Application of Nernst Equation in Potentiometric titration of Fe2+ with Ce4+ It is well Known that to find $E_{cell}$ during titration of  $Fe^{2+}$ with $Ce^{4+}$ we consider 3 domains:

Before Equivalance point ,  $E_+ = E_{Fe^{3+}/Fe^{2+}}^o+\frac{RT}{nF}log\frac{[Fe^{3+}]}{[Fe^{2+}]} \tag{1}$

At Equivalance point , $E_+=\frac{E_{Fe^{3+}/Fe^{2+}}+E_{Ce^{4+}/Ce^{3+}}}{2} \tag{2}$

After  Equivalance point ,$E_+ = E_{Ce^{4+}/Ce^{3+}}^o+\frac{RT}{nF}log\frac{[Ce^{4+}]}{[Ce^{3+}]} \tag{3}$

$E_+$ is the electrode potential of the positive electrode where the negative electrode is a fixed reference electrode.

In the book Quantitative Chemical Analysis by Daniel C Harris, Chapter 15 says that both equation 1 and 3 can be used either of the two domains and the reason why one is used instead of the other is due to the fact that the ratio of the concentration of a particular metal ion in different oxidation state is much easily found in that domain while the other ratio usually requires a complicted equilibrium equation to be solved to obtain the solution.

my questions:


Does this mean that if by some means I knew the ratio of concentrations of both metals I could substitute it in eqn 1 and eqn 3 and end up getting the same value for $E_+$.
If not does it only hold near the Equivalance point?
If one is True , The only explanation I have is that both $Fe^{2+}$ and $Ce ^{4+}$ are in equilibrium with the same electrode. But this still doesn't make sense as usually when 2 reactions are occurring at the same container the $\Delta G_{net}=\Delta G_{1}+\Delta G_{2} $ So in some sense shouldn't $ E_+ =E_{due-to-Fe}+E_{due-to-Ce}$ be the equation we require?


If I'm missing some essential concept it would be Great if someone could point it out and probably provide some links for me to look up.
It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you start the titration (Ce (IV) in the burette), the moment you add Ce(IV), it is immediately reduced by Fe(II) to Ce(III). You do not have the Ce(IV/III) couple in the solution. Before the endpoint, Fe(III/II) couple controls the potential of the indicator electrode.

Do you remember what that indicator electrode is made of? It is just a platinum wire and the reference electrode is usually silver-silver chloride electrode, all enclosed in a single unit. The platinum wire is oblivious to iron or cerium. All it "senses" is the charge. Imagine a single Pt electrode dipped in an iron solution and forget about the presence of cerium. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.