Calculate ∆Gr°xn and K for each reaction. b. The reaction of Cr3+(aq) and Cr(s) to form Cr2+(aq). [The electrode potential of Cr2+(aq) to Cr(s) is -0.91 V.]

Video transcript

Hey everyone in this example, we need to determine our equilibrium constant and gibbs free energy of our reaction at 25 degrees Celsius Using tabulated electrode potentials for the reaction of hipaa course acid with chloride to produce chlorine gas. Now we should recognize that this reaction is a redox reaction, meaning we need to write out two half reactions are first half reaction is going to come from our first reactant, which is our Hipaa Cloris acid, which produces chlorine gas as a product. So our first step is to make sure that the atoms are balanced on the product side. We have two atoms of chlorine on the reactive side. We have one atom of chlorine. So we're gonna balance the chlorine by placing a coefficient of two here. However, we should recognize that. Now we have also two atoms of hydrogen and two atoms of oxygen where we have none of hydrogen or oxygen on the product side. So we should recall that first. We want to balance out oxygen and we should recall it to bounce out oxygen. One atom of oxygen is equivalent to one mole of water, so we can use water to balance out oxygen, meaning that to get two moles of oxygen on the product side, we would go and add two moles of water to the product side. Our next step is to balance hydrogen. And so to balance hydrogen, we should recall that one mole of hydrogen is equivalent to one mole of hydride or our proton here. And so we should recognize that now we have four moles of hydrogen on the product side, where we only have two on the reactant side. And so this means that we're going to go ahead and expand our reactant side so that we would now add two moles of hydride. So that that would give us a total of four hydrogen on the reactant side and four hydrogen on the product side. So now all of our atoms are balanced as far as chlorine, hydrogen and oxygen. And our next step is to recognize that because we added these two moles of hydride, we now have a net charge of plus two on the reactant side. Whereas our product side here has a net neutral charge because we have neutral molecules. So we should recognize that we need to cancel out this plus two net charge. And we would do so by adding two electrons so we can go ahead and just add that in here. So we can say plus two electrons to get rid of that plus two charge that we would now have a net charge of zero on both sides of this half reaction. So for our second half reaction, we want to go ahead and write out from our second reaction here, chloride. So this is cl minus which produces also chlorine gas as a product. And we would recognize that we went from one atom of chlorine to two atoms of chlorine. So we need to bounce out our chlorine atoms by placing a coefficient of two in front of our chloride. An eye on our next step is to bounce out the charges we have now a coefficient of two in front of a minus one charge, meaning we have a net minus two charge on the reactive side. Whereas on the product side we have a net neutral charge because we have a neutral atom of chlorine gas, meaning that to form our two moles of the chloride an ion, we must have gained two electrons here on the product side. So we can write that in plus two electrons here which formed our two moles of our chloride an ion. So by losing these two electrons from our neutral molecule of chlorine gas, we should recognize that this is going to be our oxidation half of our reaction. Whereas our first reaction we gained two electrons on the reactant side here, so because we have two electrons gained on the reactant side, this is going to be our reduction half of our reaction. So now that we have our charges and our electrons balanced, we're going to cancel out what is the same on opposite sides of both of our half reactions. And that would be our number of electrons here. So we can cancel out the two electrons with the two electrons here on the product side from our second half reaction and adding these two reactions up. We're going to get an overall reaction where now we can say we have two moles of our hipAA course acid Plus two moles of hydride Plus two moles of chloride, producing two moles of chlorine gas Plus two moles of water. We also want to make note of these cell potentials for each of our half reactions. So we're gonna go ahead and look at our first half reaction here. And we want to recall that we stated that it occurs as a reduction half reaction. Where when we refer to our standard reduction potential table, we can find this either online or in our textbooks. We would see that for our reduction half reaction, we're going to have a self potential value equal to according to this table for hipAA course acid. So H C L O. The cell potential is 1.61. Now for our oxidation half of our reaction here where we have chloride. We're gonna refer to our standard reduction potential table either online or in our textbooks and we would see that it has a self potential value for chloride equal to a value of negative 1.36 volts. So these are in units of volts. And now we want to go ahead and calculate using these values our standard cell potential such as E. And we should recall that that is equal to our cell potential of our an ode minus or sorry, plus r cell potential of our cathode. So we should recall that our self potential of our cathode occurs at the reduction half reaction. So this would be our cathode. And we should recall that the cell potential of our A node occurs at the oxidation half reaction. And so what we would have is our standard cell potential is equal to negative 1.36V from our a node added two positive 1.61 volts from our cathode. And this will give us our standard cell potential equal to a value of 0.25 volts. So now we want to go ahead and move into calculating for our gibbs free energy of our reaction. And we should recall that we're going to use the following formula where we recall that gibbs free energy for reaction is equal to negative one times our electrons transferred at the reduction. N multiplied by Faraday's constant F multiplied by our standard cell potential. We also want to recall that our units of volts are equivalent to jules divided by columns. So we can or we can substitute jewels divided by columns for our units of volts. So now we're going to plug in what we know into this equation. So we can say that gives free energy of our reaction is equal to negative one times our electrons transferred at our redox. So we said our sorry at our reduction. So for our reduction half reaction, that was for hipaa Cloris acid. And we said that we transferred two electrons here. So n would equal to So you would say negative one times two. I'm sorry, that should be blue. And to be even more specific we can say two moles of electrons were transferred Which is multiplied by Faraday's constant f. We recall that that is equal to a value of 96, columns per mole of electrons. And then above we calculated our standard cell potential equal to a value of will plug that in as purple 0.25. And we're not going to use units of volts because we're gonna use units of joules per column. And so canceling out our units, we can get rid of columns, we can get rid of moles and we're left with jewels. And so this would tell us that gives free energy for reaction is equal to a value of negative 48, 242.5 jewels. However we should recall that gibbs. Free energy should be in units of kilo jewels. So we're going to multiply by our conversion factor where we would go from jules, tequila jewels. By recalling that our prefix kilo tells us that we have for one jewel 10 to the negative third power kilo jewels. So now we're able to cancel out jewels as our next unit leaving us with kilo jewels as our final unit for gibbs. Free energy. And this gives us a value for gibbs. Free energy of our reaction equal to 48. kg jewels which is a positive value here or sorry, that should be negative, A negative value here. And we should recall that because our value for Gibbs free energy is a negative value here. This means that therefore the reaction is spontaneous. So this would be our first answer here for gibbs free energy. And now we need to calculate our equilibrium constant according to the prompt. Now, according to the prompt, the temperature that this reaction occurs at is equal to 25 degrees Celsius. We need to convert this to kelvin. So we're going to add to 73.15. And this is gonna give us our kelvin temperature of 2 98.15 kelvin. And so now we can recall our formula here below where Gibbs free energy of our reaction is also calculated by taking the negative value of our gas constant. R multiplied by the temperature in kelvin. And then we're going to multiply this also by the Ln of our equilibrium constant. K. And so we're going to reorganize this to solve for equilibrium constant K. By saying that K. Is equal to E. To the inner exponents. We would have negative one times Gibbs free energy. So delta G of our reaction which is divided still in the exponents by r. Gas constant. R times the temperature in kelvin. So we're going to use this to calculate our equilibrium constant. So what we would have is that e is equal to negative and then in the numerator, we're gonna plug in our value for gibbs free energy, which we found above as negative 48.2 kg jewels. However, we're gonna plug in the jewel format that we calculated because we want to recall that r gas constant R uses units of jewels and we want to get rid of those units. So we're going to plug in instead -48242. jewels. And then this is divided by r gas constant. Art which we recall is equal to a value of 8. units of joules divided by moles times kelvin, Which is then multiplied by temperature, which we converted above as 298. Kelvin. And so canceling out our units. We can get rid of Kelvin, we can get rid of jewels and we're left with moles. However, we should recall that the equilibrium constant does not have units. So we're going to just cancel out moles as well. So putting all of this to an exponents or to our exponents Of E. We should get a value in our calculators equal to and sorry, this should be blue. So we're going to get a value equal to 2.83 times 10 to the eighth power for our equilibrium constant. So this would be our second and for final answer as our equilibrium constant as well as our value for gibbs, Free energy for reaction using standard reduction potentials. So I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.