# Cell voltage with different n factors

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.

Thanks!

• 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. – Safdar Faisal Aug 29 '20 at 11:37
• Conceptually related: chemistry.stackexchange.com/q/136774/95133 – Safdar Faisal Aug 29 '20 at 11:48
• 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. – Rahul Verma Aug 29 '20 at 13:56
• @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 ? – Certainly not a dog Aug 29 '20 at 14:34
• 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. – Rahul Verma Aug 29 '20 at 14:53

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.