Nernst equation
Definitions & Key takeaways
The Nernst equation is a mathematical relationship that explains how a Galvanic cell works and allows us to find the instantaneous voltage of a cell at a specific moment in time.
Its expression is:
E = E0 - (RT/zF)lnQ
Where:
E is the potential difference across the membrane (in volts) E0 is the equilibrium potential for the ion (in volts) R is the universal gas constant T is the temperature (in Kelvin) z is the charge of the ion F is the Faraday constant Q is the reaction quotient
The Nernst equation helps explain how a Galvanic cell works - so let’s start by drawing out a galvanic cell. A galvanic cell has two electrodes - so for our example, let’s say there’s one solid zinc electrode and that it’s in a 1.0 molar solution of zinc-two-plus ions, and that there’s a solid copper electrode and it’s in a 1.0 molar solution of copper-two-plus ions.
Now, copper ions turn their solution a blue color, whereas zinc ions don’t color their solution - so it’s colorless. Since the concentration of both copper and zinc ion solutions is 1.0 molar and if we assume that the temperature is 25 degrees celsius, then we’re under standard conditions.
Now, in our galvanic cell, the two electrodes are connected by a wire with an open switch and there’s also a salt bridge between the two compartments.
The salt bridge allows electrical charge to travel between the two solutions. To get the galvanic cell started, we need to close the switch, so that electrons or electrical current can flow through the wire.
Now, if we want to calculate the cell potential, or cell voltage, for this zinc-copper cell, we need to use the Nernst equation.
So here’s the Nernst equation, E refers to the instantaneous voltage of the galvanic cell, or the voltage at a specific moment in time.
Enaught or Ezero is a constant called the “standard voltage,” and it’s simply based on the two metals that we’re using in our galvanic cell.
We can look up this number for zinc and copper, and we find that it has a value of +1.10 volts. N refers to the number of moles of electrons that are transferred during the redox reaction.
Q is the reaction quotient, which is the ratio of the concentration of reactants to the concentration of products in our galvanic cell.
When we calculate this, we only include the concentration of the ions, and we leave out the pure solids like the zinc and copper electrodes.
This tells us the progress of the reaction. Now, we can look at our two half reactions.
Loss of Electrons is Oxidation, and so the solid zinc electrode is being oxidized to form zinc two plus ions. Gain of Electrons is Reduction so copper two plus ions are being reduced to form solid copper.
Remember that LEO the lion goes GER! To get the overall equation, we add the two half-reactions together and get solid zinc plus copper two plus ions yields zinc two plus ions and solid copper.
Overall, two moles of electrons are transferred during our redox reaction. So N equals two.
To get Q, we take the concentration of zinc two plus ions divided by the concentration of copper two plus ions. Since both ion concentrations are 1.0 molar, Q is equal to 1.0 divided by 1.0 which is one.
So the instantaneous voltage is equal to the standard voltage. In this case it’s positive 1.10 volts at the instant when we close the switch.
The reason why this makes sense is because we’re at standard conditions, the ion concentration in both compartments is 1.0 molar.
Now let’s change things up. Let’s keep the zinc ion concentration at 1.0 molar, but increase the copper ion concentration from 1.0 molar to 10.0 molar.
Now we have nonstandard conditions. Let’s calculate Q at the moment that we close the switch.
At that moment, Q equals the concentration of zinc two plus divided by the concentration of copper two plus. So 1.0 divided by 10.0 or 0.10.
We can plug this new Q into the Nernst equation to give us a new instantaneous voltage. We start with Ezero, which is still positive 1.10 volts.
From this number we next subtract 0.0592 divided by N, times the log of Q. N is still two, but now the reaction quotient is 0.10.
The log of 0.10 gives us a negative number, which we multiply by 0.0592 divided by 2. Because this number is negative, we end up adding this number to 1.10 in order to give an instantaneous voltage of positive 1.13 volts.
Notice what’s happened: when Q was equal to 1.0, the voltage was 1.10, and when Q was decreased to 0.10, the voltage increased to 1.13.
So more generally, when you decrease the reaction quotient, you increase the instantaneous voltage. Now, let’s change things up again.
Let’s keep the copper ion concentration at 1.0 molar, but increase the zinc ion concentration from 1.0 molar to 10.0 molar.
So we have nonstandard conditions again. Let’s calculate Q at the moment that we close the switch.
At that moment, Q equals the concentration of zinc two plus divided by the concentration of copper two plus. So 10.0 divided by 1.0, or 10.0.
We can plug this new Q into the Nernst equation to give us a new instantaneous voltage. We start with Ezero, which is still positive 1.10 volts.
From this number we next subtract 0.0592 divided by N, times the log of Q. N is still two, but now the reaction quotient is 10.0.
The log of 10.0 gives us a positive number, which we multiply by 0.0592 divided by 2. We then subtract this number from 1.10 in order to give an instantaneous voltage of positive 1.07 volts.
Notice what’s happened: when Q was equal to 1.0, the voltage was 1.10, and when Q was increased to 10.0, the voltage decreased to 1.07.
So more generally, when you increase the reaction quotient, you decrease the instantaneous voltage. Now let’s compare the Q value of 10.0 to the equilibrium constant K which is an enormous number for this reaction: 1.54 times 10 raised to the 37th power.
When Q is less than K, there are too many reactants and not enough products, so the reaction moves forward to make more products.
In other words, the mass of the zinc electrode should shrink over time and the copper electrode should gain mass over time.
Also, the concentration of zinc ions in solution increases and the concentration of copper ions decreases. This causes the value of Q to increase over time and that affects the Nernst equation.
Remember that when you increase the reaction quotient, you decrease the instantaneous voltage. For example, if Q increases to 100, the voltage will drop to 1.04.
As Q continues to increase, the instantaneous voltage continues to decrease. And that continues until Q reaches 1.54 times 10 to the 37th, because at that point Q is equal to the equilibrium constant K.
When Q is equal to K, the second term in the Nernst equation is equal to the value of the first term. So for our galvanic cell, when Q equals 1.54 times 10 to the 37th, then the second term is equal to 1.10.
Because we subtract this term from the first term, which also equals 1.10, then we wind up with an overall voltage of zero.
Zero voltage means that there’s no current flowing through the galvanic cell. Now the equilibrium constant is a huge number which means that at equilibrium we have almost all products and no reactants.
In other words, when the voltage of the galvanic cell equals zero, then the zinc electrode will have been used up, and the blue color of the copper two plus solution would have almost disappeared.
At the same time, the copper electrode will have actually grown in size. Now, it’s also possible to have a galvanic cell where the two electrodes are made out of the same metal.
For example - both may be made of nickel. This is called a concentration cell.
In a concentration cell, there’s a different ion concentration in the two different compartments. For example, let’s say that on one side there’s a dilute concentration of nickel two plus ions of 1.00 times ten to the negative third Molar.
And on the other side, there’s a more concentrated solution of nickel two plus ions of 1.00 molar. In solution, nickel two plus ions are green so the more concentrated compartment has a darker green color.
When we close the switch, current starts to flow in our cell. At the anode, the half reaction is solid nickel is oxidized to nickel two plus, giving off two electrons that flow in our wire.
At the cathode, the half reaction is nickel two plus ions gaining two electrons to form solid nickel. So when we add the two half-reactions together, we have two electrons and solid nickel on both sides of the equation, so they cancel out, giving an overall equation of nickel two plus concentrated going to nickel two plus dilute.
Let’s plug what we know into the Nernst equation. So for Q, we have the dilute nickel concentration of 1.00 times ten to the negative third divided by the more concentrated nickel concentration of 1.00 molar.
So Q is equal to 1.00 times ten to the negative third. Overall, there are two moles of electrons being transferred, and so we also know that N is equal to two.
Finally we need to know the standard voltage. Since both electrodes are the same metal, the standard voltage is zero.
When we plug all of that into the Nernst equation, we see that the instantaneous voltage is 0.0888 volts. Since the voltage is positive, we know that this reaction is spontaneous or thermodynamically favored in the forward direction.
Equilibrium in this situation is reached when there are equal concentrations of Nickel two plus ions in both compartments, which makes Q equal to one.
Since the logarithm of one is zero, when you plug Q is equal to one in the Nernst equation, the instantaneous voltage drops to zero, and you've reached equilibrium.
As a quick recap, the Nernst Equation allow us to find the instantaneous voltage of a cell at a specific moment in time.
If we know the standard voltage and the number of moles of electrons transferred, the instantaneous voltage depends on the concentration of the ions in solution.
For a galvanic or a concentration cell, as long as Q is less than K, the cell is out of equilibrium and there is a voltage at that moment in time.
When Q is equal to K, the cell is at equilibrium and the voltage is zero.
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