Use line notation to represent each electrochemical cell in Problem 43.

Video transcript

hey everyone in this example, we need to write the cell notation for the given reaction below. So according to our given reaction, we have three moles of copper two plus reacting with two moles of aluminum to give us three moles of solid copper as well as two moles of Aluminum with a three plus charge. So we should recall that to write out cell notation. We're going to use the following acronym. As easy as abc where a represents our an ode, B represents our physical barrier, which is represented by this symbol here. So be as our barrier and then C is going to represent our cathode of our cell. So what we should recognize is that based on our given reaction which is an oxidation or redox reaction? Sorry, we're going to write out two half reactions are first half reaction comes from our first reactant which is the three moles of our copper two plus carry on Which produces as a product three moles of solid copper. Now, what we should recognize is that we have balanced atoms here, we have three atoms of copper on the product side and three atoms of copper on the reactant side. However, we do not have balanced charges. We have a net charge of we have coefficient of three times plus two which would give us a net charge of plus six on the reactant side. And we have a net charge of zero on the product side. And so we need to balance this out by expanding our reactant side so that we would add Six electrons to cancel out that net charge of Plus six. And because we added electrons to our react inside, we would recognize that this is going to be the the half reaction that occurs as a reduction. Now looking at our second reactant we have according to the given equation, two moles of solid aluminum, Which forms our product two moles of aluminum as a 3-plus cat alone. And we should recognize that we do have balanced atoms. We have two atoms of aluminum on both sides. However, we do not have a balanced net charge, so we have a net neutral charge on the reactant side. However, on the product side we have two times this plus three charge, which gives us a net charge of plus six again. And so we need to cancel out this plus six charge here, meaning we would go ahead and add six electrons here on the product side. And because we added electrons on the product side, we would recognize that this half reaction here occurs as an oxidation. Now, what we should recall is our reduction half reaction occurs at our cathode, whereas our oxidation half reaction will occur at the anodes of our cell. And so now we knowing all of this information are able to fill in our cell notation. So because we've identified are an ode as our oxidation half reaction. We can write out our cell to have our oxidation half reaction, which involves aluminum, solid aluminum, which then we would place it in a bar here where we have our product being our aluminum three plus carry on. And then now we want to include our barrier. So we're going to have that barrier here. We're continuing on our cell notation. We're going to have our reduction half reaction, which occurs at our cathode of our cell. And we agree that that involves our copper two plus catalon as well as our product, which is our solid copper metal. And so this notation here would be our sell notation to complete this example as our final answer. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video.