Experimental solid-oxide fuel cells that use butane (C₄H₁₀) as the fuel have been reported recently. These cells contain composite metal/metal oxide electrodes and a solid metal oxide electrolyte. The cell half-reactions are (c) How many grams of butane are required to produce a constant current of 10.5 A for 8.00 h? How many liters of gaseous butane at 20 °C and 815 mm Hg pressure are required?

Video transcript

Hello. Everyone in this video, we're talking about multi carbonate fuel cells a previous M. C. F. C. These commonly use molten lithium and potassium carbohydrates are carbonates as electrolytes. And these have reactions are shown in these chemical reactions right over here. So of course we have our anodes and cathodes site. So in part a we're being asked what is the mass of hydrogen gas and grams needed to generate a constant of 12.4 amps for 10 hours? And the part B we're being asked what is the volumes and leaders of the hydrogen gas? If the required conditions are 650 degrees Celsius and 1406 millimeters of mercury. So in our first part here, we're gonna go ahead and find out the overall net reaction. So again, we're giving the half reactions on a note and cathode side, we can see what starting materials and products will cancel out. So what I can see right off the bat is going to be our C. 032 minus. That's gonna go ahead and cancel out. Our carbon dioxide will go ahead and also cancel out and most importantly the two electrons will cancel out. So whatever we have remaining is going to be our overall net reaction. So this reaction means that for our starting materials we have our H2 guests as well as half a mole of our auction to gas. This is going to of course just yield one mole of H20 and its liquid state. So some things we have to go ahead and remember is that one amps is equal to one column per second are fair days constant Is equal to 964, columns per moles of electrons. And then the molar mass of H two is equal to 2.16 units being grams per mole. So let's go ahead and calculate for the mass of H two. Then we'll do this in a different color here. We're gonna go and scroll down slightly for more space. So again we're calculating for the mass of our H two Starting off with our 10 hours, so 10 hours. So each we're gonna go ahead and convert this into minutes and then seconds. So of course we know that for every 60 minutes we have one hour now that we're in terms of minutes, we're gonna go ahead and say that one minute Or for every one minute we have 60 seconds to put as us. So for the unit cancelation here we see the our counseling and the minutes canceling. So we're gonna do now is convert our seconds into columns. So for every one second that we have, we have 12.4 columns. And then from here we go into moles of electrons. So for every 96485 columns, this is one mole of electrons. Just keeping everything up to date here. We see seconds cancel and columns cancel. If you recognize this, we have this being the Faraday's constant. Continue our dimension analysis here for the overall reaction for every one mole Of Our H two. We have two moles of electrons. So then the moles of electrons will cancel now that we're in terms of the moles of H two. We finally use the molar mass to go from moles into grams which is the units of mass. So continuing on the dimensional analysis here, the last conversion factor that we need is of course, like we said, the molar mass. So for every one mole Of H two we have 2.016 g of H two. So again, just making sure We can see here that the moles of H2 will cancel leaving us with just the units of g. So once I put everything into my calculator, the mass of H2 that we get is equal to 4.6636 g. And is running this to the appropriate amount of significant figures, which is three significant figures is equal to 4. g. So that's our answer for part a here. Now, working for our part to be, which again is the volumes in leaders of the hydrogen gas. All right, so that was for a Now we're moving on to part B. Alright, so we're here, we're gonna go ahead and use the ideal gas law equation that is pivoted or PV equals NRT. As already said, we are solving for volumes. So right now I'm gonna go ahead and manipulate my equation to solve for volume. So what I'm gonna do is of course just isolate my volume which is V. And divide both sides by pete. And we do. So we get the equation for volume to be equal to N. R. T. Divided by P. So we see here that what we can plug in one being our temperature. So that's T. Were given this at 650 degrees Celsius. We want this to be in the standard units for temperature. That is kelvin's. We're gonna go ahead and add to 73.15 and this gives us 923.15 Kelvin's for the pressure. We're given this in m of mercury. So that's 1406. We're gonna want to convert this into A. T. M. So for everyone a tm that we have, we have some 160 millimeters of mercury. So you see here with this conversion, the millimeters of mercury will cancel out. This gives us 2.3133 A. T. M. So now we have all the numbers, you go ahead and put into our equation right over here. So using that equation then Volume is equal to the moles is 2.3133 moles. Our our is our ideal gas law constant. So that never changes to the 0.8 to 06. The units here is leaders times A. T. M Over, moles, times kelvin's. And then for our t for temperature that is already calculated to be 9-3.15 Kelvin's. And then for our denominator, that's just one value, that's pressure. So that's 2.3133 A T. M. So for the units here we can see that the moles will cancel out, The kelvin's will cancel out and of course the A. T. M. Will cancel out. So once I put everything into my calculator, the numerical value that we get, of course we're running to three significant figures. That is 94.7 and units is going to be leaders. So this right here is going to be my final answer for this problem.