# Will the electrolyte surface become charged if one electrode is raised out of solution?

What would the potential distribution be if I were to immerse a single inert (graphite) electrode in an aqueous solution with good ionic conductivity and hold it at a high (ca. $\pu{24V}$) potential vs. an electrode that would remain in the air above the solution's surface?

I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

But, what about the air-water interface? Would the electric field induce a surface charge?

But, what about the air-water interface? Would the electric field induce a surface charge?

Yes. Basically, by locating one of your electrodes out of the solution, you've assembled an air-gap capacitor, though a fairly inefficient/low-quality one.

Per that above link, typical metal-plate "[a]ir capacitors have a small capacitance which usually lies between $\pu{100pF}$ and $\pu{1nF}$". High-level, the definition of capacitance is:

$$C = {q\over V} \quad\quad\Longrightarrow\quad\quad q = CV$$

Assuming that this water/electrode air capacitor has similar properties to the above-referenced metal-plate air-gap capacitor, the charge accumulated at the electrolyte surface under a $\pu{24V}$ applied potential should be between $2.4$ and $\pu{24nC}$. A similar amount of charge, but of opposite sign, should of course accumulate at the surface of the elevated electrode.

I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

This will be true at steady-state, as the entirety of the potential drop is going to be supported by the air gap.

However, you will have some minuscule amount of transient electrochemical activity at your submerged electrode right when you turn on your power supply, as the system adjusts from the $\pu{0V}$ to the $\pu{24V}$ applied voltage.

• Thank you! Seeing it as an air capacitor makes complete sense. Commented Apr 11, 2018 at 17:12