Standard Cell Potential & the Equilibrium Constant - Online Tutor, Practice Problems & Exam Prep

Galvanic and voltaic cells generate electricity by remaining out of equilibrium, with cell potential (E_cell) decreasing to zero at equilibrium. The Nernst equation relates E_cell to the reaction quotient (q) and the equilibrium constant (K). The connection between Gibbs free energy (ΔG) and K is expressed as ΔG = -RT ln(K), while ΔG also relates to E_cell through ΔG = -nF E_cell. At standard conditions, E_cell is determined without the Nernst equation, emphasizing the importance of understanding these relationships in electrochemistry.

Cell Potential & the Equilibrium Constant

Here we see that galvanic and voltaic cells are able to produce electricity because they are not yet at equilibrium. Recall that the chemical reaction will eventually reach equilibrium and at that point \( q \) will equal \( k \). In addition to this, once we've reached equilibrium, we would say that our cell potential under non-standard conditions would also equal 0. So here we have our initial Nernst equation. Here we have your cell potential under non-standard conditions equals cell potential under standard conditions minus \( \frac{0.05916 \text{ volts}}{n} \times \log(q) \), \( q \) being your reaction quotient. Remember, this represents your voltage at an exact. Once we've reached equilibrium, as we've said, your cell potential under non-standard conditions would equal 0 and your \( q \) would equal your \( k \). So now, we use your equilibrium constant as your new variable. Because of this, we can try to solve for what \( k \) would equal in relation to your cell potential under standard conditions.

So, here we would subtract \( E_{\text{cell}} \) from both sides which would give us a negative here and a negative here, and then you would just divide both sides by negative one to make it positive. At this point, we could then multiply both sides by \( n \) to give \( n \times E_{\text{cell}} \) under standard conditions equals \( 0.05916 \text{ volts} \times \log(k) \). Here, we're reformatting the equation to show the relationship that the equilibrium constant has towards your cell potential under standard conditions. We would then divide both sides by \( 0.05916 \text{ volts} \). \( 0.05916 \) equals \( n \times E_{\text{cell}} \) under standard conditions divided by \( 0.05916 \text{ volts} \). You want to get rid of the \( \log(k) \) so you just take the inverse log. So that would become \( k = 10^{n \times \frac{E_{\text{cell}}}{0.05916 \text{ volts}}} \). So this is the equation we'd use when we are given your cell potential under standard conditions and asked to find your equilibrium constant \( K \).

Now, we're going to say the relationship between your cell potential under standard conditions and your equilibrium constant \( K \) and Gibbs free energy can be seen in this format. So, when we have \( k \) and \( \Delta G \), we connect them by this equation here, \( \Delta G = -R \times T \times \ln(k) \). Remember that \( R = 8.314 \text{ joules/moles} \times K \). Remember here that the units could also be changed to \( \text{volts} \times \text{coulombs/moles} \times K \). Then if we have our cell potential under standard conditions and \( \Delta G \), they are connected by this formula here, \( \Delta G = -n \times \text{Faraday's constant} \times E_{\text{cell}} \) under standard conditions. And then finally, \( k \) and \( E_{\text{cell}} \) can be connected by this equation that we saw up above. Remember, at 25 degrees Celsius, this portion here we can get as a standard value. When we multiply by 2.303, that changes \( \ln \) into \( \log \). Then finally, realize that we have cell potential here and here and those cell potentials connect to this equation here. Remember, if we're dealing with standard conditions, this circle here, that would mean that our concentrations are equal to 1 molar. Therefore, we wouldn't rely on the Nernst equation to help us determine what this cell potential is at all. We would just use the cell potential of my cathode minus the cell potential of my anode. This is dealing with standard cell potential so no Nernst equation needed. Keep in mind the relationship that \( q \) and \( k \) have with each other in terms of reaching equilibrium, how there's a transition from your \( q \) value to your \( k \) value. Remember, the Nernst equation is really utilized when we have concentrations that are different from 1 molar. Here in this triangle, we're assuming that we are at equilibrium now, so \( q \) has been transitioned into \( k \). We're dealing with 1 molar concentrations and therefore, we use this simplified version to help us determine the overall standard cell potential and then look at the connection that it has to Gibbs free energy. So, keep in mind these connections as we delve deeper and deeper into looking at calculations that show the interconnectedness of these different variables.

Calculations 1

So here we need to determine the equilibrium constant K for the Fong reaction. We have 1 mole of gold plus one ion reacting with 1 mole of cerium solid to produce 1 mole of cerium 3+ aqueous. In addition to this, we're told that half reactions are determined as each one of the following with their given cell potentials. Well, from these cell potentials, we can determine what our overall standard cell potential will be. We know that we're going to figure out the standard cell potential because we're not given any concentrations, so we assume that they've approached unity where they are equal to 1 molar. Therefore, the Nernst equation isn't allowed to be used.

In addition to this, we're going to have to figure out which one of them is being oxidized versus which one is being reduced. If we take a look here at the overall reaction, we see that we have gold plus 1. So its oxidation number is plus 1 and here it's neutral. Its oxidation number has decreased in value. Therefore, it has been reduced and represents the cathode. Then if we look at cerium, we see that it's at 0 initially and then it goes up to plus 3. So its oxidation number has increased therefore it's been oxidized and represents your anode.

Next, we're gonna say at this point, we can just say that our standard cell potential equals your cathode minus your anode. So that's 1.690 volts minus a minus 2.336 volts. Minus of a minus really means you're adding them. So that's 4.026 volts. This is a very large voltage for our cell potential which means that we should expect a very large K or equilibrium constant.

Now we know that from the previous page that the relationship between your equilibrium constant and your cell potential, your standard cell potential is K=10n·E_cell0.05916. So we can plug in the values that we know, so we put 10. We have to determine the number of electrons transferred. Here, they're not the same number. You'd have to multiply this equation here by 3 to get the correct number of electrons transferred. And remember, multiplying or dividing does nothing at all in terms of the cell potential, so it still stays at that same value. And because you multiply this by 3, that means the balanced equation will be 3 here and 3 there. So we'd have 3 here times your cell potential which was 4.026 volts divided by 0.05916 volts equals 10 to 204.158, which gives us a very large equilibrium constant of 1.4410×204. This tells me that this reaction is incredibly spontaneous because it gives us a value for your standard cell potential that's much greater than 0.

And it also gives us a K that is much greater than 1. Depending on the calculator that you're using, you may get an error if you're not using a high enough or advanced enough calculator. This is an incredibly large value for K and that's the connection that we can have between your equilibrium constant K and your standard cell potential.

Calculations 1

From the two half reactions provided, the equilibrium constant is calculated as \(6.79 \times 10^{30}\). Now, from these two half reactions, we have one where the electrons are reactants, so this is our reduction reaction and represents our cathode. And then here, we have the electrons as products, which represents an oxidation, so this is our anode.

It says to determine the standard cell potential for basically the cathode compartment. They gave us the equilibrium constant \(K\). Remember that that shares a connection with our standard cell potential. Here, the overall standard cell potential equals \(0.05916 \text{ volts}\) divided by the number of electrons transferred times log of \(K\). The number of electrons transferred has to be the same. Here we have 3, and here we only have 1, so we'd have to multiply this by 3 to make it 3 electrons. Remember, multiplying our half reaction or dividing our half reaction does nothing to the cell potential value; it stays the same. The only thing that can change is if we reversed the reaction itself.

So here, we're going to say that we have \(0.05916 \text{ volts}\) divided by 3, and this is times log of \(6.79 \times 10^{30}\). When we plug that in, that gives me \(0.608 \text{ volts}\). So, we have the overall cell potential, standard cell potential, that's equal to cathode minus anode. This value here is the original cell potential of this reaction when it was written as a reduction. So, we have here \(0.608 \text{ volts} = E^{\circ}_{\text{cathode}} - 0.308 \text{ volts}\) here. So what we do here is we add this to both sides. So that comes out to \(0.916 \text{ volts}\) as the missing cell potential for the half-reaction that occurs at the cathode.

So, remember, the key to this question is realizing the connection between our overall standard cell potential and our equilibrium constant \(K\). Knowing that allows us to then isolate this equation and solve for the one missing variable, which gives us the cell potential of our cathode compartment.

Here’s what students ask on this topic:

What is the Nernst equation and how is it used to calculate cell potential under non-standard conditions?

The Nernst equation is used to calculate the cell potential (E_cell) under non-standard conditions. It is given by:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}-\frac{0.05916}{n}\times log\left(Q\right)$

Here, E_cell is the cell potential under non-standard conditions, E_cell° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient. This equation helps in determining the voltage of an electrochemical cell when the concentrations of the reactants and products are not at standard conditions (1 M concentration).

How do you calculate the equilibrium constant (K) from the standard cell potential (E_cell°)?

To calculate the equilibrium constant (K) from the standard cell potential (E_cell°), you can use the following equation:

$K={10}^{\frac{n\times E_{\mathrm{cell}}°}{0.05916}}$

Here, n is the number of electrons transferred in the reaction, and 0.05916 is a constant derived from the Nernst equation at 25°C. This equation shows the direct relationship between the standard cell potential and the equilibrium constant, allowing you to determine K when E_cell° is known.

What is the relationship between Gibbs free energy (ΔG) and the equilibrium constant (K)?

The relationship between Gibbs free energy (ΔG) and the equilibrium constant (K) is given by the equation:

$\mathrm{\Delta G}=-RT\times ln\left(K\right)$

Here, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant. This equation indicates that a negative ΔG corresponds to a spontaneous reaction and a large K value, while a positive ΔG corresponds to a non-spontaneous reaction and a small K value.

How is the standard cell potential (E_cell°) related to Gibbs free energy (ΔG)?

The standard cell potential (E_cell°) is related to Gibbs free energy (ΔG) by the equation:

$\mathrm{\Delta G}=-nF\times E_{\mathrm{cell}}°$

Here, n is the number of electrons transferred, F is Faraday's constant (96,485 C/mol), and E_cell° is the standard cell potential. This equation shows that a positive E_cell° corresponds to a negative ΔG, indicating a spontaneous reaction, while a negative E_cell° corresponds to a positive ΔG, indicating a non-spontaneous reaction.

Why is the Nernst equation not needed under standard conditions?

Under standard conditions, the concentrations of all reactants and products are 1 M, and the temperature is 25°C. In this scenario, the reaction quotient Q equals 1, making the logarithmic term in the Nernst equation zero. Therefore, the cell potential (E_cell) under standard conditions is simply the standard cell potential (E_cell°), and the Nernst equation simplifies to:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°$

Thus, the Nernst equation is not needed to calculate the cell potential under standard conditions.