When I'm studying electrochemistry, we were given an equation that all chemistry students are familiar with, which is:

                    Ecell = Eoxidation - Ereduction 

However, when I came to study electroanalytical methods, I came upon this equation:

          Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode

where Ejuntion is the uncertain potential produced from the salt bridge and the Eelectrode is the potential produced from the reference electrode used in the experiment.

My question is, is this the real equation when calculating problems for an electrochemical cells? I'm thinking that they probably cut Ejunction out of the equation because it is impossible to know the exact value for it. In addition, I think that they cut out the Eelectrode in the equation because most of the experiment data came from a SHE electrode where Eelectrode = 0.00 V.

  • $\begingroup$ I think your assumptions are right. The first equation refers to an ideal case (Ejunction=0) with SHE reference. $\endgroup$ – Buck Thorn Jul 25 '19 at 8:57
  • $\begingroup$ @BuckThorn so, is the equation correct or not? $\endgroup$ – Kent de los Reyes Jul 25 '19 at 9:20

I am afraid your equations are other way around.

Ecell = Eoxidation - Ereduction

This is not the accepted convention among electrochemists. Either the reference is outdated or there is a typo. All electrochemists in the world now follow the convention of writing

Ecell= Ereduction-Eoxidation or Ecell = Ecathode-Eanode ---Eq(1)

The reason is historical and I have written details elsewhere, but briefly the main reason is that Americans used one sign convention and the Europeans used another. It took more than 50 years to come to a consistent sign convention, which was originally used by a German chemist, Ostwald, 100 years ago. Given that all electrode potentials are now written as reduction potentials the above Eq. (1) is valid. If you look up the tables, the signs of the electrode potentials correspond to the actual electrostatic sign (+) or (-) of the electrode.

Now this simplistic equation (1) ignores any "connections" among the half cells. In reality we always need an electrically conducting connection to the half cells to keep them separated yet electrically connected. This "connection" could be a salt bridge, or a porous ceramic frit (used in pH probes and so on).

There is a general interesting electrochemical fact whenever there is an interface (the location where two surfaces meet) one can build up difference of electrical potential. This is a fundamental material property. Recall the simple voltaic cell, which has copper and zinc pieces separated by a piece of wet paper containing some common salt.

Liquid junction potential (Ej) is another type of potential which originates at the interface of two electrolytes with different concentration of ions (=charges). In that case we add the liquid junction potential to this Ecell as follows

Ecell= Ecathode-Eanode+ Ej

I have not seen your other equation. Ecell = Eoxidation + Ejunction - Ereduction - Eelectrode. Which book is using this expression and what is its year of publication?

  • $\begingroup$ The Eelectrode I saw was from an electroanalytical method chapter. The examples shown includes subtraction of the potential of the electrode from the overall potential. When they use a saturated calomel electrode, they always subtract the potential of the calomel electrode from the overall potential to determine the real potential of the cell. $\endgroup$ – Kent de los Reyes Jul 25 '19 at 13:09
  • $\begingroup$ Could you mention the name of the book? It is wrong to use Ecell=Eox-Ered. Your second part "When they use a saturated calomel electrode, they always subtract the potential of the calomel electrode from the overall potential to determine the real potential of the cell." is another story. This is done to change the scale back to the hydrogen half cell. This is just an algebraic operation which is done when your ref. electrode is not the standard hydrogen electrode. $\endgroup$ – M. Farooq Jul 25 '19 at 13:14
  • $\begingroup$ I guess I am wrong then, by your explanation I'm thinking that I will just subtract the potential of the electrode if I am determining the potential of a Half cell. $\endgroup$ – Kent de los Reyes Jul 25 '19 at 13:19
  • $\begingroup$ No don't use intuition or my explanation. You will not learn this way until and unless you see the algebra behind it on paper. Use a piece of paper and draw a number line (0 in the center, positive on the right, negatives on the left). Set hydrogen half cell to zero. Put the value saturated calomel electrode value with respect to the hydrogen half cell on the number line with its sign. As an exercise, determine the half cell of Cu(II) wrt to the saturated calomel electrode. Once you have this value, use the same number line and convert your Cu(II) (vs SCE) back to the hydrogen half cell. $\endgroup$ – M. Farooq Jul 25 '19 at 13:30
  • $\begingroup$ I see now, the subtraction is just done to "calibrate" the result to match it into a potential that will be produced if a SHE was used. $\endgroup$ – Kent de los Reyes Jul 25 '19 at 13:36

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