Were given these 2/2 reactions that occur within a galvanic cell in a party. We want to determine what the overall cell potential is for this galvanic cell. We know that four galvanic cells, the cell potential, has to come out to a positive value. And so, in order to do that we need we know that we need to reverse one of these reactions so that we cancel out the electrons on either side. So both of these half reactions represent reduction, but one of them will have to undergo oxidation In order for the cell potential to be positive, we have to keep the standard reduction potential that has a higher value, and so that will be the one that is reduced since it has a higher potential to be reduced. So the one with the lower standard reduction potential will be the one that is oxidized, since it has a lower potential to be reduced in there for a greater potential to be oxidized. So we see that since 1.50 is greater than negative 2.37 that means we have to flip the second reaction. So we're oxidizing solid magnesium and when we combine those 2/2 reactions in that way, we see that this is what that results in. And now we have to cancel out the electrons on each side. So in order to do that, we need to multiply the top half reaction by two in the bottom one by three so that we're cancelling out a total of 60 electrons on each side. And now the overall cell reaction becomes to a you three plus acquiesce was three mg solid goes to three OMG two plus a qui ists plus two solid gold. And now, since we re arrange these half reactions in this way, we see that the so potential comes out to a positive value, which is what we want for a galvanic cell. So he had those two potentials together 1.50 plus 2.37 We see that comes out to 3.87 volts, which is the answer for part A. And now, in part B, we need to use the inner city equation because we're at non standard conditions. We know that that means that they sell potential is equal to the standard cell potential minus 0.0 591 over end times The log of Q. We conform it an expression for Q. In terms of species concentrations, the quickest product is MG two plus. So the new Maria is a concentration of M G two plus and we Cuba to due to its tricky metric coefficient of three and the acquiesced reacting is a U three plus, and we square it so concentration of a U three plus squared. We're told what the value is of the nonstandard self potential, and that is 4.1 volts. This is equal to the standard cell potential, which we calculated important to be 3.87 volts and then subtract 0.0 591 over end. And we saw for the overall reaction, we had to cancel a total of six electrons on each side. So that is why N is equal to six times the log of Q. Where Q. Is this expression that we just form? We have a value for the concentration of M G two plus ions and acquia solution. You're told that that is 1.0 times 10 to the power of negative five Moeller and we cubit based on that Q expression. Then we divided by what we're solving for the concentration of a U three plus ions, and we square that. And now we rearrange this equation to isolate and solve for a U three. Plus, we subtract over these potentials and then divide by this negative ratio and we raise tend to the power of whatever that is. That result comes out to be so that we get rid of this log, and then we can easily rearrange by dividing and taking the square root to solve for the concentration of a U three plus I owns. And when we do that, we should get a final answer for the concentration of a U three plus to be about zero 0.40 Moller.