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## MCAT

### Unit 8: Lesson 17

Electrochemistry- Electrochemistry questions
- Electrochemistry
- Redox reaction from dissolving zinc in copper sulfate
- Introduction to galvanic/voltaic cells
- Electrodes and voltage of Galvanic cell
- Shorthand notation for galvanic/voltaic cells
- Free energy and cell potential
- Standard reduction potentials
- Voltage as an intensive property
- Using reduction potentials
- Spontaneity and redox reactions
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential edited
- Nernst equation
- Using the Nernst equation
- Concentration cell
- Introduction to electrolysis
- Quantitative electrolysis
- Electrolysis of molten sodium chloride edited
- Lead storage battery
- Nickel-cadmium battery

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# Calculating the equilibrium constant from the standard cell potential edited

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## Want to join the conversation?

- Is it true for most galvanic cells that k is very large?(6 votes)
- It would make sense...galvanic cells are spontaneous by definition. This means deltaG is negative and that would make K positive according to delta G= -RTlnK(4 votes)

- I used the equation with E= RT/n.F x lnK and I get a math error on my calculator, but I compute the numbers the exact same way as in the video. What am I doing wrong?(4 votes)
- This is how I solved using the natural log version:

E = 0.0257/n *lnK is the simplified equation after you plug in your constants (see the Standard cell potential and equilibrium constant video for more info on this)

We know that E is equal to 2.20V as solved in this video. I think it is easier to first simplify the constants before dividing them over. So, 0.0257/6 = 0.004283 *lnK. Divide by 0.004283 to isolate lnK. So, 2.20V / 0.004283 = 513.619. Careful with the next part! Remember that log has a base of 10 but ln does not! Natural log has a base of e. So rather than raising both sides by 10 like in the video, you should raise both sides by e. When you put e^513.619 in your calculator, you should get 1.1... * 10^223. This is equivalent to k = 10^223 since the 1.1.... number is negligible.

Hope that helped! :)(6 votes)

- I thought it was E(standard)cell = E(standard)Cath - E(standard)Ox or Ox + Cath = Ecell standard?(3 votes)
- Those are both acceptable ways of calculating the standard cell potential, except that in the first equation you would use both of the half reactions' reduction potentials as they are given, while in the second equation, the sign of the reduction potential of the oxidation half reaction must be reversed to reflect its oxidative direction. :)(2 votes)

- Can the equilibrium constant only be calculated using the standard cell potential?(1 vote)

## Video transcript

- 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 potentials, 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 of the standard cell potential in earlier videos. So, let's start with this first half reaction here, right where we see solid iodine gaining electrons, so it's being reduced to turn into iodide and ions. The standard reduction potential for this half reaction is positive .54 volts. And we can see that's what's happening in our redox reaction up here. We can see that we're going from a solid iodine on the left side, and we have iodide and ions 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, all right, and 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. So loss of electrons is oxidation, here is our oxidation half reaction. So we reversed our half reaction, and remember what we do to the standard reduction potential. We just change the sign. So for the reduction half reaction, the standard reduction potential's negative 1.66. We've reversed the reaction, so we need to change the sign, so the standard oxidation potential is positive 1.66 volts. For our reduction half reaction, we left it how it was written. So we're just gonna write our standard reduction potetial as positive .54 volts. Next, we need to look at our balanced equation. 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 though this half reaction so we don't get ourselves confused. For our first half reaction here, we have two electrons. And then over here, for our oxidation half reaction, we have three electrons, and 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. If we 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, let's do this color. So we have three times, since we have three i two now, so 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 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 intensive 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'd 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 then 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. This should be the overall reaction that we were given in our problems, let's double check that real fast. So three i two, plus two Al. 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, six i minus plus two AL three plus. So we got back our original reaction. Remember, 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, and 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 will 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, was equal to .54 plus 1.66, which is equal to 2.20 volts Now that we 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 to standard cell potential to the equilibrium constant. So 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 we would have 2.20 times six, divided by .0592. And that gives us 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. So 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 obvious a huge number. So a huge number, we got 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. With the huge value for the equilibrium constant like that, you pretty much don't have anything for your reverse reaction, 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.