How can we directly add half cell potentials to measure the EMF of a galvanic cell?

Question

We know that $E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}$. For example, in a cell $\ce{Zn(s) | Zn^2+(aq) || Cu^2+(aq) | Cu(s)}$, $$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Cu^2+/Cu}}+E^\circ_{\ce{Zn/Zn^2+}}$$

This makes sense to me. I think of $E^\circ_{\text{cell}}$ as the $E^\circ$ value of the net cell reaction i.e. of $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$.

But, consider this cell: $\ce{Zn(s) | Zn^2+(aq) || Ag^+(aq) | Ag(s)}$ According to my textbook, the $E^\circ_{\text{cell}}$ value here is again defined by:

$$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Ag+/Ag}}+E^\circ_{\ce{Zn/Zn^2+}}$$

But, I think this is not correct.
Consider the fact that this cell is made up of two half cell reactions:

$$ \begin{array}{} \text{Oxidation}&\ce{Zn(s)}&\ce{-> Zn^2+(aq) + 2e-}&E_1^\circ\\ \text{Reduction}&\ce{Ag^+(aq) + e-}&\ce{-> Ag(s)}&E_2^\circ\\ \text{Net cell reaction}&\ce{Zn(s) + 2Ag^+(aq)}&\ce{-> Zn^2+(aq) + 2Ag(s)}&E_3^\circ\\ \end{array} $$

and that we can't add $E^\circ$ values directly, since they are an intensive property. Instead, we need to say that $\Delta G_3=\Delta G_1+\Delta G_2$ and then write $2E_3^\circ=2E_1^\circ+E_2^\circ$ or $E_3^\circ=\frac{2E_1^\circ+E_2^\circ}2$, which is definitely $\neq E_1^\circ+E_2^\circ$, as my textbook says.

I believe I have correctly presented everything here, and cannot figure out my mistake. Where am I wrong?

Answer 1

You are writing $\Delta G_3=\Delta G_1+\Delta G_2$. But it should be $\Delta G_3=\Delta G_1+2\Delta G_2$ (I am considering $\Delta G_2$ as free energy change for the conversion of $\ce{Ag+}$ to $\ce{Ag}$, which is considered 2 times in the final reaction) Because $\Delta G$ also depends on the number of moles of reactants and products, if you add two times the second reaction, the $\Delta G$ will also be used as twice its value. There you've made a mistake.

Thus the equation will be $$2E_3^0 = 2E_2^0 + 2E_1^0$$ So you will have $E_3 ^0 = E_1^0 + E_2^0$.

Answer 2

It is easier to work with redox potentials and at the end use $\Delta G^\text{o}=-nFE^\text{o}$, $n$ is the number of electrons transferred in the reaction and is only used to calculate $\Delta G$ never $E$. To calculate the reaction potential take the difference in redox potentials with the reduction potential first (the one written in the same way as in the tables), $\ce{Ag+ + e = Ag ; 0.8V }$ in your example, and then subtract the other one, $\ce{Zn^{2+} + 2e = Zn; -0.76 V}$ giving $1.56$ V and $n=2$.

In your question you add potentials, in this case the same result is obtained of you reverse the sign of the second potential (written as oxidation in the reaction and is the reducing agent) and add.