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### Course: Class 12 Chemistry (India) > Unit 2

Lesson 2: Standard cell potentials- Standard reduction potentials
- Using reduction potentials
- Voltage as an intensive property
- Worked example: Identifying the best oxidising agent using Eo(red.) values.
- Worked example: Constructing a galvanic cell using reduction potential values
- Spontaneity and redox reactions
- Free energy and cell potential
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential

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# Calculating the equilibrium constant from the standard cell potential

Example problem for calculating the equilibrium constant K using the standard cell potential. Created by Jay.

## Want to join the conversation?

- Why does he use log (K), haven't we been using natural log (K)? I did it using natural log and got 6.85*10^96.(7 votes)
- Unfortunately, you will see both bases used in this chemistry. Which is used depends on who did the math, but it is more common to use log base 10.

There is no good reason for using the two bases of the logs, but either will work, you just have to use slightly different constants that incorporate the change in base of a log from the natural log.(2 votes)

- Why didn't you use G=-RTlnk to solve for K? why is the value for K different when using this equation? :((5 votes)
- He actually did use it to arrive to this "new" equation : E°= (0.0592)* log(K)

He explains it in the previous video of this series but I will try to explain :

He uses deltaG°=-RTlnK**and**deltaG°=-nFE°. With these two equations you have RTlnK = nFE°. You then leave E° on the right and bring the rest on the left : E° = (RT/nF) lnK. He then calculates RT/F, which you can because you know the value of R and F but also T since here its E° which means we are in standard conditions (25°C). This gives him a certain value we'll call b. For certain reasons in chemistry, log is more used than natural log (ln) so he gave an equation using the log. To do that, you need to divide ln(K) by ln(10) so we divide the equation by ln(10) which gives us : E°/ln(10) = (b/n)log(K). You then multiply by ln(10) to have E° instead of E°/ln(10) and you find : E° = (b * ln10/n)log(K). He calculated b*ln10 and found 0.0592V. In the end, your equation is this : E° = (0.0592/n)log(K)

When you use the equation G=-RTlnK, you might find something slightly different. That is because he calculated constants in advance, so maybe you have more precise numbers, but if you round it up, you should get the same thing.

Hope this helps, even two years later :)(2 votes)

- What happened to the units for n?(3 votes)
- n is simply a number (of electrons) and has no units(3 votes)

- Very beginning of video..You can't balance this equation with the bridge method? Why is that? This is the first time I've ever had this problem. I think it has something to do with the fact you picked an equation which has cations and anions both on product side.(4 votes)
- I am confuse with what equilibrium constant is telling us, like if it is large it only goes on a certain way (arrow part?) and if it is small the arrow to the left is larger than arrow to the right... what is equilibrium constant?(3 votes)
- you can think of k (equilibrium constant) as a kind of pool to fill in

1. if it's huge like the one in the video, your reaction should keep going for a long time to make that huge amount of products per 1 set of reactant to fill that pool (as k is proportional to "mols of product/mols of reactants")

2. on the other hand, it k is small or even less than 1 you can fill that pool quite easily and quickly

3. then the spillover (reverse reaction) starts

hope to be helpful(2 votes)

- I don't understand... Where did you determine that you originally had 2 electrons?(3 votes)
- I2(s) (2 iodine atoms in 1 solid molecule) reduces to form 2 I- ions in solution by gaining 2 e- per molecule. This is set up as a half reaction for the problem. Is this what you are asking about?(2 votes)

- at6:16isn't E*cell = Ecathode - Eanode ? why did he add those two?(1 vote)
- He changed the sign of the standard reduction potential for the oxidation half-reaction - I suggest watching the rest of Jay's videos on the subject of voltaic cells.(4 votes)

- How would you calculate the reaction quotient for this reaction?(2 votes)
- If you were to calculate the reaction quotient for this reaction, you would need the concentration of the products and the reactants, as well as the written balanced reaction. Check out this article: https://www.khanacademy.org/science/chemistry/chemical-equilibrium/factors-that-affect-chemical-equilibrium/a/the-reaction-quotient(1 vote)

- I thought Ecell was cathode (reduction) - anode (oxidation), is that only for galvanic cells? Why do you add the Ecell of the two half reactions?(1 vote)
- So Ecell is cathode - anode, but only if you're using reduction potentials since most potential tables give the potentials of the reduction reactions. So if you did this using the reduction potentials given at the start of the problem then Ecell would be: 0.54 - (-1.66), which is the same thing as 0.54 + 1.66 because of the double negative.

Jay just turned the reduction potential of the aluminum (-1.66) into an oxidation potential (+1.66) by changing the sign of the potential. And it's this oxidation potential of aluminum we can add to the reduction potential of iodine to get Ecell. But it's obviously the same math as before using only the reduction potentials.

So you can find Ecell either way, you just have to conscious of the signs. Using the reduction potentials it is: Ecell = cathode - anode. Using the reduction AND oxidation potentials it is: Ecell = cathode + anode.

Hope that helps.(3 votes)

- Couldnt you use E(cell) = E(cathode) - E(anode)(1 vote)
- That's one way to look at it. He is relating -E(oxidation) + E(reduction) to find the answer. It is -E(oxidation) because the values are given for reduction reactions, so you have to flip them around.(2 votes)

## Video transcript

- [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.