# Nernst equation

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Justin Ling, MD, MS, Evan Debevec-McKenney, William GilpinThe 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.

Zn(s) --> Zn2+ + 2 e- Cu2+ + 2 e- --> Cu(s)

During the first half-reaction, our solid zinc metal loses two electrons, and during the second half-reaction, the copper ions in solution gain two electrons.

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.

When we plug all that in to the Nernst equation, we find that the instantaneous voltage is equal to positive 1.10.

The second term in the Nernst equation turns out to be equal to zero, because the logarithm of Q is the logarithm of one, which is zero.

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.