I’m dealing with electrochemistry problems. For the following reaction $$\ce{2H+ +2e- <=>H2}\quad E^\circ=0\ \mathrm{V}$$ Does it make sense to say $$\Delta G^\circ=-nFE^\circ=0$$ Which would mean the reaction is in equilibrium?


1 Answer 1


You cannot apply the $\Delta{G}$ equation to a single electrode potential. It can be applied to a cell though so if the hydrogen electrode is connected to another electrode (say copper dipped in copper sulfate solution) then you can find the free energy.

It's better to remember the formula as:

$\Delta{G} = -nFE_{cell}^o$


$E_{cell}^o = E_{cathode}^o - E_{anode}^o$ (using all standard reduction potentials)

That way you won't accidentally apply it anywhere else.

In this case, for $\ce{\Delta G}$ to be zero you would have to take Hydrogen electrodes for cathode as well as anode.

A More Complicated Explanation

Just in case you're curious about the reasons behind this (Since I was also very confused about this when I first learned Electrochemistry. It's ok if you don't understand this),

It is very difficult to measure the absolute potential of a single electrode and so the standard reduction potentials you are using are actually based on a reference system.

This is actually very similar to the system of atomic masses. The atomic masses you use in calculations are not the actual masses of the elements in grams but they are measured in unified atomic mass units. One unified atomic mass unit is defined as 1/12th the mass of a carbon-12 atom.

[Note: Now of course, after the mole was made a fundamental SI unit, the unified atomic mass unit has become equivalent to $1 g/mol$. This is just for the sake of explanation.]

So if you come across a situation where you have to use the mass of the atom (maybe some nuclear chemistry calculations) you can't use those normal atomic masses but you would have to convert it into kg (One unified atomic mass unit is about $1.6\times10^{-27}$ kg).

Similarly, the standard reduction potentials are actually measured with reference to a standard electrode (Whose reduction potential is arbitrarily taken to be zero) which is the standard hydrogen electrode.

This standard reduction potential of an electrode can be measured easily by connecting the electrode to a standard hydrogen electrode (SHE) and then measuring the potential difference using a voltmeter and applying the $E_{cell}^o$ equation. (If $H^+$ is being reduced to $H_2$, then $E_{cathode}^o$ is zero otherwise $E_{anode}^o$ is zero).

So this just means that when you say $E^o$ of the SHE is zero, you are actually saying that "the potential of the SHE with respect to itself is zero" which is quite obvious. That doesn't mean that the potential of the electrode itself is zero.

That is why you cannot apply the first equation in this case. But this does not apply to cells. If we assume the absolute potential of the SHE to be some $x$, then,

$E_{abs}^o = E^o + x$

Then, $E^o_{cell} = E_{abs}(cathode) - E_{abs}(anode)$

$= E^o_{cathode} + x - [E^o_{anode} + x]$

$E^o_{cell} = E_{cathode}^o - E_{anode}^o$

The absolute potential of the SHE cancels out in the difference.


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