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### Course: Chemistry archive > Unit 9

Lesson 5: Cell potentials under nonstandard conditions# Galvanic cells and changes in free energy

Relationship between Gibbs free energy, reaction quotient Q, and cell voltage.

## Want to join the conversation?

- Where does the K= 1.58*10^37 come from?(21 votes)
- You can use ΔG° = -RT ln(Q) to find it

since we have:

ΔG° = -212 x 10^3 J

R = 8.314 J/Kmol

T = 298K

Although i got K = 1.45 x 10^37

(probably has to do withe rounding of ΔG°, it cant be nicely 212 KJ)

Otherwise, K is typically determined experimentally.(1 vote)

- at8:49, how do I know the number of electrons transferred? I'd like to solve for E myself to see that it's 0.98V but I don't know where to get n (number of electrons) from.(7 votes)
- n is the number of electrons in your half-cell reactions. In this case, n=2(5 votes)

- I am missing a whole bunch! I didn't understand a thing about what's going on here. So I went to the next section, Cell Potentials, which seems to give the foundation for understanding this video. So I'll come back to this one, later.

But, is it out of sequence? Should it be in the next section?(2 votes)- If I'm ever lost in the sequence of the MCAT tutorials, I switch over to the physics/chem/org sections and usually find the missing videos.(5 votes)

- Can we consider the free energy here as the energy released due to the redox reaction and that it's utilized in transferring the electrons?

I just want to relate this to voltage = work/ q.

That in delta G= -nFE°, nF gives us the quantity of charge and E° is voltage. When we multiply these we get energy or work done to transfer charges.(4 votes) - So does Q = K, the equilibrium constant in the last videos?(1 vote)
- K is the expression of [products]/[reactants] at equilibrium.

Q is the same expression but at whatever time point you are looking at, let's say 4 minutes into a reaction.

If Q=K, the reaction is at equilibrium.

If Q<K, there are more reactants than at equilibrium so the reaction toward products will be favored.

If Q>K, there are more products than at equilibrium so the reverse reaction will be favored.(6 votes)

- Why do you always do the math with a calculator? Aren't these for MCAT prep? Anyways, any help on calculating ln 10,000 w/o a calculator?(1 vote)
- These lectures are prepared to help students from around the world who are preparing to sit a range of different examinations. They're not specifically aimed at MCAT.(6 votes)

- At2:45, why do we leave copper and zinc?(3 votes)
- Is Delta G a state variable or a difference between the state variables of two states? If so, what states?(2 votes)
- At6:30why did delta G change when K increased such that delta G was no longer equal to Delta G naught? Wasn't the temperature and pressure kept the same? What change was made such that we were no longer at standard conditions?(2 votes)
- You should differentiate btw ΔG°and ΔG:

ΔG° helps to determine the work/energy (or Voltage) provided by the reaction to reach equilibrium under standard conditions ([X]= 1M; T=25 °C; P=1 atm). It is not influenced by the state or at what moment is the reaction, but solely provides information on the maximum released free energy or the maximum potential difference between the initial and final states (equilibrium) of the reaction. This is why ΔG° is only dependent to Keq (but always under standard conditions mentioned above).

On the other side, ΔG helps to identify the work (or Voltage) that must be provided by the reaction at a specific moment. For instance, at t=0, ΔG is at its maximum because the spontaneous free energy that should be released by the reaction is still awaiting the reaction to commence. Therefore, ΔG refers here to this unliberated energy (Similarly, at t=0 there is a maximum voltage/potential between reactions and products and the system (reaction) is awaiting to start to decrease this potential difference and achieve equilibrium. ΔG is identified by this maximum potential difference at this point.

At an instant t<tequilibrium, the reaction progresses, and some products are formed, causing the potential difference to decrease. This is why when Q= 10 000 in the video, the potential energy drops from its maximum at the beginning of the reaction (E0=1.1v) to E= 0.98v at Qt<Keq; at this point, the energy supplied by the reaction to attain equilibrium is reduced, and ΔG also decreases.

Once the potential difference has disappeared, all the spontaneous free energy has been released, and the reaction no longer possesses any electromotive force to alter its state. At this point, the reaction has reached equilibrium, and ΔG simply equals 0.

Hope this helps to leave your ambiguity.(1 vote)

- how can the Q be increased from 1 to 10000?(2 votes)

## Video transcript

- [Voiceover] We already know
that for this galvanic cell solid zinc is oxidized
to zinc two plus ions. Oxidation occurs at the anodes, that this electrode is our anode, and copper two plus ions
are reduced to solid copper. Reduction occurs at the cathode so this electrode must be our cathode. If we start with a one molar
concentration of zinc sulfate that means in solution
our initial concentration of zinc two plus would be one molar, and we have a one molar
concentration of copper sulfate which means we're starting
with a concentration of copper two plus of one molar. So we're under standard conditions. We're at one molar for our concentrations, we're at 25 degrees C, we have solid pure electrodes and we know that the
standard cell potentials, or the cell potential
under standard conditions for this cell is positive 1.10 volts. So we have a spontaneous redox reaction which produces a current. So electrons flow in our
wire and we get a voltage, we have a standard cell potential. In one of the videos in electrochemistry we took our standard cell
potential and from it we calculated delta G zero. So we used this equation. We said delta G zero is
equal to negative nFE zero. So we plugged in our
standard cell potential and we got delta G zero. So delta G zero is the
change in free energy under standard conditions, and we got negative 212 kilojoules. Let's look at the equation
for the change in free energy from thermodynamics and let's
analyze our galvanic cell. So the change in free energy is equal to the standard change in
free energy, delta G zero, plus RT natural log of Q. So remember, this delta G is
the instantaneous difference in free energy between your
reactants and your products. Delta G zero is the change in free energy under standard conditions. R is the gas constant, T
is temperature in Kelvin and Q is your reaction quotient. So let's think about what the reaction quotient is for this reaction. For this reaction, our
spontaneous redox reaction, Q has the same form as
the equilibrium constant, concentration of products over
concentration of reactants and you leave out pure solids. So that would be the
concentration of zinc two plus, that's our product here, we're leaving out solid copper, over the concentration of copper two plus and we're leaving out solid zinc. So over the concentration
of copper two plus. At this instant in time we
plug in our concentrations. The concentration of zinc
two plus is one molar so we have 1.0. The concentration of copper
two plus is also one molar so we're under standard conditions here. So Q is equal to one at this instant, and we plug in Q is equal
to one into our equation and we see that the natural log of one, the natural log of one is equal to zero. So this, this term goes to zero. And under standard conditions the change in free energy, delta G... The change in free energy,
delta G is equal to the change in free energy under standard
conditions, delta G zero. And that's equal to
negative 212 kilojoules, and if you want to write
per mole of reaction you could do that right here. So this makes sense because
we're under standard conditions. We're at one molar for our concentrations so the change in free energy is equal to, is equal to delta G zero. Notice that delta G is negative so we know this is a spontaneous reaction. This is a spontaneous reaction
under standard conditions. So current flows, we get a voltage. Let's go back up to here. So we get a voltage, we get a voltage at this moment in time. So the reaction goes to the right to make more of our products. We're gonna make more
of our products here. What happens to Q as the
reaction proceeds to the right? Well we're increasing the
concentration of our products, we're increasing the concentration
of zinc two plus ions. At the same time we're
decreasing the concentration of copper two plus ions. So Q increases as the reaction
proceeds to the right. So as we make more products, Q increases. So we get an increase in Q. What happens to delta G? What happens to the instantaneous
difference in free energy between our reactants and our products? Let's go ahead and plug
that in to our equation. Let's make up a number. Let's
say that Q increases from... We started over here with one. Let's say that Q increases to 10,000. We'll pick a big number here. So Q goes up to 10,000 so we have far more products than we do reactants. What is delta G? Delta G is equal to... Delta G is zero is the standard
change in free energy so under standard conditions this
is negative 212 kilojoules. So that's negative 212. And this would be negative 212,000 joules, if we convert kilojoules to joules, plus we have R, which is the gas constant. Let me go back up to here. R is the gas constant which is 8.314 joules per mole times Kelvin. So you could make this
joules per mole here so your units balance out,
moles per mole of reaction. And then we have the temperature. The temperature of our reaction, let's remind ourselves of that. So we go back up to here.
We're at 25 degrees C. 25 degrees C is 298 Kelvin
so this is 298 Kelvin. Kelvin would cancel out here. And this is times the natural log of Q. And now we've changed Q. We've said, "Okay,
let's just pick a number "that's pretty big," so Q is 10,000. So we plug in the natural
log of 10,000 so we have obviously a huge number of
products compared to reactants at this moment in time. So at this moment in time what is delta G? What is the instantaneous
change in free energy between our reactants and our products? We need to do that calculation here. Let's find the natural log of 10,000. So we have that, we're
gonna multiply that by 298 and we're gonna multiply that by 8.314, and we're going to add that number... We're going to add that
number to negative 212,000. So we get that delta G is equal to negative 189.2, let's say, kilojoules. So I'll make that into kilojoules. Delta G is equal to negative 189.2 kilojoules per mole. So I'm not really too concerned
about the exact number, I'm just trying to point out
what happens as you change Q. As we increased Q, as we
went from Q is equal to one to Q is equal to 10,000, so
what happened to delta G? We went from negative 212 kilojoules to negative 189.2 kilojoules. So we're getting closer to zero, we're getting closer to equilibrium. But at this instant, at
this instant right here when Q is equal to 10,000 we still have a negative value, We still have a negative
delta G, I should say. So the reaction is still spontaneous. The reaction is still spontaneous, we're still going to generate a current, we're still going to make
more of our products, we're still going to have a
voltage at this moment in time. What would be the voltage
at this moment in time? What would be the
instantaneous cell potential? The instantaneous cell potential is E. We can find that using our equation that relates delta G and E. Delta G is equal to negative nFE. So if we plug in this, if we plug in delta G into here we know that N is the
number of moles of electrons that are transferred,
F is Faraday's constant and then E is our
instantaneous cell potential. To save time I won't
do the calculation here but if you do that
calculation you will get that your instantaneous
cell potential is equal to .98 volts, positive .98 volts. So the reaction is spontaneous, we're still producing a voltage. We're still producing a voltage here. So notice that our voltage
has decreased a little bit. In the previous example
under standard conditions, at that moment in time
the voltage was 1.10 volts so we've lost a little
bit of voltage here, we're down to .98. But think about how large that number is. We have so many of our products
compared to our reactants and we're still getting
a pretty decent voltage. Approximately one volt, so pretty close to the original 1.1 volts. What happens at equilibrium? What happens at equilibrium? We know that at equilibrium
the reaction quotient Q is equal to the equilibrium constant K, and at 25 degrees C, K is equal to 1.58 times 10 to the 37th. So for this reaction at 25 degrees C, this is our equilibrium constant. Let's plug in our equilibrium constant into our equation for
delta G to see what we get. So delta G is equal to delta G zero, which was negative 212 kilojoules per mole plus R, which is 8.314,
times the temperature. We're still at 25 degrees
C so this is 298 K. And this time we're putting
in the natural log of K, so we're plugging in our
equilibrium constant for Q. So the natural log of
1.58 times 10 to the 37th. So let's do that math now. We have the natural log of 1.58 times 10 to the 37th. And we're going to multiply that by 298 and 8.314, and that gives us, if we round that, that's positive 212 kilojoules per mole. So positive 212 kilojoules per mole. I'm not worried about
the exact number here because I rounded this,
this is rounded as well. The point is that delta G is
equal to zero at equilibrium. We already know that, we already know delta G is equal to zero at equilibrium. This shows you that. We have negative 212 kilojoules per mole so about negative 212 kilojoules per mole and approximately 212
kilojoules per mole over here. At equilibrium these will cancel out and give you delta G is equal to zero. So at equilibrium we know that delta G is equal to zero. There is no difference in free energy between your reactants and your products. So let's think about the voltage
of our cell at equilibrium. Well if delta G is equal to
zero, we plug that into here and therefore the cell potential,
E, must be equal to zero. If this is equal to zero
then this is equal to zero, so the cell potential
is equal to zero volts. The voltage is zero when our redox reaction comes to equilibrium, and so therefore the cell
dies, your battery is dead. Hopefully this helps you
understand galvanic cells in terms of thinking about
changes in free energy.