(a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5 * 104 A flowing for a period of 24 h. Assume the electrolytic cell is 85% efficient. (b) What is the minimum voltage required to drive the reaction?

Video transcript

Hello. Everyone in this video for part a. We're going to try to determine the mass of CC informed during the electrolysis of cesium chloride. And for part B we're trying to determine the minimum voltage needed to drive this reaction. So starting off with part a were given that the sales efficiency is at 78%. Of course we can go ahead and turn that into a conversion unit. So of course we're also going to use Faraday's constant. Let's go ahead and remind ourselves what that constant is. And that is columns per mole. Okay, so starting off with that then ra wanna go ahead and cancel out this denominator unit. Of course we're going to use the cell's efficiency. So we have 0.78 moles of c. c. um at the bottom We see that our unit will cancel out, leaving us with the units of columns per mole. We put the values into our calculator, we get the value of 1.2370 times 10 to the fifth columns per mole of CCM required. Alright, so now dealing with our electric current which is not as a the union says ccM per second, We have a given value of that to be 2.5 times 10 to the third columns per second. And of course to calculate for the molar mass of CCM That is 132.9 g per mole. So having all these unit conversions, we can go ahead and start the dimensional dimensional analysis for the mass of ccm. So it says the electrolysis period is of 12 hours. So we're gonna go ahead and start with that. So have to 12 hours and we're gonna go ahead and convert this in two seconds. I don't know what the direct conversion for this. So I'll just use units that I do know. So starting off with ours we're gonna go ahead and convert that in two minutes. So for every one hour we have 60 minutes And for every one minute we have 60 seconds. So we see here that the hours cancel and the minutes cancel. Now bring in our given here for our electric current. We have 2.5 times 10 to 3rd columns per second. So now we have canceled out the units of seconds. Now for the Faraday's constant, We have one mole of ccm For 1.2370 times 10 to the 5th columns. And now finally for the last part we're gonna go ahead and use the molar mass of CCM. So again that is 132.9 g of C. C. Um for every one mole of C. C. Um we see now that the moles of CCM will go ahead and cancel out and the remaining unit will be just the grams of CCM, which is the mass of CCM. So putting all the numerical values into my calculator, I get the value of 1.2 times 10 to the fifth grams of CCM. So that's my final answer for part A. Now dealing with part B, which is the minimum voltage needed to drive our reaction. Let's go ahead and scroll down a little bit to give us more space. So dealing with the minimum voltage, that's the standard cell potential. And if we have a electrolysis of molten cesium chloride, we know we're dealing with c c. M plus, we're just going to get reduced at the cathode and then we have our chlorine, an ion which is going to be oxidized at the anodes. Let's go ahead and write the reactions for those. So for CCM it's being reduced. So we're gonna go ahead and add electron to give us the CCM in its neutral state. And here I'm just going ahead to write out our standard reduction potential. And these are just values given to us either in a textbook or may be provided to you by your professor and then further chlorine. We have cl to the diatonic molecule and then we're going to go ahead and add two electrons is to require two moles of our chlorine and ions. Again, the standard reduction potential for this is -136V is going to occur at our a note. So dealing with the reduction potential then of ourselves, that equation goes, but the S. L. Is equal to the cathode minus R E a node, plug in those values that was given to us by our textbook we see that R. S. L. Is equal to negative 303 volts minus R, negative 1.36 volts And then putting that into our calculator. We get the value of negative 1.67V. Therefore, answer then, is that the minimum voltage needed is going to be negative 1.67V, and that's going to be my final answer for part B.