the M mass of sulfur is 32.7 and we have four oxygen atoms each with a M mass of 16 so let's add up these values so you should get 159.000 milliliters we need to divide that by a th to convert it to liters so it's 0.25 lers so the unit liters will cancel and now we need to convert it to grams so we can use the M Mass there's 15962 G of copper sulfate per 1 mole of copper Sul so the unit moles cancel and we're just going to end up with grams so it's 0688 * 25 times is 2745 G of copper sulfate dissolved in 250 Mill of solution number three if the cell potential is 67 volts at 25° C what is the pH of the solution in order to find the pH of the solution we need to calculate the H+ ion concentration that's the only thing that we don't have we have the concentration of every other ion except H+ all the ions are in the Aquis phase water is in the liquid phase so we're not going to include that in a reaction quotient expression so before we calculate Q in order to find H+ let's let's find n first so we have 5 fe2+ turning into 5 fe3+ so it's an oxidation reaction the electrons will be on the right side the total charge on the left side is 10 the total charge on the right side is 5 * 3 which is 15 so we need to add five electrons to the right side to balance the charge because 15 minus 5 is 10 so we can see that n is five but let's check the other um half reaction the oxidation half reaction so all of the atoms are balanced in this reaction the total charge on the left is - 1 + 8 which is POS 7 on the right is pos2 these two numbers differ by five so we got to add five electrons to decide with the higher charge so clearly we can see that n is indeed five in this example so let's go ahead and write that here since the temperature is at 25° C we can use this equation E is equal to e minus 0591 / n * log Q so let's isolate Q in this example so let's subtract both sides by E not in the next step let's multiply both sides by negative n so it's n e minus E which equals positive 0591 * log Q so when we multiply both sides netive n on the right side the negative signs will cancel and N will cancel as well so now we got to divide both sides by 0591 so it's going to be n e minus E KN / 0591 by the way you can distribute the negative on the inside if you want and you could write it as n time positive e KN minus E if you prefer that the base of log if there's no number written it's 10 so 10 raised to everything on the left is equal to Q so therefore Q is 10 raised to the n e minus E divided by 0591 so that's the equation that you need in order to calculate Q so let's go ahead and use it so it's going to be 10 raised to the N is 5 in this example the non-standard cell potential that's 67 the standard cell potential is uh 74 and let's divide that by 0591 so all of this is on the exponent of 10 so keep that in mind now let's go ahead and plug this in if I was you I would type it exactly the way you see it on the board all at once I got a very big number I'm going to try that again just to make sure I typed in everything I got 835,000 922 let's try that one more time yep so that's Q in this example now that we have the value of Q let's write an expression that relates Q to the reactants and products so it's going to be the products over the reactants so we have the permanganate 2+ ion which has a coefficient of one and as a product we have fe3+ and the coefficient is five so we got to raise it to the fifth power as reactant we have the fe2+ ion which is also raised to the fifth power and we have the perinate ion which is raised to the first Power and the H+ ion which is raised to the eth power so all of that is equal to Q so 835,000 922 is equal to so the pregate ion is 01 and then fe3+ the concentration for that is 0.5 raised to the 5ifth power and then fe2+ that's. 25 raised to the 5ifth power and panate is just3 so our goal is to get the H+ concentration so at this point we really don't need this information anymore so let's go ahead and just make some extra space well we need to do is cross multiply so 1 time 01 * 05 to the 5th I'm just going to write it out and that's equal to Q which is this large number time 25th or 25 ra 5th * 3 * H+ ra the eth power so we need to divide both sides by everything except H+ so go ahead and type this in so I got 1.27 6 * 10 -1 is = H + raed to the 8 power so now what we need to do is raise both sides to the 1 over8 power so we can get H+ to the first power so the eth root of that number or just raises to the one over8 is going to be 043 47 so that's the concentration of H+ now that we have it we can find a pH of the solution which is negative log of the H+ concentration so it's about 1.36 it's relatively acidic so that's the answer now let's talk about how to derive the nurse equation so we need to start using this equation the non-standard Delta G value is equal to the standard Delta G Value Plus RT lnq Delta G is equal to NF so the standard Delta G value is negative NF e KN now what about the natural log of Q well you need to be familiar with the change of Base formula log base a of B is equal to log B / log a and you can add a new base which we'll call C so the natural log of Q is going to be equal to now the base of a natural log is always base e so this is going to be log Q / log e and so we're going to make it base 10 that's our C value it has to be the same in order for it to work now e is a number it's 2.71828 with some more numbers repeating so log e or log of 2.71828 if you type that in your calculator it will give you 43429 so therefore lnq is equal to log Q /4 3429 so let's let's go ahead and replace L and Q with that result so now we're going to have log Q divided by 43429 at this point we're going to divide everything by negative NF so these will cancel and those will cancel so the non-standard cell potential is equal to the standard cell potential minus so basically move this negative sign over here so it's minus RT / NF you know I'm going to use this for now Ln Q so this gives us the first version of the nurse equation so I'm not going to replace this yet I should have replaced it later so basically I put it back in this form so now to get the other version let's go ahead and replace lmq with this value R is 8.3145 T is going to be 298 Kelvin f is 9648 and then we're going to have log qide div4 3429 so if you type this in your calculator on top 8.3145 * 298 and in the bottom using parentheses 9648 * 43429 that will give you 0591 so you're going to get this form of the nurse equation it's going to be e knus 0591 log Q / n so that's the second form but to get the first form don't replace lnq until you get to this form once you get to this part then you could replace Ln Q with log Q over 43429 and that'll give you the other version of the nurse equation so that's it for this video thanks for watching and have a great day