A battery relies on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0 * 10-4 M and 1.5 M, respectively, in 1.0-liter half-cells. c. How long can the battery deliver 5.0 A before going dead?

Video transcript

hey everyone in this example we're told that a battery works on the reduction of lead two plus in the oxidation of barium in separate one liter half cells. We have 1.50 times 10 to the fourth power moller of barium two plus and to moller of lead two plus which are initially present. We need to calculate the time that the battery can deliver six ampules before it goes dead. So first beginning with our lead two plus catalon, we're going to show that it will gain two electrons to form solid lead as a product. Now because we added these two electrons to the reaction half of this reaction, this will or we should recall that this means that this reaction is occurring as a reduction. And recall that our reduction occurs at the cathode of our electrode. Now moving into our second reaction, we have to form our barium solid metal that is formed from our barium two plus kati on gaining two electrons. And so in this case we added two electrons to the product side of this reaction, meaning that this reaction is an oxidation reaction which will occur. We recall at the anodes of our cell. So now combining these two reactions, we can see that because they both have two electrons involved, we can say therefore we have two electrons transferred For our following reaction where we have solid barium reacting with lead two plus to form barium two plus and solid lead as a product. So according to the prompt we have in one liter of solution to moller Of Arkady and have led to plus. And we should recall that polarity can be interpreted in moles per liter of lead two plus where we can focus on converting from moles of lead to moles of electrons and then from moles of electrons we're going to convert from units of columns and then from columns we're going to convert two seconds from seconds to minutes and from minutes to hours for our final answer for our time. And we would recall that also for one ampule that is equivalent to one column per second. And so for our six ampules according to the prompt, we would say that that is equivalent to six or 6. columns per second. So beginning with our volume from the prompt we have one leader of solution which we will multiply by from the prompt. Our two moles of lead and sorry moles should be in the numerator. So we would have two moles of lead And this is for one liter of solution. Next we will cancel our units of leaders and focus on converting from moles have led to moles of our electron and we would recognize that for one mole of our lead two plus cat ion, we had a transfer of two moles of electrons. So now we're able to cancel out our moles of lead and now we're going to focus on converting from moles of electrons, two units of columns by recalling our Faraday's constant which has a value of 96,485 columns for one mole of electrons were able to cancel out moles of electrons and now convert from columns two seconds by taking the charge from the prompt, which is given as 6.00 columns per second, which we will now use to cancel out columns and move from seconds into minutes Where we would recall that in 60 seconds we have one minute we can cancel out seconds. And just to make more room, we just need to add our last conversion factor in to go from minutes to hours. So we would recall that we have 60 minutes and one hour. We can now cancel out minutes and we're left with ours as our final unit for time And what we're going to get here is a value equal to 17.9 hours as our final answer. So this here would be our final answer for the amount of time that it would take our battery to deliver six ampules of power before it can go dead. So I hope that everything I explained was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video