If you have an electrolyte of $\ce{Zn^2+}$ and $\ce{SO4^2-}$ (Neutral) Then you dip in it a Zinc electrode.

Zinc potential is $\pu{-0.74 V}$ (relative to SHE). So that means there is a $\pu{-0.74V}$ difference between the electrolyte and the electrode.

But When you dip in the solution the electrode makes the solution more positive and it self more negative the same amount so the overall potential of the electrode will be $\pu{-0.74 V} / 2 = \pu{-0.37 V}$, and the solution will be at $\pu{+0.37 V}$.

Is this true? (assuming the solution is not connected to any other salt bridge or porous disk)

Edit: Absolute electrode potential is really the difference between the electrolyte and the electrode potential.

And the question is just this if you have an electrode inside an electrolyte with no salt bridge will it's potential be halfed.

  • 2
    $\begingroup$ Possible duplicate of Deriving a reduction potential from two other reduction potentials $\endgroup$
    – Mithoron
    Jul 26, 2019 at 16:36
  • 1
    $\begingroup$ Please remember from physics 001 that any excess charge resides on the surface (=or in this case interface) of a body. The bulk of your electrode and solution are (in first approximation) charge-free. $\endgroup$
    – Karl
    Jul 26, 2019 at 18:37
  • $\begingroup$ In addition, you couple your electrode with an other (for example with a salt bridge) to close an electric circuit. While there are bottles of water, and grinders filled with pepper, there is no magic box in your cabinet «just so» filled with electrons. $\endgroup$
    – Buttonwood
    Jul 26, 2019 at 18:46

2 Answers 2


Those $\pu{-0.74V}$ are measured against a standard hydrogen electrode. That's the universally accepted but totally arbitrary standard.

You don't have a second electrode, hydrogen or other, so the absolute number is totally meaningless. If someone had decided to use the zinc electrode instead, as zero point for the electrochemical series, then the voltage you mean would be zero. Makes no sense.

There is an absolute cell potential (absolute electrochemical series), with a different zero point, but it's totally uncommon, because it's very tricky to measure absolute electrostatic potentials of only a few hundred millivolts. I believe the problem is the calibration, while a platinised platin electrode bubbled with reasonably pure hydrogen is probably the simplest reference standard to reproduce at all.


It is a very good question, and I have wondered about it for years, talking to highly seasoned electrochemists, only to find that nobody knows the answer! The sole reason is that the problem is overwhelmingly complex despite its deceptive simplicity. Nernst offered him theory similar to osmosis, but it did not work- 100 years ago.

However there are couples of misconceptions you have here. The first one is that when you dip a zinc electrode in a solution containing zinc, the potential difference is NOT -0.74 V. Nobody in this world can find out the potential difference between the solution and the zinc electrode. The moment you attach a probe, it introduces another potential difference, an unknown to another unknown. This is somewhat akin to (but not at related) to the Heisenberg's principle. The moment you try to probe the momentum of an electron, its position becomes uncertain.

So your arithmetic in invalid to begin with. I don't have my 2-vol Bockris, but it did have a very intuitive explanation of single electrode potentials.

What you can determine indeed, is the sign of the charge developed on the electrode wrt to the ground. Again this sign is not absolute, it is with respect to the Earth, which is "considered" to be at zero potential i.e. there is no charge build-up on the Earth. Who know if this is right or wrong, but this is the convention. There are very sensitive electroscopes now which can tell you the sign of the charge, albeit very small.

  • 1
    $\begingroup$ There is a huge voltage (~.5 MV or so) between ground and the ionosphere. "Ground" is everything, but not absolute zero. Measuring charges is however not exactly rocket science, piezo transducers are read out that way for example. $\endgroup$
    – Karl
    Jul 26, 2019 at 22:44
  • $\begingroup$ Of course, but for earthly measurements and by electrical engineering conventions, ground is considered at 0 V as a reference. Vernier markets a charge sensor, and if we measuring the "sign" charge of an isolated object, the other terminal of the electroscope is connected to the ground (say a water line etc.). In British terminology this the same as earthing. It is indeed arbitrary. $\endgroup$
    – AChem
    Jul 26, 2019 at 23:03

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.