Nernst Equation - Online Tutor, Practice Problems & Exam Prep

The Nernst equation connects cell potential (E_cell) to concentrations under non-standard conditions, where concentrations differ from 1 molar. It is expressed as E_cell = E_cell^° - (RT/nF) ln(Q). Here, R is the gas constant, T is temperature, n is the number of electrons transferred, and F is Faraday's constant. As the reaction progresses, the reaction quotient (Q) changes, affecting E_cell until equilibrium is reached, where E_cell equals zero. This relationship also ties to Gibbs free energy (ΔG) and the equilibrium constant (K).

Nernst Equation

So the Nernst equation reveals the quantitative connection between the concentrations of compounds and cell potential. The Nernst equation is utilized when the concentrations of our compounds differ from 1 molar. We're gonna say that the Nernst equation equals our cell potential here. And this cell potential represents cell potential under non-standard conditions, meaning that our concentration isn't equal to 1 molar, our temperature wouldn't be 25 degrees Celsius.

Our pH would not equal 7. Our pressure wouldn't be 1 atmosphere. All of those values represent standard conditions. So 1 atmosphere, pH of 7, a temperature of 25 degrees Celsius as well as a concentration of 1 molar. When we have all of these conditions met, that means we're dealing with cell potential under standard conditions.

So that's \( E^0_{\text{cell}} \). So this represents our cell potential under standard conditions. Minus \( \frac{RT}{nF} \times \ln \frac{a}{a} \). So here again we said that this represented our standard cell potential. \( R \) here is our gas constant.

It is equal to 8.314 J/(mol\cdot K). Also remember here that when we talk about joules we're talking about energy. A joule is equal to volts times coulombs. Remember a volt is equal to joules over coulombs. So you substitute joules over coulombs times coulombs and that's how it equals joules.

So we can say this or 8.314 V\(\cdot\)C/(mol\(\cdot\)K). \( n \) equals the number of electrons transferred within our redox reaction. \( F \) equals Faraday's constant which is 96485 C/mol of electrons. \( A \) represents our activities. So activity here would just be our activity coefficient times the concentration of the ion if necessary.

Many times we may not be given some type of coefficient so you can just say activity, you could substitute in the concentration for that value. Now, we're going to say that this expression here represents products over reactants in terms of our redox reaction and it's equal to \( Q \), our reaction quotient that's \( R \cdot T \) divided by \( F \), here at 25 degrees Celsius, remember we have our \( R \) constant, we have our temperature of 25 degrees Celsius. We add 273.15 to this. So that gives me 298.15 K. \( F \) is our Faraday's constant.

Here we would see that from this moles would cancel out, kelvins would cancel out. What we'd have left at the end is joules over coulombs which is equal to volts. So \( \frac{RT}{F} \) reduces all the way down to 0.0257 V. So that means that our Nernst equation now becomes cell potential under non-standard conditions equals cell potential under standard conditions minus 0.0257 V/n, the number of electrons transferred, times \( \ln Q \). Remember \( Q \) is just your equilibrium expression.

Now here, if we multiply \( ln \) by 2.303 we can attain the log function. Now when we multiply this portion here by 2.303, we get a new value of 0.05916 V/n, now \( \log Q \). This is true because we say that \( \log(x) = \ln(x) / \ln(10) \). \( \ln(10) \) equals 2.303. So when I multiply both sides by 2.303, we can see that multiplying by 2 point potential calculated from the Nernst equation is the maximum potential at the instant the circuit the cell circuit is connected.

So that's the moment that the current or the flow of electrons moves from the anode to the cathode. So we're talking about the transferring of electrons from one electrode to another electrode. And we're gonna say as the cell discharges and current flows, the electrolyte concentrations will change. What's going to happen here is that \( Q \) will begin to increase and as a result, our cell potential over time will decrease until it gets to a point where the cell potential of my electrochemical cell equals 0. That's when we have a dead battery because at that moment, the battery has reached equilibrium.

So over time, the reaction will reach equilibrium and then \( Q \) which is our reaction quotient will equal \( K \) which is our equilibrium constant. The cell potential like I said would equal 0. We'd have a dead battery at that moment. Now as a result of this, once we've reached equilibrium we can substitute in \( K \) instead of \( Q \). So now our new equation can become \( e_{\text{cell}} \) equals \( e_{\text{cell}} \) under standard conditions minus \( RTNF \times \ln(K) \).

So here, if we were to work this out we could have 0 equals \( e_{\text{cell}} \) minus remember this value here would be 0.0257 volts/n, now times \( \ln(K) \). So what we can do here is subtract cell potential from both sides. So that'd be negative standard cell potential equals negative 0.0257 volts/n times \( \ln(K) \). Multiply both sides by \( n \). Then divide both sides now by negative 0.0257 volts.

So we get at this point \( \ln(K) = n \) times your cell potential under standard conditions divided by 0.0257 volts. To get rid of this \( \ln \), we take the inverse of the natural log so that means that \( K = e^{(n \times \text{your cell potential under standard conditions} / 0.0257 \, \text{volts})} \). And that's when we're dealing with \( \ln \). If we're to substitute in \( \log \) instead, so if we're dealing with a log function then, it would be 0.05916 volts/n times \( \log(K) \). In this instance, if we did the same exact, mathematical conversions.

In this case, because we're dealing with \( \log \), we'd find that \( K \) at the end equals \( 10^{(n \times \text{your cell potential under standard conditions} / 0.05916 \, \text{volts})} \). So this is how we connect our equilibrium constant to our standard cell potential with these two formulas. 1 when we're dealing with \( \ln \) and 1 where we're dealing with \( \log \). Now we could also say that when you're at equilibrium, we can talk about the connection to Gibbs free energy and your equilibrium constant \( K \). So here we'd say that \( \Delta G \) under nonstandard conditions equals \( \Delta G \) under standard conditions minus \( RT\ln(K) \).

When we've reached equilibrium, this \( \Delta G \) equals 0. So that means 0 equals \( \Delta G0 \)-\( RT\ln(K) \). Subtract \( \Delta G0 \) from both sides. Subtract \( \Delta G0 \) from both sides. So negative \( \Delta G0 \) equals negative\( RT\ln(K) \).

Divide both sides now by negative\( RT \). So \( \ln(K) = \Delta G0 / RT \). And so \( K = e^{(\Delta G0 / RT)} \). So we get this value at the end in terms of our connection between Gibbs Free Energy and your equilibrium constant \( K \). So just keep in mind some of the connections that we've seen here in terms of how things are connected to one another.

These two equations are just a way of us connecting cell potential to \( K \), and then from \( K \) to \( \Delta G \). And we know that from earlier we could also connect to cell potential from \( \Delta G \) as well when we have \( \Delta G = -n \times \text{Faraday's constant} \times \text{cell potential} \). Later on, we'll talk about this connection between the three variables. But for now, just realize we deal with the Nernst equation when we have concentrations that are not equal to 1 molar. And remember, the Nernst Equation can be written in 2 different ways.

We can write it in this version when we're dealing with \( \ln \) or we can have the Nernst Equation in this version when we're dealing with \( \log \). To go from \( \ln \) to \( \log \), you just use 2.303 to go from your \( \ln \) function to your \( \log \) function. So keep in mind the intricate dynamics involved in the Nernst equation in how it connects your cell potential in standard conditions to non-standard conditions.

Nernst Equation Calculations 1

So here we need to determine the cell potentials of the following concentration cells. Here it's written in line or cell notation for us. So what we're going to do here is we're going to deconstruct it and give us both of our half reactions. So, here we have our anode, our physical break, and our cathode. For the anode, we have our silver solid producing silver ion.

Here the charge is 0 and the overall charge here is plus 1, so I add 1 electron to this side. For my other half reaction, I have copper 2 ion producing copper solid. The overall charge here is plus 2 and the overall charge here is 0. I add electrons to the more positive side. So I add 2 electrons to this side so that it becomes 0 overall just like this side is 0 overall.

Notice that my electrons do not match. They need to match because we have the transferring of an equal number from one electrode to another. So I'd have to multiply this half reaction times 2 which gives me 2 silver solids producing here we should do reversible arrows gives me 2 silver ions plus 2 electrons. So, these are reversible reactions so we put double arrows. Alright.

So now, we have our two half reactions. So the first one deals with the anode where oxidation occurs. So here that would represent this potential for the silver. Now we're going to say that \( e^- \), which is the potential of our anode, equals the potential under standard conditions minus \( \frac{0.05916}{n} \) times log of \( q \). So notice here we're using the Nernst equation because we have a concentration at least one concentration that is not equal to 1 molar.

So these are non-standard conditions. So we're going to plug in \( 0.79 \, \text{volts} - \frac{0.05916 \, \text{volts}}{2} \). And then remember, \( q = \frac{\text{products}}{\text{reactants}} \). So our product is 2 moles of silver ion. Remember we ignore solids and liquids so this reactant we would ignore it because it's a solid.

So here it would just be log of \( (\text{Ag}^+)^2 \) which would be 0.0010 Because again \( q = \frac{\text{products}}{\text{reactants}} \). It ignores solids and liquids. So \( q \) here would just be \( (\text{Ag}^+)^2 \). When we punch all that in, that gives us a number of 0.976 volts.

Now we do the same thing for the compartment that is our cathode. So that's going to be \( e^+ = e^+ \) under standard conditions minus \( \frac{0.05916 \, \text{volts}}{n} \times \log(q) \). Here my positive under standard conditions is 0.339 volts minus \( \frac{0.05916 \, \text{volts}}{2} \). And then \( q \) is products over reactants. So based on this half reaction, our product is a solid so we're going to ignore it.

So that's 1 over. And then we have copper 2 ion. There's just a coefficient of 1 here so we don't have to raise that concentration to any power. So it would be log of 1 over the concentration of \( \text{Cu}^{2+} \) which is 1 molar. Realize here that this just becomes log of 1 divided by 1.

So this is 1 inside of here. Log of 1 is equal to 0 so all of this drops out. So this is just 0.339 volts. So we've found the cell potential of the anode and the cathode. Now, we can find out the overall cell potential if we want.

So overall cell potential equals cathode minus anode or positive minus negative here. So that's \( 0.339 \, \text{volts} - 0.976 \, \text{volts} \) which gives me negative 0.637 volts as our final answer. What this value is telling me is because my cell potential overall is less than 0, this is a non-spontaneous process. Which would mean that the only way that silver would be oxidized by copper is if a battery were used because that battery would force the electrons to leave the silver and go towards the copper electrode. So this would represent an electrolytic cell because it's non-spontaneous.

Now that you've seen the basic setup for a question like this, attempt to do example 2. Try to approach determining which one is the cathode and which one is the anode. Try to set up the Nernst equation for each compartment of the electrochemical cell. And from there, determine what the overall cell potential is and describe what it means in terms of spontaneity based on the overall cell potential that you determine. Once you do that, come back and see how your answer matches up with mine.

Nernst Equation Calculations 1

So here we need to determine the spontaneity and cell potential based on the given cell notation. Alright. So here we have our platinum solid which represents our inert electrode, and we're using this because here in our anode compartment we have the presence of KBr. So we have bromide becoming Br2. And here they're not separated by a phase diagram because this is bromide aqueous, it's dissolved in a solvent, and the solvent happens to be bromine.

Then on this side where we have the cathode, we have iron ion becoming iron solid. So that is the particular ion that's undergoing reduction. So from this information, we're going to write our two half reactions. So we have Br- aqueous produces Br2 liquid here. Again, they're not separated by a phase boundary because they're in direct contact with one another.

They're mixed together because bromide ion is just dissolved in bromine bromine liquid. Here we have Fe2++2So we add electrons with a more positive side so that's 2 electrons. Then on this side here it's minus 1 and here it's 0. Here that would mean 1 electron on this side but the electrons don't match so I'd have to multiply, by 2.

So this here is actually 2 electrons and this is 2. What this is saying is that we have 2 bromide ions and they connect together to give me Br2 liquid. And they do this by releasing an electron each. This is my actual balanced half reaction. Alright now, we have this as our anode so that's e- equals e-under standard conditions minus 0.05916 volts divided by n times log of q.

And then this is e+ equals e+under standard conditions minus 0.05916 volts divided by N times log of Q. Alright, so now we plug in the values that we know. So E-under standard conditions would be 1.078 volts minus 0.05916 volts divided by number of electrons transferred which is 2. And then log of q. Q equals products over reactants.

It ignores solids and liquids. The product is a liquid so it's ignored. So that's just one over Br-2 squared. So that'd be 1 over 0.01 which is going to be squared. Here, when we plug that in, that gives me 0.960 volts.

Over here, e_positive under standard conditions is negative 0.440 volts minus 0.05916 volts divided by number of electrons transferred which is 2. And then q here would equal products over reactants, ignore the product because it's a solid, divided by Fe2+. So that'd be 1 divided by by 1. All of these cancels out because log of 1 is equal to 0. Now that we have those values, we can find the overall cell potential.

So overall cell potential equals e+ minus e-. So that's negative 0.440 volts minus 0.960 volts which gives me negative 1.40 volts. So as in the example above, we got a value that's less than 0. So we can say that this process is a non-spontaneous process and this represents an electrolytic cell. Remember, for it to be spontaneous, you would want your overall cell potential to be greater than 0.

By being greater than 0 would be spontaneous and represent a galvanic or voltaic cell.

Nernst Equation Calculations 2

Consider a standard voltaic cell based on the reaction of 2 moles of H⁺ ions reacting with 1 mole of tin solid to produce 1 mole of tin (II) ion plus hydrogen gas. It mentions which of the following actions would change the EMF of the cell? Here, EMF refers to our electromotive force, which is essentially our cell potential. Since we are discussing increasing or decreasing the pH, this implies that we are adjusting the concentration of H⁺. Therefore, we are discussing cell potential under non-standard conditions.

Using the Nernst equation, our cell potential under non-standard conditions equals our cell potential under standard conditions minus 0.05916N volts divided by N times the log of Q. Remember, Q is our reaction quotient. It equals products over reactants. We ignore solids within our expression, so it would be the concentration of Sn²⁺. Hydrogen here is a gas, so we normally put p there to represent pressures involved for that gas, divided by H⁺, and because there's a 2 here, it'd be squared.

We don't include the tin because it's a solid. Now, we are discussing adjusting the pH; we are talking about changing concentrations and pressures. So, for the first one, it says increasing the pH at the cathode. Realize here that whether we are increasing or decreasing the pH, that means that we are having a direct impact on the concentration of H⁺. Remember that there is an inverse relationship between our reaction quotient and the overall cell potential.

If our Q increases, that means that our overall cell potential will decrease and vice versa. They are inversely related to each other. So, doing either one of these things will affect the concentration of H⁺, therefore changing the value of Q, and therefore changing the value of our cell potential or our electromotive force.

If we were increasing the pH, that would mean that we are dropping the concentration of H⁺ because, remember, if you increase the pH, it becomes more basic, so this H⁺ would decrease which would cause an increase in Q, which would therefore cause a decrease in your cell potential. Lowering the pH would increase the concentration of H⁺, which would then decrease the reaction quotient and increase our cell potential. Next, increasing the concentration of Sn²⁺ at the anode, so you're increasing the numerator here in the expression, which will increase Q and therefore decrease your cell potential. Increasing the hydrogen gas pressure at the cathode, so again here, you're increasing H₂ now, so you're increasing Q which will lead to a decrease in your cell potential. So here the answer would be e because all of these actions would result in a change in your EMF value.

Just realize here because we're changing concentrations, we're dealing with non-ideal conditions or non-standard conditions. Therefore, we introduce the Nernst equation and try to relate our reaction quotient Q to the overall cell potential.

Nernst Equation Calculations 2

So, here it says consider the following half-cell reaction at 25 degrees Celsius. What will be the value for E, the half-cell potential for static conditions for the reaction? Here we need to notice that we have reversed the reaction. Our product is now a reactant and our reactants are now products. So we've reversed the reaction.

In addition to that, realize that we've multiplied by 2. So we've multiplied this reaction by 2. Now when it comes to these changes, we need to understand what will the effect be on my standard cell potential here. Realize that you can multiply or divide your half-reaction by any number and that will not cause a change in your cell potential. Your cell potential would still stay the same.

But if you reverse your half-reaction, that reverses the sign of your cell potential. So by reversing it, now it's going to become positive 0.260 volts giving us option B as an answer. Multiplying or dividing by any number does nothing to the cell potential, so we would say that this could not be an answer, nor this. So, again, reversing the reaction reverses the sign for your cell potential. Multiplying and dividing by any number causes no effect.

Here’s what students ask on this topic:

What is the Nernst equation and how is it used in electrochemistry?

The Nernst equation is used to calculate the cell potential (E_cell) of an electrochemical cell under non-standard conditions. It is expressed as:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(Q\right)$

Here, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of electrons transferred, and F is Faraday's constant (96485 C/mol). Q is the reaction quotient. The equation helps in understanding how the cell potential changes with varying concentrations of reactants and products.

How does the Nernst equation relate to Gibbs free energy?

The Nernst equation is closely related to Gibbs free energy (ΔG). The relationship is given by:

$\mathrm{\Delta G}=-\mathrm{nF}{E}_{\mathrm{cell}}$

At equilibrium, ΔG = 0, and the Nernst equation can be rearranged to connect the equilibrium constant (K) with the standard cell potential (E_cell°):

$E_{\mathrm{cell}}=\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(K\right)$

This shows how the cell potential under standard conditions is related to the equilibrium constant and Gibbs free energy.

What are the standard conditions in the Nernst equation?

Standard conditions in the Nernst equation refer to a set of specific conditions under which the cell potential is measured. These include:

• Concentration of 1 M for all reactants and products
• Temperature of 25°C (298.15 K)
• Pressure of 1 atm
• pH of 7

Under these conditions, the cell potential is denoted as E_cell°. When conditions deviate from these standards, the Nernst equation is used to calculate the cell potential.

How do you convert the Nernst equation from natural logarithm (ln) to common logarithm (log)?

To convert the Nernst equation from natural logarithm (ln) to common logarithm (log), you multiply the ln term by 2.303. The Nernst equation in terms of ln is:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(Q\right)$

Multiplying by 2.303, it becomes:

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

This conversion is useful for calculations involving common logarithms.

What is the significance of the reaction quotient (Q) in the Nernst equation?

The reaction quotient (Q) in the Nernst equation represents the ratio of the concentrations of products to reactants at any point in time. It is given by:

$Q=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}$

Q affects the cell potential (E_cell) as the reaction progresses. When Q = 1, the cell potential is at its standard value (E_cell°). As Q changes, it indicates the shift in the reaction towards equilibrium, affecting E_cell until it reaches zero at equilibrium.