Main content

## 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Using the Nernst equation

Using the Nernst equation to calculate the cell potential when concentrations are not standard conditions. Created by Jay.

## Want to join the conversation?

- Once we find the cell potential, E how do we know if it is spontaneous or not?(10 votes)
- For a reaction to be spontaneous, ΔG should be negative.

Since ΔG= -nF*E (n=>n-factor or number of electrons transferred, F=96500 C= 1 Faraday

=>If E is positive, ΔG is negative. If E is negative, ΔG is positive. So, E should be positive for the reaction to be spontaneous.(20 votes)

- I still don't understand about the n. What does it represent?(8 votes)
- It's when you're doing redox reactions and trying to cancel out the number of electrons to balance each side. That number would be n. In other words, it would be the number of electrons you're transferring, as Andrews had said.(6 votes)

- What if we are dealing with an equation like 3 moles of Solid Iodine reacting with 2 moles of Aluminum(3+) giving 6 moles of Iodine(-) and 2 moles of Solid Aluminum

Would the Reaction Quotient be

[Iodine(-)]^6/[Aluminum(3+)^2(5 votes)- You got it. Remember the reaction quotient only depends on aqueous ions, not solids, so your equation, after looking through it, seems correct.(3 votes)

- why do leave uot concentration of pure solids while writing nernst equation??(3 votes)
- Using concentrations in the Nernst equation is a simplification. In reality, what we care about is the activity. For solutions, the activity is equal to the concentration, which is why we can get away with just writing concentrations for these species. Pure solids and liquids have an activity of 1, so we can ignore them (since multiplying by 1 doesn't change the value). If you're interested in learning more about activity, it is sometimes also called "chemical activity" or "thermodynamic activity".(5 votes)

- What happens to the cell potential if the temperature is increased and vice versa?(5 votes)
- For those of you who are thinking about this:

E_cell = E_standard_cell_potential - (RT)/(nF) * ln([products]/[reactants])

So it depends on the sign of the log term.

If concentrations of the products are greater than the concentrations of the reactants, then the cell potential will become more negative as temperature increases. (the log bit keeps the second term negative).

If the concentrations of the products are less than the concentrations of the reactants, then the cell potential becomes more positive as temperature increases (the log bit makes the second term positive).(1 vote)

- What is the cell potential when Q is greater than 0 and less than 1, or the concentration of zinc ions is smaller than the concentration of copper ions? Is this cell potential greater than the standard potential? What would happen if there is no zinc ion in the beginning of the reaction (the concentration of zinc ions is 0)? That means Q is 0, and cell potential will be infinite. How could that be?(4 votes)
- Where does the number above n come from ? This wasn't shown. I have tried multiplying R by T and I do not get the same answer.

Thanks(2 votes)- The number has been obtained from thermodynamic relationship (RT)/F and then multiplied by ln(10) to convert it to a log base 10. It is explained in the previous video called 'Nernst equation.' I hope this helps!(4 votes)

- What if we have a galvanic cell with 1-molar zink and copper solutions, but are working at a tempetature not equal to 25 degrees celcius? If we plug everything into the Nernst-equation, we would still get 1.1 V. But is this correct? Because i thougt the voltage depends on the temperature too?(3 votes)
- If you are not at 25*C,

Nernst says: E_cell = E_standard_cell_potential - (RT)/(nF) * ln([products]/[reactants])

so when Jay made the simplification of the second term to 0.0592/n * log_10(q), it is at 25*C. This is a good generalization if working with standard conditions, but you should use the first equation, plugging in all the conditions that are required.(1 vote)

- when you write the equation with log, do you mean ln acturally?because the calculated value indicated this way.(2 votes)
- When he writes
*log*he means log base 10. Notice in the video that log 10 is equal to 1 and log 100 is equal to 2.(1 vote)

- What will be the emf if of the cell if you given all the concentration except one and it is 0.5atm(2 votes)

## Video transcript

- [Voiceover] You can
use the Nernst equation to calculate cell potentials. Here we need to calculate
the cell potential for a zinc-copper cell, where the concentration
of zinc two plus ions and the concentration of copper
two plus ions in solution is one molar, and we're at 25 degrees C. So we're talking about
standard conditions here. Just to remind you of the
reduction half reaction and the oxidation half reaction, copper two plus ions are reduced. They gain electrons to form solid copper. And solid zinc is oxidized,
so zinc loses two electrons to form zinc two plus ions. Those two electrons, the
electrons lost by zin, are the same electrons
gained by copper two plus, so they cancel out when you
write your overall reaction. So down here we have our
overall redox reaction, and the standard cell potential is equal to positive 1.10 volts, so you just add the
standard reduction potential and the standard oxidation potential. So all of this we've
covered in earlier videos and now we're gonna see how to calculate the cell potential using
the Nernst equation. So let's go ahead and write
down the Nernst equation, which is the cell potential is equal to the standard cell potential, E zero, minus .0592 volts over n, times the log of Q. So this is the form of
the Nernst equation, this is one of the forms that we can use when our temperature is 25 degrees C. So let's think about
what these things mean in the Nernst equation. The standard cell potential, E zero, we've already found
that, that's 1.10 volts. So this 1.10 would get plugged in to here in the Nernst equation. And it's the number of
moles that are transferred, number of moles of electrons that are transferred in our redox
reaction, and that's two. Two moles of electrons are transferred. So n is equal to two. Q is the reaction quotient, so Q is the reaction quotient, and Q has the same form as K but you're using
non-equilibrium concentrations. So think about writing an
equilibrium expression. To write Q think about
an equilibrium expression where you have your
concentration of products over the concentration of your reactants and you leave out pure solids. So we're gonna leave out,
we'll leave out solid copper and we have concentration
of zinc two plus, so concentration of our product, over the concentration of our reactants. We're gonna leave out the solid zinc so we have the concentration
of copper two plus. We know what those concentrations are, they were given to us in the problem. The concentration of zinc
two plus is one molar, the concentration of copper
two plus is one molar. So we have one over one. So the reaction quotient for
this example is equal to one. So let's go ahead and plug in everything. So we have the cell
potential E is equal to the standard cell potential. That was 1.10 volts, minus .0592 over n, where n is the number
of moles of electrons, that's equal to two, times the log of the reaction quotient. Well, log of one, our reaction quotient for this example is equal to one, log of one is equal to zero. So we have the cell
potential is equal to 1.10 minus zero, so the cell
potential is equal to 1.10 volts. So we know the cell potential is equal to the standard cell potential, which is equal to 1.10
volts, positive 1.10 volts. So this makes sense, because E zero, the standard cell potential, let me go ahead and
highlight that up here, the standard cell potential E zero is the voltage under standard conditions. And that's what we have here,
we have standard conditions. Our concentrations, our
concentrations are one molar, we're at 25 degrees C, we're dealing with pure
zinc and pure copper, so this makes sense. The Nernst equation
should give us that the cell potential is equal to
the standard cell potential. Let's find the cell potential
again for our zinc copper cell but this time the concentration of zinc two plus ions is 10 molar, and we keep the concentration of copper two plus ions the same, one molar. So if we're trying to
find the cell potential we can use our Nernst equation. So the cell potential E is equal to the standard cell potential E zero minus .0592 volts over n times the log of Q where Q
is the reaction quotient. Let's plug in everything we know. We know the standard cell
potential is positive 1.10 volts, so we have 1.10 volts. We're trying to find the cell potential E, so E is equal to 1.10 minus .0592 over n. So n is the number of
moles of electrons that are transferred, so
that was two electrons. So n is equal to two so
we plug that in here. Times the log of Q, and
from the previous example, remember, Q, the reaction quotient is the concentration of zinc two plus... Concentration of zinc two plus over the concentration of copper two plus. So concentration of
products over reactants, ignoring your pure solids. So for this example the concentration of zinc two plus ions in
solution is 10 molar. So that's 10 molar over--
Copper two plus is one molar, so 10 over one. So Q is equal to 10 for this example. So now let's find the cell potential. The cell potential is E. So E is equal to 1.10 minus-- You can actually do all
of this in your head. So .0592, let's say that's .060. So this is .060, divided
by two which is .030. So we have .030. Log of 10 is just equal to one, so this is .030 times one. So 1.10 minus .030 is equal to 1.07. So the cell potential
is equal to 1.07 volts. I like to think about this as the instantaneous cell potential. So when your concentrations
are 10 molar for zinc two plus and one molar for copper two plus, 1.07 volts is your
instantaneous cell potential. What happens to the cell potential as the reaction progresses? Well let's think about that, let's go back up here
to our overall reaction. What happens as we make more
and more of our products? Well, the concentration
of zinc two plus ions should increase and we're losing, we're losing our reactants here so the concentration of copper
two plus should decrease. So what happens to Q? If we're increasing the
concentration of zinc two plus and decreasing the concentration
of copper two plus, Q should increase. And what does that do
to the cell potential? So, in the Nernst equation,
if we're increasing Q what does that do to E? Well let's go ahead and
let's just plug in a number. Let's just say that Q is equal to 100. So let's say that your Q is equal to 100. Let's plug that into the Nernst equation, let's see what happens
to the cell potential. So E is equal to E zero, which, we'll go ahead and plug in 1.10 there. So 1.10 minus .0592 over two times log of 100. So now we're saying
that Q is equal to 100. So we have more of our products
as the reaction progresses. So what is the cell potential? E is equal to 1.10, log
of 100 is equal to two. So log of 100 is equal to two, that cancels out this two here so we have one minus .0592. One minus .0592. I'll just say that's equal to .060, just to make things easier. So 1.10 minus .060 is equal to 1.04. So the cell potential
is equal to 1.04 volts. So notice what happened
to the cell potential. So we increased-- Let
me change colors here. We increased Q. We went from Q is equal to
10 to Q is equal to 100. What happened to the cell potential? The cell potential went from
1.07 volts to 1.04 volts. So as the reaction progresses, Q increases and the instantaneous cell
potential, E, decreases. So Q increases and E decreases. What happens at equilibrium? What is the cell potential at equilibrium? What is the cell potential at equilibrium. If you remember the equation
that relates delta G to the cell potential, so
we talked about this one, delta G is equal to negative nFE, and from thermodynamics, at equilibrium, delta G is equal to zero. So if delta G is equal
to zero at equilibrium, what is the cell potential at equilibrium? E must be equal to zero, so the cell potential is
equal to zero at equilibrium. Let's think about that. If the cell potential is
equal to zero at equilibrium let's write down our Nernst equation. The Nernst equation is E is equal to E zero minus .0592 over n, times the log of Q. Well at equilibrium, at
equilibrium E is equal to zero, so we plug that in. So we have zero is equal to
the standard cell potential, E zero, minus .0592 over n, times the log of Q. But at equilibrium,
remember, Q is equal to K. So we can plug in K here. Now we have the log of K, and notice that this is the equation we talked about in an earlier video. The standard cell potential
E zero is equal to .0592 over n times the log of K. So just an interesting way to think about the Nernst equation. The Nernst equation is
very useful for calculating cell potentials when you have
different concentrations.