Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li1s2 + Ag2CrO41s2 ¡ Li2CrO41s2 + 2 Ag1s2 (b) Choose the two half-reactions from Appendix E that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions?

Video transcript

Hello everyone. So in this video we're given this overall reaction right over here and we're trying to calculate the standard E. M. F. Generated by the Baltic cell. So let's go ahead and actually rewrite this overall reaction. So we can kind of break this down in a more accessible way. All right. So let's go ahead and look at each starting material as well as the products SARS ink or solid zinc. That is at its natural elemental state, which means that the oxidation number is equal to zero. Now, looking at the second starting material here, we know that the oxidation state or number of oxygen is always negative too. We have two moles of this. So that's going to be of course -4. And to balance out this um molecule, the oxidation number then of MN will be plus for our positive four. So of the oxygen that is negative to each and then of R. N. That's going to be positive for now looking at the products then. So this is the reactant. Then we can go ahead and look at the products. Okay, so for the products we have is the N. O. The oxidation number of auction of course, like we said, is -2. So the oxidation number of our medal, the zinc should be positive too. So the oxidation number again of our oxygen is negative two and the oxidation number of our zinc is positive two. Now, for our second products, the M. N. 203. Alright, so again the oxidation number of oxygen is -2. And because we have three of our oxygen atoms, that's going to be a total of negative six. So we expect that the cannon should have an equal number for the oxidation number to kind of balance it out. Because we have two moles of our metal, we just divide the six by two, giving us positive three value. So we have plus three. Okay, so we can see here that there is a change in oxidation state for first of all our zinc. So right here, so we're going from an oxidation state of zero to positive two. So we see an increase in our oxidation number, meaning that it's being oxidized. And this occurs at the note. Let's go ahead and actually put this in writing. So for our zinc, the oxidation number has increased. So it's being reduced. And this happens at the cathode or the anodes. I apologize. It's being oxidized. Okay, now, for the mm and you can see here that the oxidation state again is changing from a positive four to a positive three. So the oxygen number has decreased is being reduced. And this occurs at the cathode again, this oxidation number has decreased is being reduced. And of course the reduction occurs at our cathode. Now coupling for R. E. So then using the values that are given to us in either the textbook are given to you by Professor. I'm using the values I have. So R. S. L. Is equal to R. E. Cathode minus R. E an ode. And at the cathode we get a value of 0.15V -. Well, from a note, it's negative 1.28V. So putting all of this into my calculator. My sl will give me a voltage of 1.43V and that right there is going to be my final answer for this problem. Of course. We're also being asked, Let's see. We're being asked To provide two half reactions. Let's go ahead and take a look at this then. So because we already know that the channel is where the oxidation occurs and that's when we lose electrons. We go ahead and write this for our zinc. So let's go ahead and break this down to a node. And this is the zinc Reacting with two moles of our hydroxide. To give us cno of course H 20 and our two electrons. Okay, now, at the cathode then We have to MNO. two with water and adding the two electrons. Because again at the cathode is where reduction occurs and we're trying to gain electrons Then reproduce MN And two moles of our hydroxide. So here we have it. We have the reaction the half reaction of the node and the half reaction for our cathode. Because we found out through analyzing our overall reaction the oxidation number like what's occurring to that. And we are able to go ahead and write out the or predict the half reactions that occurs at the annual and the cathode, as well as our sl voltage.