I was studying electrochemistry from my school textbook.

The cell potential in the book is defined as the difference between the electrode potentials of the cathode and the anode.$$E_\text{cell}=E_\text{right}-E_\text{left}$$

We know that electrode potential of an individual half-cell can't have an absolute value and thus we measure it taking the standard hydrogen electrode ($\ce{Pt(s) | H2(g) | H+(aq)}$, SHE) in relation that is assigned a zero potential at all temperatures.

As the electrons flow from the anode to the cathode (higher potential to lower potential), the cathode should possess a lower electrostatic potential than the anode.

That means that the electrode potential of a half cell when measured against the SHE should always be negative if the reaction is feasible, i.e., electrons flow from SHE to the cathode. Thus, the more negative the value of the electrode potential of the cathode, the more electrons will from the SHE, and the more the tendency of the substance of the cathode to get reduced.

But the values of standard electrode potentials of different half cells in my textbook and Wikipedia show that the electrode potentials of substances that get easily reduced is positive, e.g., $E^\ominus = 2.87$ V for $\ce{F_2(g) +2e^-->2F^-}$. Isn't this the complete opposite?

Shouldn't the values of electrode potentials of substances that are easily reduced be negative as we just discussed? What am I missing here?

  • 1
    $\begingroup$ @Poutnik I'm unable to understand what you're trying to say. I have a simple question: electrons flow from higher potential to lower potential, in our case from anode (SHE) to the cathode, i.e., cathode has a lower potential. As the potential of SHE is taken to be zero, the potential of the cathode should be negative (lower) if the reaction is feasible. But for such half cells, we see that the electrode potential is positive. Why so? $\endgroup$
    – Silica19
    Aug 9, 2022 at 13:21
  • $\begingroup$ I have moved my comments to the answer. $\endgroup$
    – Poutnik
    Aug 10, 2022 at 8:22

4 Answers 4


A large number electrochemicals misconceptions will be solved if we take the electrode potential sign as the electrostatic sign of the galvanic cell with respect to the hydrogen electrode. Thus this sign will not be affected by how the half cell is written. This was the European understanding of the leading 20th century top-electrochemists. The Germans decided to set the hydrogen electrode at 0 V, the reference point. The Germans also decided to set the hydrogen electrode at 0 V, as the reference point. This is the take of many leading modern electrochemists including A.J. Bard.

Now let us take an electrode such as copper. Its half-cell electrode potential is +0.34 V vs. SHE. Now interpret the sign this way: If we make a galvanic cell of copper and SHE, the copper electrode will be *positively charged" = the cathode, and the hydrogen cell will behave as an anode.

In the same way, take the fluorine half-cell (hypothetical), and if we make a galvanic cell of fluorine and SHE, the fluorine electrode will be *positively charged" = the cathode, because its half-cell electrode is + 2.87 V.

In a galvanic cell, the current is from A(node) -> C(athode) i.e., A to C.

  • $\begingroup$ One nagging detail that has bothered me: shouldn't the Faraday constant (and by extension the elementary charge of the electron) be negative? It would seem easier to visualize this distinction under that circumstance, but I can't recall seeing the Faraday constant written that way. $\endgroup$ Aug 10, 2022 at 0:44
  • $\begingroup$ @MikeSerfas, Good question. Please post a new question. I have not historically explored this but will search about it. $\endgroup$
    – AChem
    Aug 10, 2022 at 1:02
  • $\begingroup$ @AChem Does that mean that in actuality, the cathode has a lower potential than SHE (thus negative), but just to make it easier to understand, we use the positive sign for feasible reactions? $\endgroup$
    – Silica19
    Aug 10, 2022 at 7:44
  • $\begingroup$ No, this will be the actual electrostatic sign. $\endgroup$
    – AChem
    Aug 10, 2022 at 13:35
  • 1
    $\begingroup$ From the modern electronic current point of view, electrons flow from "lower" potential to "higher potential. Positive current flow (so called conventional current) flows from higher to lower potential. $\endgroup$
    – AChem
    Aug 11, 2022 at 18:46

Blame Ben Franklin or physicists for not knowing what + and - are and having terms like current flow! We chemists know better and think only of electrons altho we do lapse into positive ion flow in Li ion batteries and in acids. No one is perfect.

Oxidation = Loss of electrons; Reduction = Gain of electrons

Oxidation is at the ANODE; Electrons LEAVE the anode into the wire. Anions move to the anode.

Reduction is at the CATHODE; Electrons enter the cathode from the wire. Cations move to the cathode.

Those are the simple rules. Unfortunately the physicists still get a say! They defined volts backwards and we are stuck with it [but some references are unstuck so it still can be confusing, so be careful]. A positive voltage means a reaction proceeds as written and its deltaG is negative. Electrode potentials are written [usually] as reduction half reactions. To appease the physicists a positive voltage means the reaction proceeds as written; a negative voltage means it proceeds in reverse. These half reactions are for a reaction under standard conditions with a hydrogen electrode. To get any reaction simply reverse one of the half reactions with its sign and add the equations and the voltages, then correct for differences from standard conditions. If the voltage is positive the reaction goes as written; if negative reverse the equation and change the sign. [or else change the concentrations and the cell voltage. If the cell voltage is close to zero and the electrode reactions reversible the direction of reaction or the cell voltage is easily manipulated.]

The pertinent equation is: DeltaG = -EnF = -E[0]nF +RTlnQ Q is the reaction quotient with the form of the equilibrium constant Keq. At equilibrium deltaG = 0 so
E[0]nF = RTlnKeq = -deltaG[0]. As can be seen all these + and - signs can get confusing. [LOOK THIS UP IN A PCHEM TEXT or two] [The font this site uses substitutes a lower case o for the numeral zero.]

To sum it up a positive cell voltage corresponds to an equilibrium constant greater than 1. and a negative deltaG. The reaction proceeds as written. How far it goes depends on the voltage and the amount of material [the size of the battery]. At equilibrium the cell voltage is zero and the various concentrations satisfy the equilibrium constant

To react lithium and fluorine: galvanic cell: Li = Li+ + e- E= +3v; F- = 1/F2 +e- E= -3. We have two reduction reactions and need one oxidation and one reduction so must reverse one. Lets do Lithium e-+ Li+ = Li E= -3v. Put the two half reactions together Li+ + F- = Li + 1/2F2 E = -6V. Wow it is going nowhere. So we reverse the reaction. Li + 1/2F2 = Li+ + F-; E = 6V! an explosion!! Not so Good so we set up a cell with an anode, Li, a wire to a motor then to a cathode [who knows what, it can't react with F2], surrounded with F2 and a magic electrolyte to carry the Li+ and F- ions formed remotely and desperately trying to get to the cathode and anode respectively. To reverse this a potential greater than 6V is applied in reverse. I do not think that this particular cell has been made not even by accident.

Work out some simple cells such as a hydrogen ion concentration cell or electrolytic deposition of copper to get an idea what happens to cell voltages as concentrations change. If you get stuck on conventions or signs remember Li is the strongest reductant and F2 is [almost] the strongest oxidant use these two to determine if a different convention is being used for signs.


Easily reduced redox systems "suck" electrons from an electrode, charging it to the more positive equilibrium potential, at which the electron transfer rate is zero.

And vice versa for easily oxidized redox systems: They are pushing electrons to an electrode, charging it to the more negative equilibrium potential.

In galvanic cells, cathodes have higher potential than anodes, in contrary to electrolytic cells. By other words, when switching the cell mode, electrode names switch places, as the current switches direction.

E.g. Imagine the notoriously known Daniell cell:


If we perform electrolysis on it, with the negative contact attached to $\ce{Zn}$, the cathode is the left, more negative electrode, as electrons come to $\ce{Zn}$ and reduce $\ce{Zn^2+(aq)}$.

If we use it as the galvanic cell, the cathode is the right, more positive electrode, as electrons come to $\ce{Cu}$ and reduce $\ce{Cu^2+(aq)}$.

SHE has the absolute alectrode potential $\pu{+4.44 \pm 0.02 V}$.

For Ox/Red redox pair and a formal reaction $\ce{Ox + n/2 H2 -> n H+ + Red}$, there is direct relation of the standard reaction Gibbs energy and the respective standard electrode potential $\Delta_\mathrm{r} G^{\circ}=−nFE^{\circ}$.


This is because of Gibbs free energy. The equation $$ E^{\circ}=-\frac{\Delta_{\text{r}}G^{\circ}}{nF} $$ makes it clear that for spontaneous reactions $(\Delta_{\text{r}}G^{\circ}<0)$, the electrode potential is positive. Highly electronegative elements such as fluorine undergoes reduction very spontaneously and hence, the positive value. In contrast s- and p-block elements have negative electrode potentials as they tend to oxidize easily.

The reference of SHE allows one to determine if the given element can reduce $\ce{H+}$ to hydrogen gas.


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