So now we'll work on problem 52 from chapter 20 And this problem, we're asked for each of the following reactions. Right. A balanced equation. Calculate the standard IMF calculate Delta G at 2 98 Calvin and calculate the equilibrium constant at 2 98 Calvin. So our first reaction is obvious. I died ion oxidized. I too by Mercury two plus So we have this eye minus our quiz ion, which is going to be oxidized by Mercury two plus to produce I too. And then the mercury will be reduced. Teo neutral artery and there should be a two here is well to make to be behind on molecule so we can look in appendix d e excuse me and get the reduction potentials for the these 2/2 reactions. And so we'll put the reduction reaction first of the reduction of mercury and we get zero point 79 minus zero point 0.5 for and that's equal to 0.25 boats. So that's our EMF. So then the calculate don't g we need to look no are, uh, equation is negative. End very days, constant times this imf. So this end here represents the number of electrons that are transferred. And since we see Mercury to going to neutral mercury, we should say that there's two electrons transfer, so we have negative, too. And we have Faraday's constant, which is 99 1,000 six. 96,485 units for Faraday's Constant are, uh, Jules. So then we need to multiply by the IMF, which we just generated, which is 0.25 So here we multiply through on we get negative, too. Uh, 0.34 times 10 to the fifth Jules Purple, which is also equal to 234 kill Jules. And the way is correct, but we'll need to use Jules promotes to calculate the K value. So now our equation for K involves Delta Gee being equal to negative r t a line of K. So if we put this around to solve for Ellen of K, we get lnk is equal to Delta G over our tea. Uh, negative. So here we can put. So I actually wrote down the wrong number here. Okay, this here is the for the next one, and I got them mixed up So for the Delta Ji, for part A is negative. 4.82 times 10 to the fourth. And this is negative. 48.0. If, uh, yes. So Delta. Gee, here we have this negative 4.8 times. 10 to the fourth Jules Permal, divided by Artie. So the value of our that we're going to use is 8.314 because it's in units of jewels per children more. And then we're going to multiply that by 298 Kelvin. And that gives value of 19 point for seven. So then the Ellen, uh, this is equal to Elena K. So then, K, if we take the inverse of the Ellen or E, we get 2.86 times 10 to the eh? So now we can move on to Part E where the reaction which we're looking at is, uh, acidic solution in acidic solution copper one ion is oxidized a cover to buy nitrate. So we have cover one, see you plus plus nitrate and then we're in acidic solution. So we had protons present and it's copper. One is being oxidized to cover two plus Plus, you know and water. So now we can balance his equation. We know that cover this nitrate tonight. You know, from the half reaction there, Appendix E will be a transfer of three electrons over this copper. Ah, elect copper One to cover, too, is a transfer of won elections. So we'll need to balance this batting three here and then we also need for protons for this nitrate reaction and that will produce two waters. So now first, calculate the IMF, we first put the reduction. So nature it to an old measures oxide. And we get 0.96 from the appendix. And then for the oxidation, we get 0.15 So we have a value of 0.81 votes. So now we can get you a Delta G, which is equal to negative and f e. In this case, we're transferring three electrons. Saturday's concert is 96 45 and R E, which we just calculated a 0.81 So we go ahead and calculate 2.3 four times 10 to the Yes, Jules, her mall. And then for Kay, we can solve the same way we did before Alan K. Is equal to Delta G over negative, Artie. So this is equal to negative. 2.34 times 10 to the fifth Trooper over in a native, 8.314 Jules Kelvin Moore and to 98 couples, which is the temperature were given in the problem. This gives the value of 94.6. And when we solve for Kay, we get 1.25 times 10 to the 41. Very merry large value. So now, in part, see, yeah, we're given the reaction of again an acidic and basic solution from him hydroxide. It's oxidants crummy by Cielo minus. So for this reaction, here we have crewmate criminal hydroxide uh, we're in. It's going to be oxidized by seal Oh, minus and we're in basic solution. So this produces will leave the sub scripts for the second part. So we have Ah, crummy cr 04 two minus. We produce water and Lorraine chloride. So the chromium hardrock said reaction the oxidation of chromium Jack said that the crew mate involves a transfer of two electrons. The reduction of er co write involves the transfer of three elections. So we're going to need to multiply the chromium by two and the hydroxide in the chlorine by three. So if we had and 1/2 reactions up, some of the hydroxide will cancel from both sides, so we get 400 oxides in five water. So the IMF, the reduction value here is 0.89 for the reduction of the correct and then the reduction or the reduction half reaction. The chromium is zero from 13 and that gives the value of 1.2 bolts for Delta G calculate. In the same way we again have a transfer of six electrons. Faraday's constant in our mess get a value of negative 1st 0.90 times 10 to the fit. Well, Jules promote. Excuse me. So then we calculate lnk. We get negative 5.90 times 10 to the fifth over, our gas constant and our temperature, and we get a value of 238. When will these take the Ellen? We get kay this week. The interest Ellen or the E 2.25 times 10 to the 1 0/3 So very, very large.