 Okay, so lead can display silver from solution by the following reaction. Calculate K and delta G at standard for this reaction at 25 degrees Celsius. And you remember at 25 degrees Celsius the standard potential equation reduces to a more manageable equation to use. Okay, so first things first, let's figure out what the half reactions are. So 2 Hg plus Hg is goes to Hg solid plus there, Eb solid goes to Eb 2 plus Hg is plus 2 electrons. Okay? So now, remember what I just said, you have to look at your table and find those electron potentials, okay? Or those cell potentials. So, the electro potential for this one is 0.80 volts and for this other one, okay? So the E cell, the potential of the cell is going to be the potential of the cathode minus potential of the anode. Okay, so for this cell to work, it's got to be a positive number, so 0 volts and minus volts. Okay, so that's how you get the cell potential. So the number of moles of electrons transfers is something we need to know, so that's n in this case is 2. So in order to do this problem, so we're going to have to remember how we relate the cell potential here to k. So I'm going to erase this part of the equation here, volts divided by k is the log. So we're looking for k here, so if we rearrange this, we would get the log of k, get k from that E cell over there, so 5, 9, 2 volts. Notice volts cancel there and there, k is going to be unitless as always. Reaction does it go or not? Yeah, super, go. Okay, so this over here was, okay, so the last thing it wants us to do is figure out delta G, so delta G equals negative N F times E cell, like that. Remember N is that, E cell is that and F is going to be 6.5 kilojoules per volt, all of the electrons, like that. That's given to, so negative 2 squared. You guys can ask questions about that one. Okay.