Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K: (c) 10 Br -1aq2 + 2 MnO4-1aq2 + 16 H+1aq2 ¡ 2 Mn2+1aq2 + 8 H2O1l2 + 5 Br21l2

Video transcript

Welcome back everyone in this example, we're given the below electrode reduction potential. We are told that we have the following two reactions where in the first reaction we have the formation of chlorine gas recognize that that occurs when we have our chloride an ion which gains two electrons, which gives us our neutral chlorine gas. Now we want to recall our oxidation numbers and when we begin with our an ion that is going to have an oxidation number equal to its charge. So we would have an oxidation number of minus one. Now for the oxidation number of our chlorine gas in its standard state, we want to recall for anything in its standard state will have an oxidation number of zero. And so we went from a negative one oxidation number to a neutral oxidation number. So we have an increased oxidation number 2, 0. And so we would say that therefore the reaction is an oxidation and that is also due to the fact that this occurs at the anodes. If we're if we're undergoing an oxidation reaction, so we'll say that this is our an ode of our voltaic cell. Moving on to the second reaction, we have our Permanganate, an ion which gains eight protons and a total of five electrons. We form our manganese, two plus catalon and four moles of water. So we should recognize that we have oxygen in this poly atomic ion here, which has a oxidation number of negative two. And this is something that we memorize from our oxidation rules. Now. As far as manganese, we don't know its oxidation number. So we'll call it X, meaning we're going to need to solve for its oxidation number. And we would recall that we have X plus our oxidation number for oxygen, which we said is minus two. But we need to multiply by that subscript of four. So we would have negative two times four equal to our overall charge of our poly atomic ion, which is minus one. And we need to solve for X to get the oxidation number of manganese. And so we're going to say that we have X plus negative eight equal to negative one, meaning that we can say that X is equal to positive seven. And so this would be therefore the oxidation number Of Manganese is going to be Plus seven. And so if we know that, we can say that we went from plus seven to plus two as the oxidation number, Since we have the charge of plus two. And so we have a decrease in oxidation number of our manganese. And so we would say that therefore the reaction is an reduction, meaning that this standard electrode potential value occurs at our cathode of our voltaic cell. Now we want to recognize that outside of our atoms being balanced in both of these given reactions here, we need to balance out the electrons, we have two electrons in the first reaction and five electrons in the second reaction. And so to balance out the electrons, we're going to take both of these formulas or equations here and multiply them by the appropriate number of electrons. So we would multiply the first equation by five. And we're going to multiply the second equation by two. So that's going to give us 10 electrons for both of our reactions, meaning we can cancel out the electrons to get the overall reaction given below. But let's show how that works out. So we have when we multiply everything in the first equation by five, we have five moles of cl two gas plus 10 electrons yields 10 moles of our chloride an ion. And so moving on to our second reaction, we would have two moles of our permanganate an ion with 16 moles of our protons and electrons which gives us two moles of our manganese, Cagayan. And just to make more room, let's scoot this over With eight moles of our water and we can see that we can cancel out our 10 electrons here and here. And actually, since we know our first reaction occurs as an oxidation, we can actually write it out flipped. So we would actually write it out as 10 moles of our chloride. An ion where we have five moles of our chlorine gas and 10 electrons. So again we cancel the 10 electrons. And now we want to add up these two equations together to give us our overall reaction given in the prompt. And that's going to give us our 10 moles of our chloride an ion plus our two moles of our permanganate, an ion plus r 16 moles of protons yields us two moles of our manganese caddy on eight moles of water. And just to make room for our last free agent, we have eight moles of water and then we have our five moles of our cl two. So that's C L two here. And when we bounced out our electrons, we determined that we had a transfer of a total of 10 electrons. So we would say that n equals 10. Now, going back to the prompt, we need to figure out the value of our equilibrium constant. Right now we know the values of our anodes and cathodes of our self potential of our electrode with these two reactions. And so we're going to need to figure out our standard cell potential which is our term Enoch, which we should recall, is found by taking the cell potential of our cathode, subtracted from the standard cell potential of our a node. And so plugging in what we know from above. We're going to say that we have our standard cell potential equal to the cell potential of our cathode, which we determined above is 1.51 volts subtracted from our cell potential of our an ode which we determined has a charge of 1.36 volts. And so the difference here gives us our standard cell potential as the value 0.15 volts. Now that we have our standard cell potential value. We want to recall our formula to calculate standard cell potential where it's related to our gas constant. R multiplied by r temperature in kelvin and divided by are electrons transferred, multiplied by Faraday's constant, which is a capital F. And then we take the natural log of our equilibrium constant K. And we need to solve for our equilibrium constant K here. So we're going to plug in everything that we know we determined above that are standard cell potential is 0.15 volts. We're setting this equal to r gas constant. R. Which we should recall is the value 8.314 with units of jewels times moles divided by kelvin Plugging in our temperature. I believe we were given that from the prompt as to 98 Kelvin. So we'll plug that in. And then in our denominator we're going to plug in. Sorry, that's kelvin down there. So in our denominator we're going to plug in our electrons transferred, which we determined above to be our 10 electrons that are transferred overall. And then this is multiplied by Faraday's constant, which we should recall is the value 96,485 with units of columns per mole. This is then multiplied by the natural log of our equilibrium constant. So that's L N of K. Now let's go ahead and cancel our units. We can get rid of moles with moles in the denominator and kelvin here with kelvin and denominator here, leaving us with jewels per column as our final units. We want to recall that our equilibrium constant case should have no units. So we want to cancel all units. And if we recognize That we have jewels per column, we would recall that one jewel per column is equivalent to one volt. And so we can actually reinterpret what we have here and say that we have .15V taking the products of our quotient and dividing everything out. We're going to get a value of 0.002568. And we can say volts now, since we understand that a jewel per column Is equivalent to a vault and then this is still multiplied by the natural log of our equilibrium constant. So to isolate further, we're going to divide both sides by the value. We just came up with 0.002568V. So it cancels out on the right hand side 002568V. So this is going to give us our natural log of our equilibrium constant equal to a value of 58. because our units of volts cancel out here. So we have no units left. And sorry, this is a zero. So now we want to get rid of that Ln term by recalling that. We can cancel out Ln by using Mueller's number. So we're going to make both sides exponents to Euler number which is this term here. This gets rid of that Ln term. And now we can say that our equilibrium constant K is equal to the value of 2.33 times 10 to the 25th power. Now we want to pay attention to Sig figs and recognize that in our calculation our smallest number of Sig figs was one Sig fig which came from our electrons transferred being 10. And we recognize that 10 just has this one Sig fig from our number one here. And so that means our final answer should have only one sigfig as far as the decimal place that we allow. So we're going to put one sig fig and say that we have an equilibrium constant where k is equal to 2.3 times 10 to the 25th power since we just want one decimal place of one Sig fig. And so this here would be our final answer to complete this example as the value of our equilibrium constant for our overall reaction above. I hope that everything I reviewed was clear. If you have any questions, lead them down below and I'll see everyone in the next practice video