I’m a bit confused regarding cell potentials. If I have two half reactions each involving a different number of electrons, there are two ways I can see to calculate the net emf of their cell:

  1. Simply add the potentials across the two theoretical half cells, that is, treat each half reaction as a voltage source in series
  2. Calculate the value for the net cell reaction as a whole, that is, make consideration for the energetics of the reaction

As a more specific example, say I have a metal $\text M$, and $$\ce{E^0_{M^{3+}/M}=2V, E^0_{M^{2+}/M}=1V}$$ and would like to figure the standard cell potential, say $x$ of $\ce{E^0_{M^{3+}/M^{2+}}}$.

With Method 1 we have $$x= \ce{{E^0_{M^{3+}/M}}-{E^0_{M^{2+}/M}}}=2-1=\mathbf{1V}$$

With Method 2 we have $$\ce{\Delta G^0_{M^{3+}/M}=-3*F*2=-6F, \Delta G^0_{M^{2+}/M}=-2*F*1=-2F}$$ $$\implies \ce{\Delta G^0_{net} = -6F - (-2F) = -4F}$$ $$\implies x=\frac{-\ce{4F}}{\ce{-1*F}}=\mathbf{4V}$$

How do I rectify these inconsistent values? Does it make any difference here whether I say I am looking for $\ce{E^0_{M^{3+}/M^{2+}}}$ or $\ce{M|M^{2+}(1 molar)||M^{3+}(1 molar)|M}$? What if I use an inert electrode like $\ce{Pt}$ in the latter representation?

Further, if I have another metal $\ce Y$ for which $\ce{E^0_{Y^+/Y}}=\ce{3V}$, which is a stronger oxidizing agent? $\ce{M^{3+}}$ or $\ce Y^+$?

In various sources it seems some use method 1 while others use 2. In this answer on CSE it seems 1 is the right one. But I maybe missing some conceptual context.


  • 2
    $\begingroup$ Cell potential is not additive but free energy is.. and so the second method is accurate/correct.. Try using method 2 in that reaction and it would be correct as well. $\endgroup$ Aug 29, 2020 at 11:37
  • 1
    $\begingroup$ Conceptually related: chemistry.stackexchange.com/q/136774/95133 $\endgroup$ Aug 29, 2020 at 11:48
  • $\begingroup$ The first method is actually derived from the second one. In case of "different-elements" Galvanic cell, the first method works, but in case of disproportionation, it doesn't. $\endgroup$ Aug 29, 2020 at 13:56
  • $\begingroup$ @RahulVerma Whats the difference in such a cell? It is the same physical process, so why should there be different rules? And does this have any thing to do with the electrodes—I.e. if I use an inert electrode will it behave like a normal galvanic cell ? And I did figure where the first method comes from, it’s rather the order of operations, if you will, that I find confusing. @ Safdar But won’t that also give a mathematically different result since in one case it is a simple difference and the other is a weighted average of sorts ? $\endgroup$ Aug 29, 2020 at 14:34
  • $\begingroup$ Take any example of "different elements" cell (say Cu-Ag Galvanic cell), and then use the second method to calculate E(cell). You'll always end up with same values from both methods. $\endgroup$ Aug 29, 2020 at 14:53

1 Answer 1


If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.

This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$

\begin{array} {rlllc} \require{cancel} \ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\ \ce{ M^x+ +xe- &-> M} &E^\circ = E_m^\circ &\quad|\times(y) \\ \hline \ce{yM^x+ + xN +\cancel{(xye-)} &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ \\ \end{array}

So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$

Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.

For just one mole of M being reduced , we get

$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$

For just one mole of N, we get

$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$

Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to

\begin{align} \Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\ &= xyF(E^\circ_m -E^\circ_n) \tag{3} \end{align}

Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:

$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$

Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get

\begin{align} \cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\ \implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n \end{align}

For a case where you have the same compound disproportionate, this question and its answers should be enough.


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