What is the electrode potential at the equivalence point if concentrations of titrant and titrand are going to be zero?

Let's assume that the titrand is $$\ce{Fe^2+}$$ and the titrant is $$\ce{Cr2O7^2-}$$. Then

$$E = E^\circ_\ce{Fe^{2+}/Fe^{3+}} - \frac{RT}{nF}\ln\frac{\ce{[Fe^{2+}]}}{\ce{[Fe^{3+}]}}$$

$$E = E^\circ_\ce{Cr^{6+}/Cr^{3+}} - \frac{RT}{nF}\ln\frac{\ce{[Cr^{3+}]}}{\ce{[Cr^{6+}]}}$$

Adding both the above equations, we get

$$2E = E^\circ_\ce{Fe^{2+}/Fe^{3+}} + E^\circ_\ce{Cr^6+/Cr^3+} - \frac{RT}{nF}ln\frac{\ce{[Fe^{2+}][Cr^{3+}]}}{\ce{[Fe^{3+}][Cr^{6+}]}}$$

Since $$\ce{[Fe^2+]=[Cr^6+]}$$ and $$\ce{[Fe^3+]=[Cr^3+]}$$

They would all be getting cancelled, ending in

$$E = \frac{E^\circ_\ce{Fe^{2+}/Fe^{3+}} + E^\circ_\ce{Cr^{6+}/Cr^{3+}}}{2}$$

But, at equivalence point, the titrant ions would have entirely reduced and the titrand ions would have entirely oxidised, leaving

$$\ce{[Fe^{2+}]=[Cr^{6+}]}=0$$

And as we know, in mathematics, $$\frac00$$ is infinite or not defined. So, how do can we actually mathematically derive the electrode potential at the equivalence point?

• "ions would have entirely reduced" is not true, there eq. constants are finite. – Mithoron Oct 29 '18 at 18:16
• @Mithoron So does that mean there will be a little (close to 0) amount of unreacted Fe 2+ ions even after the supposed complete oxidation? – Sashank Sriram Oct 31 '18 at 15:54