For the following half-reaction, E° = 1.103 V: Calculate the formation constant Kf for Cu(CN)2-.

Video transcript

Hello everyone in this video talking about when hex a floral illuminate reacts with three electrons to form aluminum solid and six moles of f minus. The standard reduction potential is given to be negative 2.69 volts. So let's first recognize that the formation equation of the heads of flora illuminate is that we have the A. L three plus, Which is a quick reacting with six moles of RF -5. And this gives our product to be the hexafluoride illuminates. So A L F six with a three minus charge. Alright, so for oxidation side which of course occurs at the anodes. This is when the aluminum solid reacts with six moles of RF to give us our main product here Followed with three electrons. And then for the reduction side which occurs of course at the cathode. This when the aluminum three plus cat ion reacts with our three electrons which yields are neutral and solid aluminum. So we're gonna go ahead and see what cancels out. And this will be our overall net reaction. All right, so I can see canceling is going to be our aluminum solid as well as the three electrons. So, I can see being on my star material side for the overall reaction is going to be our aluminum three plus caddy on as well. There's six moles of f minus is a quest state and this yields are hexafluoride illuminate, which is of course a L F six with a negative three charge. And now we're going to go and sell for our K F value. We're gonna go ahead and scroll down for more space here. So let's go ahead and recognize that our E cell value is equal to the center reduction potential given to us on our cathode side, minus the standard reduction potential given to us on the an outside. And these values are given to us. So for the catholic side that is negative 1.662 volts and we're minus ng a negative 2.69 volts. So once you put that into the calculator, we get this value to be equal to positive 0. volts. Okay, I'm gonna go ahead and scroll down from our space here. So we know that n is going to be equal to the moles of electrons transferred according to our two half reactions. We see that being three electrons. So using my equation or formula that the E value equal to 0.592 volts over the moles of electrons transferred times the log of Cape or rather K F. So of course we wanna isolate our K F value. So first want to do some simplification and of course we already know the value. So we're gonna go ahead and plug that in and that's just this value that we just saw for So that is 0.407V equaling to the same exact side on the right hand. That's 92V over N log of K F. And now we can go ahead and actually do some simplification. So scrolling down again for more space, we can get that the log Of K. F is equal to 20.625. I was gonna do was trying to undo the log here. So we get that the K. F. Value is unequal to 10, raised to power 20.625 and we're simply undoing our log and once you put that into the calculator, we get that the K. F constant is equal to 4. times 10 to the positive 20. So this right here is going to be my final answer for this problem.