Calculate the standard cell potential for each of the electro- chemical cells in Problem 43.

Video transcript

hey everyone in this example, we're given the below electrochemical cell and we need to calculate the standard cell potential. So looking at our given cell, we have manganese for oxide as an an ion Which reacts with four moles of hydride to produce manganese to oxide, Which is a solid here as well as two moles of liquid water. Now we want to recognize that we would have needed to add three electrons here on the reactant side to form our manganese to oxide. And so we would look up the oxidation of manganese to oxide from our standard electrode potential table, which we can find either online or in our textbooks. And we would see that has a value equal to 1. volts for the oxidation of manganese to oxide. Now looking at our solid potassium that forms on the product side that is going to be formed from our potassium carry on gaining an electron. And we would see for the oxidation of our solid potassium, we have a cell potential reduction value or electrode value equal to -2.92V for the oxidation of solid potassium. And we would agree that this is an oxidation because we have electrons, we have one electron rather that we added to our product side. And when we add electrons to the product side of our reaction, this is going to be an oxidation. Whereas for our first reaction, we added electrons to the reactant side here, specifically three electrons. And so this would be our reduction. So what we want to make sure is that our electrons are balanced for both of our equations here. So we want to go ahead and multiply this second equation By a value of three. And we also want to make note of the fact that since we've identified this second reaction as an oxidation reaction, we would recall that this occurs at the anodes and for our reduction reaction above here. So this was our reduction. We agreed this would occur at our cathode of our cell. So when we multiply the second reaction by three, we're going to get three moles of solid potassium that is formed from three moles of potassium catalon gaining three electrons. And so now we have the electrons that are balanced in both our an ode and our cathode reactions. And now we're going to be able to add up these two reactions. So we're gonna add our catholic reaction to are an odd reaction here. And what we would produce as our final reaction is manganese for oxide Plus four moles of hydride, Plus three moles of solid potassium produced from or produces manganese to oxide. Solid plus two moles of water in liquid form Plus three moles of potassium for our product here. And sorry, that says a Q. And so now we can go ahead and calculate our standard cell potential degrees cell by recalling that. We would take our cell potential of our cathode and subtract that from our cell potential of our an ode. And so from above, we would plug those values in for our standard cell potential. Where for our cathode, we said that we have a value of for the oxidation of manganese to oxide 1.68V from our reduction reaction. And then minus our self potential value for our node, which occurs at our oxidation reaction, which we set above, has a value of negative 2.92V for the oxidation of solid potassium. And so in our calculators, we're going to get a value equal to 4.60V as our final answer for our standard cell potential value here. So what's highlighted in yellow represents our final answer. I hope that everything I explained was clear. But if you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video.