4
$\begingroup$

I'm studying electrochemistry without a physics background--never taken it for extenuating reasons--for a research project using Bard's Electrochemical Methods. I'm having trouble understanding the concept of open-circuit potential (OCP): the electric potential measured in a cell when disconnected from a circuit. First off, why can you measure OCP at all? It's a result of thermodynamics--i.e. the sum of the standard potentials of the half reactions at the given electrodes. But wouldn't the lack of a current preclude any measurement of potential? I mean, a voltmeter, with its very high resistance, still allows some current flow. Since you can measure OCP, does that mean the half-reactions are taking place in small quantities?

Second, Wikipedia describes OCP as "the voltage that must be applied to a solar cell or a battery to stop the current." If a given cell has an OCP of 0.5V, that means I apply 0.5V to stop the current once a circuit is completed? But if there exists 0.5V without a current, applying an external 0.5V won't then create a total potential of 1V? In my head I want to apply -0.5V to negate the current flow.

Thank you in advance. The inclusion of mathematical formulae in your answer would be much appreciated!

$\endgroup$
3
  • 1
    $\begingroup$ The answer to your first question is yes. The higher the input impedance of the voltmeter or electrometer, the better for minimizing having the half reactions take place. For the rest, see my answer here: chemistry.stackexchange.com/a/153076/79678. You apply the external potential to oppose the cell potential, not augment it. $\endgroup$
    – Ed V
    Jul 10, 2023 at 22:04
  • 1
    $\begingroup$ @EdV Ahh ok. So if you have two electrodes just sitting there, nothing's going on. But measuring the OCP requires completing the circuit. Without a complete circuit, the difference in potential still exists, but no current = no rxn. It's as if I have a zipline at the top of a hill. The potential to zipline down exists, but without actually connecting the thing, I can't ride it. $\endgroup$
    – GMoss
    Jul 10, 2023 at 22:19
  • 3
    $\begingroup$ Well said! With the typical voltmeter mode in a digital multimeter, the input resistance is 10 M ohm. Using this to measure an approximation of the open circuit potential is fine for most purposes since, e.g., 0.5 V divided by 10 M ohms is only 50 nA. But for a standard reference electrode, e.g., Ag/AgCl, no more than a couple of pA can be drawn without polarizing the electrode. So then you use an electrometer voltmeter, commonly used with pH electrodes. They typically have 1 T ohm input resistance. $\endgroup$
    – Ed V
    Jul 10, 2023 at 22:26

1 Answer 1

5
$\begingroup$

Basically you've have the right ideas.

The open-circuit potential of an electrochemical cell is the theoretical voltage that one would measure as the current flow approaches zero. Think of plotting small current flows against measured voltage and then extrapolating to zero current flow. Pragmatically though common voltmeters will only measure a few significant figures. So a microamp current flow is essentially zero current flow. This isn't math where pi can be calculated to millions of digits.

For the solar cell you're right. If the cell output is +0.5 volts then you'd have to apply -0.5 volts to stop the current flow if the voltage is applied across the cell itself. Two +0.5 sources in parallel would only give +0.5 volts output however. You'd have to arrange the solar cell and an additional +0.5 volts in series to get +1.0 volts output.

$\endgroup$
2
  • 1
    $\begingroup$ Ok so OCP is kind of like the limit as i -> 0. In reality you can never reach 0 b/c you need at least a tiny bit of current to measure the voltage. $\endgroup$
    – GMoss
    Jul 10, 2023 at 22:26
  • 1
    $\begingroup$ GMoss - Yes but you can't measure voltage to some huge number of significant figures. So to accurately measure about 0.5 volts of a battery to 3 or 4 significant figures a microamp of current would be small enough to be essentially zero current. $\endgroup$
    – MaxW
    Jul 10, 2023 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.