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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?

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    $\begingroup$ "ions would have entirely reduced" is not true, there eq. constants are finite. $\endgroup$ – Mithoron Oct 29 '18 at 18:16
  • $\begingroup$ @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? $\endgroup$ – Sashank Sriram Oct 31 '18 at 15:54

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