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Current time:0:00Total duration:9:43

- [Voiceover] Our goal is to calculate the equilibrium constant
K for this reaction, so for this reaction right here. Now we're gonna use the standard reduction
potentials to do so. So in the previous video, we
talked about the relationship between the equilibrium constant K and the standard cell potential. So E zero. So if we can find E
zero for this reaction, we can calculate the
equilibrium constant K. And we've seen how to find E zero, the standard cell potential
in earlier videos. So, let's start with this
first half reaction here, where we see solid
iodine gaining electrons, so it's being reduced to
turn into iodide anions. The standard reduction
potential for this half reaction is positive .54 volts. Now we can see that's what's happening in our redox reaction up here, right? We can see see that we're going from solid iodine on the left side and we have iodide
anions on the right side. So we're gonna keep this, we're gonna keep this
reduction half reaction. If we look at what's
happening with aluminum, we're going from solid aluminum
to aluminum three plus, so solid aluminum must be losing electrons to turn into aluminum three plus. So aluminum is being oxidized here. But this half reaction is written as a reduction half reaction. So we need to reverse it, right? We need to start with solid aluminum. So we reverse our half reaction to start with solid aluminum, so aluminum turns into aluminum three plus and to do that it loses
three electrons, right, so loss of electrons is oxidation. Here is our oxidation half reaction. So we reversed, we reversed
our half reaction, all right. And remember what we do to the
standard reduction potential, we just change the sign, all right. So, for the reduction half reaction, the standard reduction
potential is negative 1.66. We reversed the reaction, so
we need to change the signs. So the standard oxidation
potential is positive 1.66 volts. For our reduction half reaction, we left it how it was written, all right. So we're just gonna write our
standard reduction potential as positive .54 volts. Next, we need to look at our
balanced equation, all right. We need to make the
number of electrons equal for our half reactions. So for this first half reaction, I'm just gonna draw a line through, I'm just gonna draw a line
through this half reaction so we don't get ourselves confused. For our first half reaction
here, we have two electrons. Then over here, right, for
our oxidation half reaction, we have three electrons. We need to have the same
number of electrons. So we need to have six electrons
for both half reactions, because remember the
electrons that are lost are the same electrons that are gained. So we need to multiply our
first half reaction by three. All right, if you multiply our
first half reaction by three, we'll end up with six electrons. And our second half reaction,
we would need to multiply the oxidation half reaction by two, in order to end up with six electrons. So let's rewrite our half reactions. So, first we'll do the
reduction half reaction, so we have, let me
change colors again here. And let's do this color. So we have three times, since
we have three I two now, We have three I two, and three times two
gives us six electrons. So three I two plus six electrons, and then three times
two, all right, gives us, three times two gives us six I minus. All right, so we multiplied
our half reaction by three, but remember, we don't
multiply the voltage by three, 'cause voltage is an intense of property. So the standard reduction potential is still positive .54 volts. So we have positive .54
volts for this half reaction. Next, we need to multiply our oxidation half reaction by two. So we have two Al, so this is
our oxidation half reaction, so two Al, so two aluminum, and then we have two Al three plus, so two Al three plus, and then two times three
gives us six electrons. So now we have our six
electrons and once again, we do not multiply our standard
oxidation potential by two, so we leave that, so the
standard oxidation potential is still positive 1.66 volts. Next we add our two
half reactions together. And if we did everything right, we should get back our overall equation. So our overall equation here. We have six electrons
on the reactant side, six electrons on the product side, so the electrons cancel out. And so we have for our
reactants three I two, so we have three I two, plus two Al, and for our products right here, we have six I minus, so six I minus, plus two Al three plus,
plus two Al three plus. So this should be our
overall reaction, all right, this should be the overall reaction that we were given in our problem. Let's double check that real fast. So three I two plus two Al, all right, so right up here, so
three I two plus two Al, should give us six I minus
plus two Al three plus. So six I minus plus two Al three plus. So we got back our original reaction. Remember our goal was,
our goal was to find the standard cell potential E zero, because from E zero we can calculate the equilibrium constant K. So we know how to do that,
again, from an earlier video. To find the standard cell
potential, all right, so to find the standard cell potential, all we have to do is add our
standard reduction potential and our standard oxidation potential. So if we add our standard
reduction potential and our standard oxidation potential, we'll get the standard cell potential. So that would be positive .54 volts, so positive .54 plus 1.66, plus positive 1.66 volts. So the standard potential for the cell, so E zero cell is equal to .54 plus 1.66 which is equal to 2.20 volts. All right, now that we've found
the standard cell potential, we can calculate the equilibrium constant. So we can use one of the
equations we talked about in the last video that relates
the standard cell potential to the equilibrium constant. So I'm gonna choose, I'm gonna
choose one of those forms, so E zero is equal to 0.0592 volts over n times log of the
equilibrium constant. So again, this is from the previous video. So, the standard cell
potential is 2.20 volts, so we're gonna plug that
in, we're gonna plug that in over here, so now we have 2.20 volts is equal to 0.0592 volts over n. Remember what n is, n is the number of moles transferred
in our redox reaction. So we go back up here and we
look at our half reactions and how many moles of
electrons were transferred? Well, six electrons were lost, right, and then six electrons were gained. So n is equal to six. So we plug in n is equal
to six into our equation. So n is equal to six. And now we have the log of
the equilibrium constant K, log of the equilibrium constant K. So we just need to solve for K now. So let's get out the calculator and solve for K. So we would have 2.20
times six, all right, divided by .0592 and that gives us, let's say 223. So this gives us 223. So 223 is equal to, so this is equal to log of
our equilibrium constant K. So we need to get rid of our log. And we can do that by taking
10 to both sides, all right. So if we take 10 to the
223 and 10 to the log of K, then that gives us K,
the equilibrium constant. So K, the equilibrium constant, is equal to 10 to the 223rd power, which is obviously a huge number. So, huge number, we get a huge value for the equilibrium constant, which is a little bit
surprising, because we only had 2.20 volts, which doesn't
sound like that much. So from only 2.20 volts,
we get a huge number for the equilibrium constant. So the reaction goes to
completion, all right, with a huge value for the
equilibrium constant like that, you pretty much don't have anything for your reverse reaction, which is why, which is why we're not drawing any kind of arrow going backwards here. We only have this arrow
here going forwards. So because of our huge number
for the equilibrium constant.