During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (b) How many coulombs of electrical charge are transferred from Zn to MnO2?

Video transcript

welcome back everyone if 3.25 g of lithium is consumed at the batteries and no during the discharge of a lithium ion battery. With an overall reaction of lithium solid and cobalt oxide solid to produce lithium cobalt oxide solid. What amount of electrical charge and columns is transferred from lithium to cobalt oxide. So our first step is to write out our two half reactions. We have our lithium solid which is oxidized to our lithium carry on with the release of one electron. And so because we have an electron on our product side, we recognize that this will be our node or this reaction will occur at our an ode as an oxidation. Now for our second half reaction we have our lithium catalon And this says a queue here. This is reacting with our cobalt oxide solid which is going to be gaining an electron on our reactant side to form as a product. Our lithium cobalt oxide solid. Now because we have an electron on our reactant side, we would recognize this to occur at the cathode as a reduction and this is a queue here just so that's clear. We also want to make sure we have cobalt written out correctly. So that's not confusing because we next want to check to see whether each of our atoms are balanced in both of these reactions. We can see in our first reaction we have all of our atoms balanced and in our second reaction we also have all of our atoms balanced as well as electrons in both of these reactions the same amount here being one electron in each reaction. And so we can now add these two reactions together to form an overall reaction which is what we were given in the prompt as lithium solid plus our cobalt oxide solid, producing our lithium cobalt oxide. Now we want to make note of the fact that for one mole of electrons transferred, we have one mole of lithium. So writing that out, we have one mole of electrons transferred per one mole of lithium based on our reaction here. So beginning with that massive lithium consumed given from the prompt as 3.25 g of lithium. We're going to multiply to go from the molar mass of lithium where we have the molar mass of lithium from the periodic table which we see as 6.941 g of lithium for one mole of lithium. This is then multiplied by our conversion factor. Where we recognize that we have for one mole of lithium, one mole of electrons transferred according to our reactions above. Where we will then multiply by our constant. Which we should recall that for third is constant is for one mole of electrons equivalent to 96,485 columns. So 485. This is a five here columns. So canceling out our units. We can get rid of our units of gramps moles moles of electrons. And we're left with columns as our final unit. And in our calculators, plugging everything in carefully. We're going to yield the results of 45,177 which we can write in scientific notation as three sig figs because we have the minimum amount of sig figs in our prompt being three sig figs. So 23 sig figs. This would be written out as 4.52 when we round up times 10 to the fourth power columns. And this is our amount in electrical charge in columns that is transferred from our lithium to cobalt oxide. So I hope that everything I explained was clear what's highlighted in yellow is our final answer. If you have any questions, please leave them down below and I will see everyone in the next practice video.