Does the Nernst equation also apply when there is a net current flow through the electrode?

1 Answer
Jul 26, 2016

Answer:

Not really. Just as you need to apply voltage across a resistor run current through it, so you have to apply a potential different from the equilibrium value to run current through an electrode.

Explanation:

That potential difference, in relation to an electrochemical reaction, is called overpotential. For example, suppose an electrochemical reduction has an equilibrium potential, calculated with the Nernst Equation, of #-0.2 V# versus the standard hydrogen electrode (SHE) at equilibrium. However, to make the cathodic reduction actually go you can't just apply a potential of #-0.2 V# vs SHE. You have to force a more negative potential to make the reduction go at the rate you want.

Say you need an overpotential of #0.2 V# beyond the equilibrium potential, then instead of #-0.2 V# vs SHE you apply #-0.4 V#.

When it comes to overpotential, you need to remember two important facts:

1) Overpotential is a more negative potential for reduction at a cathode, as in the above example, and more positive for oxidation at an anode.

2) The amount of overpotential can be very different for different possible reactions at an electrode, because overpotential depends on the kinetics of a given reaction as well as the current you are applying. So you might see electrochemical reactions take place "out of order" compared with equilibrium.

An example of the second point: We can electrogalvanize steel with zinc in an acidic plating bath. Hydrogen can be reduced at a more positive potential than zinc, so we "should" be evolving hydrogen instead of plating zinc on the steel cathode. But the kinetics of hydrogen evolution is slow on steel and zinc, so evolving hydrogen requires a lot of overpotential. That overpotential is enough to enable zinc to plate first, since zinc plating is kinetically much faster and requires significantly less overpotential.

See https://en.wikipedia.org/wiki/Electrochemical_kinetics for a more detailed discussion.