Topic 9.10: Cell Potential Under Nonstandard Conditions

Nonstandard cell potential: why a battery’s voltage changes

On a windy December afternoon in 1884, five hundred and fifty-five feet above Washington, workers set one last piece on top of the Washington Monument. It was a little pyramid about nine inches tall, and with it in place the monument was finished, the tallest structure on Earth. And the pyramid is made of aluminum.

That’s the same metal as a soda can. You probably threw one away last week. In 1884 that wasn’t funny. Aluminum is the most common metal in the ground under your feet, but getting it out of rock was so hard that the metal cost a fortune. So before this pyramid went up, it went on display at Tiffany’s in New York, like a piece of jewelry.

On the very day the tip went up, in Oberlin, Ohio, a college student named Charles Martin Hall turned twenty-one. Fourteen months later, in the woodshed behind his family’s house, Hall ran an electric current through a crucible of molten minerals and pulled out little beads of shiny aluminum. There was no power grid to plug into, so his electricity came from batteries he built himself, rows of glass jars full of zinc and acid, wired together.

Within a few years aluminum was cheap, and today rooms like this make it by the ton. Today we’ll find out why a battery’s voltage sags as it runs. Then we’ll work out how long one modern cell takes to make the metal for this tip. Hours or days? Make a guess and hold on to it, because the answer is kind of ridiculous.

So let’s start where Hall started, with a battery. We’ve set one up on his bench, a zinc plate in zinc sulfate on the left and a copper plate in blue copper sulfate on the right. A wire and a salt bridge join them.

With both solutions at one molar, the meter reads one point one oh volts. That’s the standard cell potential, E naught, from the last lesson, the voltage when every solution is one molar at twenty-five degrees. And the AP exam asks what happens to a cell like this under nonstandard conditions, when somebody changes the concentrations, so let’s change them. Before we close the switch, make a prediction. As this battery runs, what happens to its voltage?

Let’s close the switch and let it do some work. The zinc gets thinner, the copper gets thicker, the blue fades, and the needle slowly drifts down. Down at the metal surfaces, zinc atoms let go of two electrons each and drift off as zinc ions, and on the other side copper ions grab two electrons each and join the copper. The electrons travel only through the metal and the wire, never through the liquid.

So the mixture keeps changing, more zinc ions and fewer copper ions, and that’s what pulls the voltage down. Then, when the copper ions are nearly gone, the needle drops all the way to zero, and that’s a dead battery. To see why the voltage follows the mixture, and why it ends at exactly zero, we need one number that describes the mixture.

Q against 1: reading the Nernst equation

That number is the reaction quotient, Q, built like an equilibrium expression, products over reactants, with the solid metals left out. So here Q is the zinc ion concentration over the copper ion concentration, and in a fresh battery that’s one over one, exactly one. The Nernst equation links Q to the voltage. E equals E naught minus RT over nF, times the natural log of Q, with n equal to two electrons for this cell, and everything in front of the log is just a positive number.

You’ll rarely need to plug numbers into it, because what matters is reading it, so let’s read it like a sentence. If Q is exactly one, the log of one is zero, nothing gets subtracted, and E is exactly E naught. Standard conditions give you Q equal to one, but so does any other mixture where the ratio comes out to one.

If Q is bigger than one, the log is positive, we subtract, and the voltage comes out lower than E naught. That’s our battery running down, with zinc ions piling up on top of Q. And if Q is smaller than one, the log is negative, so subtracting it adds, and the voltage comes out higher than E naught.

So compare Q to one, and you know how E compares to E naught. Or think about the meter, where more zinc ions or fewer copper ions push the needle down, and the opposite pushes it up.

In this cell, every tenfold change in Q moves the voltage by only about three hundredths of a volt. Now give it a hundred times more zinc ions than copper ions. That’s just two of those steps. The voltage only drops to about one point oh four volts.

On a log scale the whole story is one straight line, through E naught at Q equals one, and down to zero volts where Q equals K, the equilibrium constant, which for this cell is about ten to the thirty-seventh. A fresh battery starts unbelievably far from equilibrium. Use up ninety-nine percent of its copper ions. Copper drops to a hundredth of a molar, and zinc climbs to about two molar.

Two over a hundredth is about two hundred. That’s still nowhere near K, and the voltage barely moves. The cell potential is the reaction’s push toward equilibrium, and the farther away, the harder it pushes. And at equilibrium there’s no push left at all, which is exactly what zero volts means.

Now I’d like you to use only the two concentration dials to make this battery push harder than one point one oh volts. Suppose we dilute both solutions from one molar to zero point one molar. How does the voltage compare to one point one oh volts now?

Here’s an answer students write all the time. Add copper ions, and the student says the equilibrium shifts right by Le Châtelier’s principle, so the voltage goes up. The voltage does go up, but the reasoning falls apart, because Le Châtelier’s principle is about a system at equilibrium, and a working battery isn’t at equilibrium, or it would read zero volts. The argument that earns the point uses Q. More copper ions make the bottom of Q bigger, so Q drops, the cell sits farther from equilibrium, and the voltage rises.

Now let’s flip it around, with a student who adds zinc ions to the zinc side and says the voltage drops. Which justification would earn the point?

Concentration cells: one metal, two strengths

Here’s a battery that looks like it shouldn’t work at all. It has copper on both sides. The left jar holds a very dilute solution, a thousandth of a molar, and the right jar holds a one molar solution, with a liter in each. Both sides run the same half-reaction, so E naught is zero, and yet the meter reads about eighty-nine millivolts.

That’s because Q, the dilute concentration over the concentrated one, is only a thousandth, while equilibrium here just means the two solutions are equally strong, so K is one. So the cell is far from equilibrium, and it pushes. The only way it can get there is for the two solutions to even out. So which copper plate gets heavier as this battery runs? Tap it.

Let’s let it run. The pale side gets bluer and the dark side gets paler. And the needle sinks as they close in on each other. And when the two blues match, the needle sits at exactly zero.

On paper, the dilute side is the anode and the concentrated side is the cathode. Q starts at a thousandth, far below K, so E is positive, and it ends with both sides at half a molar, Q equal to K, and E at zero. And this dead cell isn’t empty, because each jar still holds half a mole of copper ions. A cell at zero volts is a cell at equilibrium, and it can still be full of chemicals.

Here’s one more pair of batteries, two zinc–copper cells with a liter in every jar, one with one molar solutions and one with half-molar solutions, and both read one point one oh volts, because in both of them Q is one. We’re about to run both at the same current. Which one dies first, and why?

The AP exam asked this in 2014 about a tin and copper cell with both concentrations halved, and the answer was the same voltage and a shorter life. So voltage tells you how hard each electron gets pushed, but not how many electrons a cell can move. For that we have to count electrons, which is exactly how Hall’s batteries made aluminum.