A rechargeable battery is constructed based on a concentration cell constructed of two Ag/Ag+ half-cells. The volume of each half-cell is 2.0 L, and the concentrations of Ag+ in the half-cells are 1.25 M and 1.0 * 10-3 M. b. What mass of silver is plated onto the cathode by running at 3.5 A for 5.5 h?

Video transcript

Hey everyone in this example we're told that a concentration cell made up of two iron two plus and two iron three plus half cells serves as the foundation for the construction of a rechargeable battery. Each half cell has a 1.5 liter capacity. There are two point oh five molar and 1.4 times 10 to the negative three solar of iron three plus concentrations inside each half sell how much iron is plated onto the cathode when it is operated at 1.9 ampules for 6.7 hours. So according to the ions mentioned in our prompt, we're going to show the redox reaction where we take our iron three plus kati on and we will add one electron to this cat ion so that as the product we would form the iron two plus cat ion And we would refer to our standard electrode potential table either online or in our textbooks and see that for the oxidation of iron two plus we have a self potential value equal to negative or positive. 0.77V. Now moving into the info from the prompt, we are given a time of 6.7 hours. So we're going to cancel out the units hours to get two minutes and we want to recall that in one hour. We have 60 minutes. So now we're able to cancel out ours and we're going to convert from minutes to units of seconds. So we should recall that for one minute. We have 60 seconds. We're going to cancel out minutes and now we're going to focus on converting from seconds into units of columns. By recalling that from the prompt. Were given a charge or Yeah, a charge of 1.9 ampules. And we would recall that because one ampule is equal to one column, sorry, Is equal to one column per second. We would interpret this as 1.9 columns per second. So we would have 1.9 columns in one second. Now we're able to cancel out seconds and convert from columns two moles of electrons by recalling our Faraday's constant, which is equal to a value which we recall as 996,485 columns for one mole of electrons. And so now we're going to cancel out units of columns. And just to make more room for our next conversion factor, we're going to convert from moles of electrons, two units of moles of iron. And because we recognize that we only transferred one electron, we would say that as our conversion factor, we have one moles of electrons transferred for one mole of iron. And so now we want to go ahead and cancel out moles of electrons and convert from moles of iron, two g of iron. And so we would recall our molar mass from the periodic table for one mole of iron which we c equals a mass of 55.85 g of iron. So now we can cancel out moles of iron. We're left with grams of iron and this is going to give us our quantity of iron that is plated onto the cathode as our final answer, which is equal to a value of 26.53 g of iron. And so this would be our final answer here for the massive iron plated onto the cathode when it's operated at 1.9 ampules for 6.7 hours. So if you have any questions, just leave them down below. Otherwise, I will see everyone in the next practice video.