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 (b) Use the thermodynamic data in Appendix B to calculate the values of E° and the equilibrium constant K for the cell reaction at 25 °C. Will E° and K increase, decrease, or remain the same on raising the temperature?

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Welcome back everyone. Molten carbonate fuel cells. MC FC are newly developed fuel cells that commonly use molten lithium and potassium carbonates as electrolytes. The half reactions in the cell are shown below. We have an anote and cathode half reaction using the thermodynamic values below determine the cell potential and the equilibrium constant K. For the overall reaction at 25 Celsius identify whether the values of eno and K will increase, decrease or remain unchanged as the temperature decreases. We're also given a table with our entropy formation for hydrogen oxygen and water as well as our entropy values for each of those reagents. Looking at first our anote half reaction, we have hydrogen gas reacting with carbonate or two carbonate two minus anion to form liquid water and carbon dioxide gas. And the release of two electrons. For our cathode half reaction, we have one half mole of oxygen gas reacting with carbon dioxide gas and one elect or rather two electrons to form one mole of our carbonate two minus anion. Notice that because electrons are on the reactant side for the cathode reaction. This reaction is a reduction and because electrons are on the product side for our a node reaction, this is our oxidation. And this would make sense since we recall that oxidation occurs at the A node and reduction always occurs at the cathode electrode. We need to come up with an overall reaction. And since these half reactions are already balanced with the same number of electrons being two in each reaction, we can just simply add them together. So this would allow us to then cancel out the compounds that we see repeat. We see that we have one mole of carbonate. For our A node reaction on the reactant side and one mole of carbonate on our product side of our cathode reaction. We also find carbon dioxide gas on the product side in our A node reaction. And then for our cathode reaction, we find carbon dioxide on the reactant side. So we can cancel those out as well as the two electrons on the product side in our first reaction and the two electrons on the reactant side in our cathode reaction. So now adding everything that's left, we have hydrogen gas plus a half mole of oxygen gas forms one mole of water plus. And rather that would be our only product, just one mole of water. This is our overall reaction. Let's next recall that we can calculate our standard cell potential eot or standard reduction potential and our equilibrium constant K using Gibbs free energy change delta G because one of our formulas for Gibbs free energy change involves entropy change. We're going to need to calculate our entropy change delta H for this overall reaction. And so recall that for the entropy change of our reaction, we would take our entropy formation of our products subtracted from our anthropic formation of our reactants. So beginning with the entropy of formation of our products for our overall reaction, our only product is liquid water. And according to the chart, it has an entropy of formation of negative 2 52 85.8 kilojoules per mole. So we have one mole of water on our product side multiplied by its enro of formation given as negative 2 85.8 kilojoules per mole. This completes the sum of entropy formation of our products. We're going to subtract by the entropy formation of our reactants. In which for our first reactant, we have one mole of hydrogen gas multiplied by its standard entropy formation. Given in the chart as zero kilojoules per mole. Adding to this, we have our second reactant and I'll just have to make more room. So our second reactant is our one mole of or sorry, not one mole. But from our overall reaction, we have one half mole of oxygen gas multiplied by its elia formation. Also as zero kilojoules per mole as given in our chart notice that we can cancel our units of moles with moles in the denominator. So this should say one half moles of oxygen that's canceled out there. And in our next line, we'll find that our entropy change of our reaction overall is equal to negative 285.8 kilojoules again because moles cancels out. Recall that we also use entropy within one of our formulas of gives gibbs free energy. So we're going to need to calculate our entropy change for our reaction. So that's delta s for our overall reaction. and recall that it's calculated by taking our standard entropy of our products subtracted from. So we subtract from the entropy change of our reactants. And specifically, this is the sum of the entropy change of products minus sum of entropy change of our reactants. So plugging in our values from our chart, we'll find that our entropy change of our reaction overall is equal to beginning with the sum of entropy change of our product. We have just one product being our one mole of water multiplied by its entropy change given as 69.9 Jews per moles times Kelvin. This completes the sum of entropy change of our products subtracted from the entropy change of our reactants in which we begin with our first reactant, one mole of hydrogen gas multiplied by its standard entropy change of 69.9 Jews per mole times Kelvin. Then adding to this, we have our one half mole of oxygen gas just making more room. We will multiply this. Sorry. So we have one half more here of oxygen gas multiplied by its standard entropy change. Given in the chart as 205.0 units of jewels per mole times Kelvin canceling out our units of mol. We're left with jewels per Kelvin as our final units for entropy change. And in our next line, our first bracket for the sum of entropy change of our products, we still have 69.9. Again, jewels per Kelvin subtracted from the result of the sum of entropy change of our products, which is minus 172.4 jewels per Kelvin. And this will result in our entropy change for the reaction of end apologies. I just would like to make a correction because our entropy change of hydrogen gas is given in the chart as rather 130.6 Jews per mole times Kelvin. So this would change the sum of entropy change of our reactants to a value of 233 0.1 joules per Kelvin. And so now taking the difference between our entropy change of our product, 69.9 jules per Kelvin minus our entropy change of our reactants. 2 33.1 Jules per Kelvin, we're left with the value of the difference as negative 163.2 Jews per Kelvin as our entropy change of our overall reaction. Now, so far take note that our calculated entropy change of our reaction was negative. And so because this is negative, we can say that therefore, the reaction is going to be exothermic because again, we have a negative delta H value 2 85 point negative 2 85.8 kilojoules. Now, with our entropy change and entropy change values, we can recall our first formula for gibbs free energy change in which our Gibbs free energy change of our reaction overall is equal to our entropy. A formation of our reaction or sorry, not entropy of formation. But our entropy change for our reaction delta H which is then subtracted from our temperature of our system multiplied by our entropy change of our reaction. So plugging in our values that we solved for above, we would find that our gibbs free energy change of our reaction is equal to our entropy change which we just calculated above as negative 2 85.8 kilojoules. This is then subtracted from our temperature of our system in which according to the prompt, we have a temperature at 25 Celsius. So if our temperature is 25 Celsius, we would convert this to Kelvin by adding 2 73.15 to get 2 98.15 Kelvin. So we would subtract from 2 98.15 Kelvin and then multiplying this temperature by our entropy change of our reaction, which we just calculated to be negative 163.2 Jews per Kelvin. So it makes sense to convert our temperature to Kelvin so that we can cancel out our units of Kelvin with Kelvin in the denominator. And next we call that our unit for Gibbs free energy change is in jewels. So we need to cancel out kilos and we're going to have to expand our equation so that we multiply our entropy change of our reaction by a conversion factor to go from kilojoules in the denominator to Jews in the numerator. So this comes before our subtraction sign and recall that our prefix kilo tells us that we have an equivalent of 10 to the third power of our base unit. Jeel. So canceling out kilos, we're left with our units of jewels as our final units which we're subtracting from. And this is going to simplify to a value of negative 237,141 0.92 jewels as our gives free energy change of our reaction. Now we're going to recall our next formula for calculating gibbs free energy change, which is related to our value for Eno where negative one is multiplied by N or moles of gas, which is then multiplied by fair constant and then multiplied by eno because we need to find Eno as one of our final answers. We're going to isolate Eno by dividing both sides by negative N times F so that they will cancel it on the right, and then we'll have negative N times F division on the reactant side or on the left side. And so now we will find that our term for Eno, which is our salt potential is equal to our GPS free energy change of our reaction divided by negative one times our moles of gas times fa constant. So plugging in our numerator, we found that our GPS free energy change is equal to negative 237,141 0.92 Jews. And just to make a correction, recall that N here represents not moles of gas, but rather moles of electrons transferred, which is going to come from our reaction above where we saw that N is equal to two electrons because two electrons were released in our A node reaction and two electrons were gained in our reduction reaction. So here and equals two, we can say note that N is equal to two electrons. So in our denominator, we'll plug in negative one times two electrons for N multiplied by fair constant, which we should recall from our textbooks of the value 96,485 colum per moles of electrons recall the conversion factor that one jewel, sorry, one jewel per colum is equivalent to one volt. So when we simplify our quotient, we're going to come up with a value equal to 1.2289. Notice that we can cancel out our units of electrons with multiple electrons in the denominator leaving us with jewels per colum as our final units in which we understand is equivalent to 1.2289 volts. And so for our cell potential eno we can say that it's equal to a value as three sig figs being 1.23 volts. So this would be our first answer so far as eno the cell potential of our reaction being 1.23 volts. Now we're going to need to recall our next equation for Gibbs free energy change delta G which is found from taking negative one times our gas constant R times our natural log Ln of our equilibrium constant K, we need to isolate for K and apologies. So this equation is not correct. As written, it should be gives three energy change equal to negative one times our gas constant R times our temperature in Kelvin times our natural log of our equilibrium constant K. So now in this case, to isolate for the Ln of K, we're going to divide both sides by negative RT. This will tell us that our natural log of our equilibrium constant K is equal to or gives free energy change divided by negative RT. So plugging in again, our gibbs free energy change in our num. We found that above as negative 237,141 0.92 Jews, we're going to interpret it as jewels per mole of our reaction. Since it's the free energy change per mole of our overall reaction, this is then divided by negative one times our gas constant R which from our textbooks we should recall is the value 8.314 jewels divided by moles times Kelvin multiplied by our temperature in Kelvin which again, we have converted above to 2 98.15 Kelvin. So canceling out our units. Notice we can cancel out Jews with Jews in the denominator moles of our reaction with moles in the denominator. And Kelvin, we're left with no units and recall that this makes sense because our equilibrium constant does not have any units. So simplifying this quotient, we're going to come up with a value equal to 95.66729577. Now, in order to cancel out our Ln term recall that we can utilize Ehlers number on both sides of our equation to cancel out Ln. And what we will find is that our equilibrium constant K is equal to a value of 3.53 times 10 to the positive 41 power. So this is going to be our second answer as our equilibrium constant of our reaction. Notice that temperature being decreased is going to impact our two terms, which is our standard cell potential and our equilibrium constant. Because in our second Gibbs free energy equation, we see temperature, on the right hand side of the equation where we have our cell potential and our equilibrium constant. On the left hand side of our equation, we can say that our cell potential eno and our equilibrium constant have an inverse relationship with temperature. And so therefore, as temperature increases our cell potential and equilibrium constant will decrease. On the other hand, as temperature decreases our cell potential and equilibrium constant will increase. And so because the prompt says that temperature is going to be decreasing, that will mean that we should expect our cell potential and equilibrium constant to increase. So what's highlighted all in yellow represents our three final answers which will correspond to choice D and the multiple choice to complete this example, I hope this made sense and let us know if you have any questions.

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Key Concepts

Electrochemical Cell Reactions

Standard Electrode Potential (E°)

Equilibrium Constant (K) and Temperature Dependence