Consider a galvanic cell that utilizes the following half-reactions:

(b) What are the values of E° and the equilibrium constant K for the cell reaction at 25 °C?

Video transcript

Hello in this primary told a certain party Excel uses the half reaction shown below. Using the thermodynamic values below 25 degrees Celsius were asked to calculate the standard cell potential and the equilibrium constant for the overall cell reaction, we can relate the change in Gibbs free energy to the standard cell potential and we can relate the change in gibbs free energy to our equilibrium constant. We can find the change that gives free energy from the entropy and the entropy. We're gonna need to know our temperature in Kelvin. So we have 25°C plus 273.15. Our temperature is 298.15 Kelvin. In order to find the change in Gibbs strategy, we need to calculate change in entropy and entropy. And this requires having a balanced reaction equation. So if we look at our two half reactions, we see that the number of electrons lost does not equal that which are gained. And so we need to multiply then the reaction taking place at the cathode by two So that our moles of electrons being lost is equal to those being gained. We'll combine these two to get our overall reaction. So we have then zirconium and two moles the manganese too Plus two moles of water point of form the core team oxide plus for most of Hibernians plus two moles of my agonies. So we can calculate then the entropy or heat of reaction. Using the values in the table. The entropy reaction then is the sum of the please of formation more products minus the sum of the entities of formation for reactant. So this then works out to we have one mole zirconium oxide times it's entropy of formation Plus formals of Hydra nine times its entropy of formation which is zero. Class two moles Of Manganese Times. Its entropy of formation which is also zero. Then we subtract off, we have one mole zirconium Times its entropy information which is zero plus two moles, Manganese, two plus times its entropy of formation. Oops and then we have two moles of water times is entropy information. So this works out then to negative 87.4 killed jules. So our moles will cancel. We were left with little jewels. You can similarly find the entropy the reaction people to them, the some of the entropy is of formation for our products minus the sum of the entropy s formation for reacting. This then works out to one mole of chromium oxide times its entropy of formation plus formals of Hydra nines times its entropy of formation which is zero Plus two moles of manganese and its entropy of formation then dragged off our reactant. We have one mole of zirconium times it entropy of formation Lost two moles of manganese too times its entropy information And then two moles of water times its entropy formation. So our moles will cancel. We'll be left with jewels per kelvin. This will work out to then 211. tools for Calvin, you can now calculate the gibbs free energy for the reaction. We have our change in entropy minus temperature times the change in entropy. So we have 87.4 Phil jules minus 2, 15 Kelvin Times are entropy of two or 11.6 joules per Kelvin. We'll convert then from jules kill jules. So our units of kelvin will cancel, jewels will cancel. We're left with kill jules combining these. Then this works out to - .489 bill jules. We can now use this to find the standard cell potential rearranging equation to get the standard cell potential by itself We then have negative 150.49. Kill jules divided by negative four moles of electrons. Number of electrons that were transferred. We have a fair day which is the charge on one mole of electrons Is equal to 96,485 columns. And then we're going to convert our pill jewels, two jewels, one kill jules equal to 1000 jewels. So we have killed jules, cancels our moles of electrons cancels and a jewel per column is a volt. So this will work out to then 0.390V. So this is our standard cell potential. Then we're asked to find the equilibrium constant again. Making use of our changing gibbs free energy. This is then equal to negative rt on of k. So we can then isolate our equilibrium constant. They then will be equal to E to the negative change in gibbs free energy over gas constant divided by temperature We have and the negative of our Gibbs free energy, which is negative. 150 .489 killed Jules. Bye bye. Are A .314 times 10 to the -3 Phil jewels. For kelvin Times are temperature of two or 98 15 kelvin. Sorry, units have killed jules cancels, kelvin cancels. And so our equilibrium constant. Okay Then works out to 2.3, 2 times 10 to the 26. So our equilibrium constant and our standard cell potential correspond. Then to answer a thanks for watching. Hope to help.